Vectors

Directed Line Segments and Vectors

A directed line segment is defined as an initial point, P, and a terminal point Q.

Example 

        P = (2,3)     and     Q = (-1,4)

Definition of a Vector

A vector is the equivalence class of all directed segments of the same length and direction.

We can represent a vector by writing the unique directed line segment that has its initial point at the origin. 

 

Example

        P = (2,3) and Q = (-1,4)

is equivalent to the directed line segment 

        "Q - P"  =  <-3, 1>

When we write the <> we mean that the vector has initial point at the origin and terminal point at (-3,1). This notation is called the component form of the vector.

The length of the vector <x,y> is called the norm or magnitude.

We can find it by the formula:

 Length of a Vector

Example:  

       

We also use the notation 

        -3i + j  

to denote the vector <-3,1>.

Example

A vector that has length 10 makes an angle of /6 with the x-axis.  Find its components.

Solution:  

        x  =  r cos ,     y  =  r sin

So that 

        x  =  (10)( / 2),      y  =  10 (1/2)  =  5

We can write the vector as 

        5   i + 5j

Unit Vectors in the Direction of v

A vector is called a unit vector if it has magnitude = 1. If 

        v = <a,b>

then the unit vector in the direction of v can be found by

 The Unit Vector in the Direction of v

                     1          u  =            v                   || v ||

 

Example:  

The unit vector in the direction of <-3,1> is

       

We can use the <> notation and the i  j notation interchangeably.

Algebra of Vectors

If 

        v = <a,b>     and      w = <c,d> 

and k is a constant, then we can define the sum an scalar multiplication as follows

       Sum and Scalar Multiplication of Vectors

          v + w = <a + c,b + d>

and

          kv = <ka, kb>

 

Example

        3 <2,1> - 2<-1,3>  =  <6 + 2,3 - 6> 

        =  <8,-3>  =  8i - 3j

Geometrically v + w is the vector that corresponds to the diagonal of the parallelogram with two sides v and w.

The appropriate diagram can also be drawn to show how 

        v - w = v + (-w).

 

Prope

rties of Vector Addition and Subtraction

We have the following four properties of vectors:  If u, v ,and w are vectors and a and b are numbers then

1. (u + v)  + w  =  u + (v + w)2. a(u + v)  =  au + av

3. a(bv)  =  (ab)v

4. u + v  =  v + u

Applications

A boat captain wants to travel due south at 40 knots.  If the current is moving northwest at 16 knots, in what direction and magnitude should he work the engine?  

Solution

We have 

        u = v + w

where u corresponds to the velocity vector of the boat, v corresponds to the engine's vector, and w corresponds to the velocity of the current.  We have

     

   u = -40j     and     w = -8 i + 8 j

Hence

        v  =  u - w  =  -40j - (-8 i + 8 j)  =  8 i - (40+8 )j

The magnitude is 

        [(8 )2 +  (40+8 )2]1/2  =  52.5

The direction is 

       

3 Dimensional Coordinates

To generalize the plane to 3 dimensions, we draw a third axis, called the z-axis at a right angle from the plane so that if you grab on to the z-axis with your right hand your hand will curl from the positive x-axis to the positive y-axis.  To plot a point in the xyz-space

We first plot a point in the xy-plane and then draw a segment parallel to the z-axis of length equal to the z coordinate. 

Example:  

Plot (1,2,3) 

Solution:We first draw the x,y, and z-axes.  Then we plot the point (1,2) in the xy-plane.  Finally move up three units and plot the point.

Exercise

Plot  (2,4,3)

The Distance Formula

The distance formula is derived from the three dimensional version of the Pythagorean theorem, which is displayed below.

The distance between two points (x1,y1,z1) and (x2,y2,z2) and is given by

        

     Distance Formula in Three Dimensions

 

         

Algebra of vectors in 3D

A vector in space is given by 

        <x,y,z> = xi + yj + zk

The algebra rules are similar to those in two dimensions.

The Dot and Cross Product

The Dot Product

                 Definition We define the dot product of two vectors

     v = ai + bj  and w = ci + dj 

to be

          v . w = ac + bd 

Notice that the dot product of two vectors is a number and not a vector.  For 3 dimensional vectors, we define the dot product similarly:

Dot Product in R3 

If 

     v = ai + bj + ck and w = di + ej + fk

then           v . w = ad + be + cf

 

 

Examples:

If 

        v  =  2i + 4j  

and 

        w  =  i + 5j

then

        v . w  =  (2)(1) + (4)(5)  =  22

 

Exercise

Find the dot product of 

        2i + j - k     and     i + 2j

The Angle Between Two Vectors

We define the angle theta between two vectors v and w by the formula

                            v . w                   cos   =                                           ||v|| ||w|| 

so that        

v . w = ||v|| ||w|| cos

Two vectors are called orthogonal if their angle is a right angle.  

We see that angles are orthogonal if and only if 

        v . w  =  0

Example  

To find the angle between 

        v =  2i + 3j + k 

and 

        w  =  4i + j + 2k 

we compute:

        and

       

and

        v . w  =  8 + 3 + 2 = 13

Hence 

       

1 13 cos14 21

Direction Angles

     Definition of Direction Cosines

Let 

          v = ai + bj + ck  

be a vector, then we define the direction cosines to be the following:

1.                   acos   =                            ||v||

2.                   bcos   =                            ||v||

3.                   ccos   =                            ||v||

 

Projections and Components

Suppose that a car is stopped on a steep hill, and let g be the force of gravity acting on it. We can split the vector g into the component that is pushing the car down the road and the component that is pushing the car onto the road.  We define

Definition

 Let u and v be the vectors.  Then u can be broken up into two components, r and s such that s is parallel to v and r is perpendicular to v.  s is called the projection of u onto v and r is called the component of u perpendicular to v.  

We see that 

                                                   ||u|| ||v|| ||s||            u . v  =  ||u|| ||v|| cos    =                          =  ||v|| ||s||                                                         ||u||

hence 

               u . v  ||s||  =                                 ||v||

We can calculate the projection of u onto v by the formula:

                  u . vprojvu =                v                    ||v||2 

 

Work

The work done by a constant force F along PQ is given by 

W = F . PQ 

 

Example

Find the work done against gravity to move a 10 kg baby from the point (2,3) to the point (5,7)? 

Solution

We have that the force vector is 

        F  =  ma  =  (10)(-9.8j)  =  -98j

and the displacement vector is

        v  =  (5 - 2) i  +  (7 - 3) j  =  3i + 4j

The work is the dot product

        W  =  F . v  =  (-98j)

. (3i + 4j)  

        =  (0)(3) + (-98)(4)  =  -392

Notice the negative sign verifies that the work is done against gravity.  Hence, it takes 392 J of work to move the baby.

 

Torque

Suppose you are skiing and have a terrible fall.  Your body spins around and you ski stays in place (do not try this at home).  With proper bindings your bindings will release and your ski will come off.  The bindings recognize that a force has been applied.  This force is called torque.  To compute it we use the cross produce of two vectors which not only gives the torque, but also produces the direction that is perpendicular to both the force and the direction of the leg.

The Cross Product Between Two Vectors

                          Definition   Let u = ai + bj + ck  and v = di + ej + fk be vectors then we define the cross product v x w by the determinant of the matrix:

                                

We can compute this determinant as

       

        =  (bf - ce) i + (cd - af) j + (ae - bd) k

Example

Find the cross product u x v if

        u  =  2i + j - 3k            v  =  4j + 5k

 

Solution

We calculate

       

        =  17i - 10j + 8k

        

Exercises

Find u x v when

A. u = 3i + j - 2k,      v = i - kB. u = 2i - 4j - k,      v = 3i -  j + 2k

Notice that since switching the order of two rows of a determinant changes the sign of the determinant, we have

         u x v  =  -v x u

Geometry and the Cross Product

Let u and v be vectors and consider the parallelogram that the two vectors make.  Then

        ||u x v|| = Area of the Parallelogram

and the direction of u x v is a right angle to the parallelogram that follows the right hand rule 

Note:   For i x j the magnitude is 1 and the direction is k, hence i x j = k.

 

Exercise

Find j x k and i x k

Torque Revisited

We define the torque (or the moment M of a force F about a point Q) as

M = PQ x F

 

Example

A 20 inch wrench is at an angle of 30 degrees with the ground.  A force of 40 pounds that makes and angle of 45 degrees with the wrench turns the wrench.  Find the torque.

Solution  

We can write the wrench as the vector

        20 cos 30  i  +  20 sin 30  j  = 17.3 i  +  10 j

and the force as

        -40 cos 75 i - 40 sin 75 j = -10.3 i - 38.6 j

hence, the torque is the magnitude of their cross product:

       

        =  -564 inch pounds

Parallelepipeds

To find the volume of the parallelepiped spanned by three vectors u, v, and w, we find the triple product:

Volume = u . (v x w)

This can be found by computing the determinate of the three vectors:

   

 

Example

Find the volume of the parallelepiped spanned by the vectors

        u  =  <1,0,2>        v  =  <0,2,3>        w  =  <0,1,3>

Solution

We find

      

 

Lines and Planes

Lines

Our goal is to come up with the equation of a line given a vector v parallel to the line and a point (a,b,c) on the line.  The figure (shown in 2D for simplicity) shows that if P is a point on the line then 

        <x,y> = P + tv

for some number t.  

The picture is the same for 3D.  The formula is given below.

        

Parametric Equations of a Line

The parametric equations for the line through the point (a,b,c) and parallel to the vector v are

      <x,y,z> = <a,b,c> + tv 

Example:  

Find the parametric equations of the line that passes through the point (1, 2, 3) and is parallel to the vector <4, -2, 1>

Solution:

We write:

        <x, y, z> = <1, 2, 3> + t <4, -2, 1> = <1 + 4t, 2 - 2t, 3 + t> 

or

        x(t) = 1 + 4t,   y(t) = 2 - 2t,   z(t) = 3 + t

Exercise

Find the parametric equations of the line through the two points (2,1,7) and (1,3,5).

Hint: a vector parallel to the line has tail at (2,1,7) and head at (1,3,5).

Planes

If S is a plane then a vector n is normal (perpendicular) to the plane if it is orthogonal to every vector that lies on the plane.  Suppose that n is a normal vector to a plane and (a,b,c) is a point on the plane.  Let (x,y,z) be a general point on the plane, then 

        <x - a, y - b, z - c> 

is parallel to the plane, hence

n . <x - a, y - b, z - c> = 0

this defines the equation of the plane.

Example:  

Find the equation of the plane that contains the point (2,1,0) and has normal vector <1,2,3>

Solution:

We have

        <1,2,3> . <x - 2,y - 1,z - 0>  =  0

so that 

        1(x - 2) + 2(y - 1) + 3z  =  0

or

        x + 2y + 3z  =  4

Example 

Find the equation of the plane through the points 

        P  =  (0,0,1)      Q  =  (2,1,0)     and      R  =  (1,1,1)

Solution

Let 

        v  =  Q - P  =  <2, 1, -1> 

and 

        w = R - P = <1, 1, 0> 

then to find a vector normal to the plane, we find the cross product of v and w:

       

or 

        <1, -1, 1>

We can now use the formula:

        <1, -1, 1> . <x, y, z - 1>  =  0

or

        x - y + z - 1  =  0

or

        x - y + z  =  1

Distance Between a Point and a Plane

Let P be a point and Q be a point on a plane with normal vector n, then the distance between P and the plane is given by

 Distance Between a Point P and a Plane With Normal Vector n

Let Q be a point on the plane with normal vector n.  Then the distance from the point P to this plane is given by

                                   ||PQ . n||              ProjnPQ  =                                                         ||n||

Example  

Find the distance between the point (1,2,3) and the plane 

        2x - y - 2z = 5

Solution

The normal vector can be read off from the equation as

        n  =  <2, -1, -2>

Now find a convenient point on the plane such as Q  =  (0, -5, 0).  We have 

        PQ  =  <-1, -7, -3>

and 

        n . PQ  =  -2 + 7 + 6  =  11

We find the magnitude of n by taking the square root of the sum of the squares.  The sum is

        4 + 1 + 4  =  9

so 

        || n ||  =  3

Hence the distance from the point to the plane is 11/3.

        

The Angle Between 2 Planes

The angle between two planes is given by the angle between the normal vectors.

Example

Find the angle between the two planes 

        3x - 2y + 5z  =  1      and      4x + 2y - z  =  4

We have the two normal vectors are

        n = <3,-2,5>      and      m = <4,2,-1>

We have

        n . m  =  3, 

        

        , 38 21 n m

 

 hence the angle is

       

       

3arccos 1.46

38 21

 

Surfaces

Cylinders

Let C be a curve, then we define a cylinder to be the set of all lines through C and perpendicular to the plane that C lies in.

We can tell that an equation is a cylinder is it is missing one of the variables.

 

Quadric Surfaces

Recall that the quadrics or conics are lines, hyperbolas, parabolas, circles, and ellipses. In three dimensions, we can combine any two of these and make a quadric surface.  For example

       

is a paraboloid since for constant z we get a circle and for constant x or y we get a parabola.  We use the suffix -oid to mean ellipse or circle.  We have:

   

       x2          y2          z2

             +           +            =  1 is an ellipsoid           a2          b2          c2        

          x2         y2         z2

    -          -           +            =  1 is a hyperboloid of 2 sheets while             a2         b2          c2       

       x2          y2          z2

             +           -            =  1 is a hyperboloid of 1 sheet           a2          b2          c2        

        

Surface of Revolution

Let y = f(x) be a curve, then the equation of the surface of revolution abut the x-axis is

        y2 + z2 = f(x)2

Example

Find the equation of the surface that is formed when the curve

        y   =   sin x          0  <  x  < /2

is revolved around the y-axis.

Solution

This uses a different formula since this time the curve is revolved around the y-axis.  The circular cross section has radius sin-1 y and the circle is perpendicular to the y-axis.  Hence the equation is

        x2 + z2 = (sin-1 y)2

 

 

Cylindrical Coordinates

We can extend polar coordinates to three dimensions by

     x = rcosy = rsinz = z

Example  

We can write (1,1,3) in cylindrical coordinates.  We find 

       

and

         

so that the cylindrical coordinates are

        ( , /4, 3) 

Spherical Coordinates

An alternate coordinate system works on a distance and two angle method called spherical coordinates.  We let denote the distance from the point to the origin, represent the same as in cylindrical coordinates, and denote the angle from the positive z-axis to the point.  The picture tells us that

        r  =  sin  

and that 

        z  =  cos  

From this we can find

     x = rcos = sin cos     y = r sin = sin sin     z = cos

Immediately we see that 

        x2 + y2 + z2 = 2

We use spherical coordinates whenever the problem involves a distance from a source.

 

Example

convert the surface 

        z  =  x2 + y2

to an equation in spherical coordinates.  

Solution

We add z2 to both sides

        z + z2  =  x2 + y2 + z2 

Now it is easier to convert

        cos + 2 cos2   =  2 

Divide by to get

        cos + cos2   = 

Now solve for .

                      cos                     cos            =                            =                     =  csc   cot                       1 - cos2                 sin2   

 

 

 

Vector Valued Functions

Definition of a Vector Valued Function

A vector valued function is a function where the domain is a subset of the real numbers and the range is a vector.  

In two dimensions

        r(t) = x(t)i + y(t)j 

In three dimensions

        r(t) = x(t)i + y(t)j + z(t)k

You will notice the strong resemblance to parametric equations.  In fact there is an equivalence between vector valued functions and parametric equations.

 

Example  

        r(t) = 3i + tj + (sin t) k

To graph a vector valued function we can just graph the parametrically defined function

 

Example

Sketch the graph of 

        r(t)  =  (t - 1)i + t2 j 

 

Solution

We draw vectors for several values of t and connect the dots.  Notice that the graph is the same as 

        y  =  (x + 1)2 

       

 

Limits

We define the limit of a vector valued function by taking the limit of each of the components.  Formally

 

The Limit of a Vector Valued Function

         

 

Example

Find the limit

       

if

                         sin t          r(t)  =  et i   +              j     + (t ln t) k                             t

Solution

We take the three limits one at a time

The first function is continuous at t = 0, so we can just plug in to get

        e0  =  1

For the second function, we get 0/0, so we use L'Hospital's rule to get

        cos t                               1

Now plug in to get

        1/1  =  1

For the kth component, we rewrite as 

        ln t                        1/t

Now use L'Hospital's rule to get

        1/t                  =  t        -1/t2

Plugging in 0 gives 0.  Finally, gathering our results gives a limit of 

        i + j

 

Continuity

We define continuity of vector valued functions in a similar way to how continuity of real valued functions was defined.  

 

Definition of Continuity

A vector valued function is continuous at t0 if it is defined at t0 and 

              

The practical way to investigate continuity is to look at each of the components.

 

Example

Determine where the following vector valued function is continuous.

        r(t)   =   ln(1 - t)i  + 1/t j  + 3t k

 

Solution

The first component is continuous for all values of t less than 1, the second component is continuous for t nonzero, and the third component is continuous for all real numbers.  We can conclude that r(t) is continuous for all t less than 1 but not equal to 0.

 

 

Differentiation and Integration of Vector Valued Functions

Calculus of Vector Valued Functions

The formal definition of the derivative of a vector valued function is very similar to the definition of the derivative of a real valued function.

 

The Derivative of a Vector Valued Function

Let r(t) be a vector valued function, then

              

Because the derivative of a sum is the sum of the derivative, we can find the derivative of each of the components of the vector valued function to find its derivative.

 

Examples

        d/dt (3i + sintj) = costj

        d/dt (3t2 i + cos(4t) j + tet k)  =  6t i -4sin(t)j + (et + tet) k

 

 

Properties of Vector Valued Functions

All of the properties of differentiation still hold for vector values functions.  Moreover because there are a variety of ways of defining multiplication, there is an abundance of product rules.

 

Properties of Vector Valued Functions

Suppose that v(t) and w(t) are vector valued functions, f(t) is a scalar function, and c is a real number then

1.  d/dt(v(t) + w(t)) = d/dt(v(t)) + d/dt(w(t))

2.  d/dt(cv(t)) = c d/dt(v(t))

3.  d/dt(f(t) v(t)) = f '(t) v(t) + f(t) v'(t)

4.  (v(t) . w(t))' = v'(t)

. w(t)+ v(t) . w'(t)

5.  (v(t) x w(t))' = v'(t)  x  w(t)+ v(t) x w'(t)

6.  d/dt(v(f(t))) = v'(f(t)) f '(t)

 

Example

Show that if r is a differentiable vector valued function with constant magnitude, then

        r . r'  =  0

 

Solution

Since r has constant magnitude, call it k,

        k2  =  ||r||2  = r . r

Taking derivatives of the left and right sides gives

        0  =  (r . r)'  =  r'

. r + r

. r' 

        =  r . r' + r

. r'  =  2r

. r'  

Divide by two and the result follows

Integration of vector valued functions

We define the integral of a vector valued function as the integral of each component.  This definition holds for both definite and indefinite integrals.

 

Example

Evaluate

        (sin t)i + 2t j - 8t3 k dt

 

Solution

Just take the integral of each component

        ( (sin t)dt i) + ( 2t dt  j) -  ( 8t3 dt k)

        =  (-cost + c1)i  + (t2 + c2)j  +  (2t4 + c3)k 

 

Notice that we have introduce three different constants, one for each component.        

 

 

Velocity and Acceleration

 

Definition of Velocity and Speed

In single variable calculus the velocity is defined as the derivative of the position function.  For vector calculus, we make the same definition.

 

Definition of Velocity

Let r(t) be a differentiable vector valued function representing the position vector of a particle at time t.  Then the velocity vector is the derivative of the position vector.

               v(t)  =  r'(t)  =  x'(t)i + y'(t)j + z'(t)k

 

Example

Find the velocity vector v(t) if the position vector is 

        r(t)  =  3ti + 2t2j - sin t k 

 

Solution

We just take the derivative

        v(t)  =  3i + 4tj + cos t k 

 

When we think of speed, we think of how fast we are going.  Speed should not be negative.  In one variable calculus, speed was the absolute value of the velocity.  For vector calculus, it is the magnitude of the velocity.

 

Definition of Speed

Let r(t) be a differentiable vector valued function representing the position of a particle.  Then the speed of the particle is the magnitude of the velocity vector.

          Speed  =  || v(t) ||  =  || r'(t) ||

 

 

Example

Let 

        r(t)  = 3i + 2t j + cos t k

Find the speed after /4 seconds.

 

Solution

We first find the velocity vector

        v(t)  =  r'(t)  =  2 j - sin t k

We have 

        v(p/4)  =  2 j - /2 k

Its magnitude is the square root of the sum of the squares or

        Speed  =  || v ||  = 

 

Acceleration

In one variable calculus, we defined the acceleration of a particle as the second derivative of the position function.  Nothing changes for vector calculus.

 

Definition of Acceleration

Let r(t) be a twice differentiable vector valued function representing the position vector of a particle at time t.  Then the acceleration vector is the second derivative of the position vector.

               a(t)  =  r''(t)  =  x''(t)i + y''(t)j + z''(t)k

 

Example

Find the velocity and acceleration of the position function

        r(t)  = 4t i + t2 j 

when t  =  -1.  Then sketch the vectors.

 

Solution 

The velocity vector is

        v(t)  =  r'(t)  =  4 i  + 2t j

Plugging in -1 for t gives

        v(-1)  =  4 i - 2j

Take another derivative to find the acceleration.

        a(t)  =  v'(t)  =  2j

Below is a picture of the vectors.

       

 

Projectile Motion

Since the velocity and acceleration vectors are defined as first and second derivatives of the position vector, we can get back to the position vector by integrating.

 

Example

You are a anti-missile operator and have spotted a missile heading towards you at the position

        re  =  1000i + 500j

with velocity 

          ve  =  -30i + 3j 

You can fire your anti-missile at 100 meters per second.  At what angle should you fire it so that you intercept the missile.  Assume that gravity is the only force acting on the projectiles.

 

Solution

The acceleration vector of the enemy missile is 

        ae(t)  =  -9.8 j 

Integrating, we get the velocity vector

        ve(t)  =  v1 i + (v2 - 9.8t) j

Setting t  =  0 and using the initial velocity of the enemy missile gives

        ve(t)  =  -30 i + (3 - 9.8t) j

Now integrate again to find the position function

        re(t)  =  (-30t + r1) i + (-4.9t2 + 3t + r2) j

Again setting t  =  0 and using the initial conditions gives

        re(t)  =  (-30t + 1000) i + (-4.9t2 + 3t + 500) j

The acceleration of your anti-missile-missile is also

        ay(t)  =  -9.8 j 

Integrating, we get the velocity vector

        vy(t)  =  v1 i + (v2 - 9.8t) j

Since the magnitude of our velocity is 100, we can say

        vy(0)  =  100 cos i + 100 sin j

So that

        vy(t)  =  100 cos i + (100 sin - 9.8t) j

Now integrate again to find the position function

        ry(t)  =  (100t cos + r1) i + (-4.9t2 + 100t sin + r2) j

Our anti-missile-missile starts out at base, so the initial position is the origin.  All the constants are zero.

        ry(t)  =  (100t cos ) i + (-4.9t2 + 100t sin ) j

Since we want to intercept the enemy missile, we set the position vectors equal to each other.

        (100t cos ) i + (-4.9t2 + 100t sin ) j  =  (-30t + 1000) i + (-4.9t2 + 3t + 500) j

Equating coefficients gives

        100t cos =  -30t + 1000

        -4.9t2 + 100t sin   =  -4.9t2 + 3t + 500

The first equation gives 

                         1000        t  =                                                    100cos   +  30

Simplifying the second equation and substituting gives

             100000 sin                     3000                                      =                                  +  500                    100cos   +  30          100cos +  30        

Clear denominators to get

        100000 sin   =  3000 + 50000 cos + 15000

At this point we use a calculator to solve for to

        =  .62535 radians  

 

 

The Unit Tangent and the Unit Normal Vectors

 

The Unit Tangent Vector

The derivative of a vector valued function gives a new vector valued function that is tangent to the defined curve.  The analogue to the slope of the tangent line is the direction of the tangent line.  Since a vector contains a magnitude and a direction, the velocity vector contains more information than we need.  We can strip a vector of its magnitude by dividing by its magnitude.  

 

Definition of the Unit Tangent Vector

  Let r(t) be a differentiable vector valued function and v(t) = r'(t) be the velocity vector.  Then we define the unit tangent vector by as the unit vector in the direction of the velocity vector.

                                 v(t)                T(t)  =                                                      ||v(t)||

 

 

Example

Let 

        r(t)  =  t i + et j - 3t2 k

Find the T(t) and T(0).

 

Solution

We have 

        v(t)  =  r'(t)  =  i + et j - 6t k

and 

       

To find the unit tangent vector, we just divide

       

To find T(0) plug in 0 to get

       

 

The Principal Unit Normal Vector

        A normal vector is a perpendicular vector.  Given a vector v in the space, there are infinitely many perpendicular vectors.  Our goal is to select a special vector that is normal to the unit tangent vector.  Geometrically, for a non straight curve, this vector is the unique vector that point into the curve.  Algebraically we can compute the vector using the following definition.

 

Definition of the Principal Unit Normal Vector

Let r(t) be a differentiable vector valued function and let T(t) be the unit tangent vector.  Then the principal unit normal vector N(t) is defined by

                                 T'(t)                N(t)  =                                                      ||T'(t)||

 

 

Comparing this with the formula for the unit tangent vector, if we think of the unit tangent vector as a vector valued function, then the principal unit normal vector is the unit tangent vector of the unit tangent vector function.  You will find that finding the principal unit normal vector is almost always cumbersome.  The quotient rule usually rears its ugly head.  

 

Example

Find the unit normal vector for the vector valued function

        r(t)  =  ti + t2 j 

and sketch the curve, the unit tangent and unit normal vectors when t = 1.

 

Solution

First we find the unit tangent vector

       

Now use the quotient rule to find T'(t)

       

Since the unit vector in the direction of a given vector will be the same after multiplying the vector by a positive scalar, we can simplify by multiplying by the factor

       

The first factor gets rid of the denominator and the second factor gets rid of the fractional power.  We have

       

Now we divide by the magnitude (after first dividing by 2) to get 

       

Now plug in 1 for both the unit tangent vector to get

                   

The picture below shows the graph and the two vectors.

           

 

Tangential and Normal Components of Acceleration

Imagine yourself driving down from Echo Summit towards Myers and having your brakes fail.  As you are riding you will experience two forces (other than the force of terror) that will change the velocity.  The force of gravity will cause the car to increase in speed.  A second change in velocity will be caused by the car going around the curve.  The first component of acceleration is called the tangential component of acceleration and the second is called the normal component of acceleration.  As you may guess the tangential component of acceleration is in the direction of the unit tangent vector and the normal component of acceleration is in the direction of the principal unit normal vector.  Once we have T and N, it is straightforward to find the two components.  We have

 

Tangential and Normal Components of Acceleration

The tangential component of acceleration is 

              

and the normal component of acceleration is 

              

and

                   a   =   aNN + aTT

 

Proof

First notice that 

        v  =  ||v|| T        and        T'  =  ||T'|| N

Taking the derivative of both sides gives

        a  =  v'  =  ||v||' T + ||v|| T'  =  ||v||' T + ||v|| ||T' || N 

This tells us that the acceleration vector is in the plane that contains the unit tangent vector and the unit  normal vector.  The first equality follows immediately from the definition of the component of a vector in the direction of another vector.  The second equalities will be left as exercises.

 

Example 

Find the tangential and normal components of acceleration for the prior example

                r(t)  =  ti + t2 j 

 

Solution

Taking two derivatives, we have

        a(t)  =  r''(t)  =  2j

We dot the acceleration vector with the unit tangent and normal vectors to get

       

       

 

 

Arc Length Curvature

Arc Length

For a parametrically defined curve we had the definition of arc length.  Since vector valued functions are parametrically defined curves in disguise, we have the same definition.  We have the added benefit of notation with vector valued functions in that the square root of the sum of the squares of the derivatives is just the magnitude of the velocity vector.  

 

Definition of Arc Length

Let 

               r(t)  =  x(t) i + y(t) j + z(t) k

be a differentiable vector valued function on [a,b].  Then the arc length s is defined by 

              

 

Example

Suppose that 

        r(t)  =  3t i + 2j + t2k

Set up the integral that defines the arc length of the curve from 2 to 3.  Then use a calculator or computer to approximate the arc length.

 

Solution

We use the arc length formula

       

Notice that we could do this integral by hand by letting t  =  9/2 tan , however the question only asked us to use a machine to approximate the integral.  A TI 89 calculator gives

        s  =  5.8386

 

Parameterization by Arc Length

Recall that like parametric equations, vector valued function describe not just the path of the particle, but also how the particle is moving.  Among all representations of a curve there is a "simplest" one.  If the particle travels at the constant rate of one unit per second, then we say that the curve is parameterized by arc length.  We have seen this concept before in the definition of radians.  On a unit circle one radian is one unit of arc length around the circle.  When we say "simplest" we in no way mean that the equations are simple to find, but rather that the dynamics of the particle are simple.  To aid us in parameterizing by arc length, we define the arc length function.

 

Definition of the Arc Length Function

If r(t) is a differentiable vector valued function, then the arc length function is defined by

         

 

Remark:  By the second fundamental theorem of calculus, we have

                    s'(t)  =  ||v(t)||

If a vector valued function is parameterized by arc length, then 

        s(t)  =  t

If we have a vector valued function r(t) with arc length s(t), then we can introduce a new variable 

        s  =  s-1(t)

So that the vector valued function r(s) will have arc length equal to 

        s(s-1(t))  =  t

so that r(s)  will be parameterized by arc length.  This process is usually impossible for two reasons.  Firstly, the integral that defines arc length involves a square root in the integrand.  This integral is usually impossible to determine.  Secondly, even if the integral is possible to evaluate, finding the inverse of a function is often impossible.  There are a few special curves that can be parameterized by arc length.  We will demonstrate one of them.

 

Example:     Parameterizing by arc length

Find the arc length parameterization of the helix defined by 

        r(t)  =  cos t i + sin t j + t k

 

Solution

First find the arc length function

       

Solving for t gives

        t   =  s /

Now substitute back into the position equation to get

        r(s)  =  cos(s / ) i + sin(s / ) j + s / k

 

Definition of Curvature

Consider a car driving along a curvy road.  The tighter the curve, the more difficult the driving is.  In math we have a number, the curvature, that describes this "tightness".  If the curvature is zero then the curve looks like a line near this point.  While if the curvature is a large number, then the curve has a sharp bend.  

More formally, if T(t) is the unit tangent vector function then the curvature is defined at the rate at which the unit Tangent vector changes with respect to arc length.  

        Curvature = k = ||d/ds (T(t)) || =  ||r''(s)||

As we stated previously, this is not a practical definition, since parameterizing by arc length is typically impossible.  Instead we use the chain rule to get

         ||d/ds (T(t)) ||  =  ||T'(t) dt/ds|| 

                  ||T'(t)||              ||T'(t)||          =                      =                                        ||ds/dt||              ||r'(t)||

This formula is more practical to use, but still cumbersome.  T'(t) is typically a mess.  Instead we can borrow from the formula for the normal vector to get the curvature

 

                    ||r'(t) x r''(t)||     K(t)   =                                                    ||r'(t)||3

 

Example  

Find the curvature at /2 if

        r(t) = cos t i - 1/t j + sin t k

 

Solution

We take derivatives

        r'(t)  =  -sin t i + 1/t2 j + cos t k

        r''(t)  =  -cos t i - 2/t3 j - sin t k

Plugging in /2 gives

        r'(/2)  =  -i  +  4/2 j

        r''(/2)  =  -16/3 j - k 

Now take the cross product to get

        r'(/2)  x  r''(/2)  =  -4/2 i - j + 16/3 k

Finally, we plug this information into the curvature formula to get

       

 

Curvature of a Plane Curve

If a curve in the xy-plane is defined by the function y = f(t) then there is an easier formula for the curvature.  We can parameterize the curve by 

        r(t)  =  t i + f(t) j 

We have

        r'(t)  =  i + f '(t) j 

        r''(t)  =  f ''(t) j 

Their cross product is just

        

        r'(t)  x  r''(t)  =  f ''(t) k

which has magnitude 

        ||r'(t)  x  r''(t)||  =  |f ''(t)|

The curvature formula gives

 

Curvature for a Plane Curve

         

 

Example

Find the curvature for the curve

        y  =  sin x

 

Solution

We have 

        f '(x)  =  cos x

        f ''(x)  =  -sin x

Plugging into the curvature formula gives

       

 

The Osculating Circle

In first year calculus we saw how to approximate a curve with a line , parabola, etc. Instead we can find the best fitting circle at the point on the curve.  If P is a point on the curve, then the best fitting circle will have the same curvature as the curve and will pass through the point P.  We will see that the curvature of a circle is a constant 1/r, where r is the radius of the circle.  The center of the osculating circle will be on the line containing the normal vector to the circle.  In particular the center can be found by adding

OP + 1/K N  

       

 

Exercise  

Find the equation of osculating circle to y = x2 at x = -1.  

Solution

 

The Normal Component of Acceleration Revisited

How is the normal component of acceleration related to the curvature.  If you remember, the normal component the acceleration tells us how fast the particle is changing direction.  If a curve has a sharp bend (high curvature) then the directional change will be faster.  We now show that there is a definite relationship between the normal component of acceleration and curvature.

        a(t) = aTT(t) + aNN(t)

We have

        a(t) = r''(t) = d/dt(r'(t)) = d/dt(||r'(t)||T(t)) = d/dt(||r'(t)||)T(t) + ||r'(t)||T'(t)

        = s''(t)T(t) + s'T'(t) = s''(t)T(t) + s'||T'(t)||N(t) = s''(t)T(t) + ks'2N(t)

So that the tangential component of the acceleration is s''(t) and the normal component is k(t)s'2(t)  

Exercise

Find the tangential and normal components of

        r(t) = ti - 2tj + t2k

 

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Functions of Several Variables

Definition of Functions of Several Variables

A function of several variables is a function where the domain is a subset of Rn and range is R.

Example:  

        f(x,y) = x - y 

is a function of two variables

                             x - y        g(x,y,z) =                                              y - z

is a function of three variables.

 

Finding the Domain

To find the domain of a function of several variables, we look for zero denominators and negatives under square roots:

Example

Find the domain of 

       

First, the inside of the square root must be positive, that is 

        x - y  >  0

second, the denominator must be nonzero, that is 

        x + y      0 

hence we need to stay off the line 

        y  =  -x

Putting this together gives 

        {(x,y) | x - y > 0 and y -x}

The graph to the right shows the domain as the shaded green region.

.

Exercise

Find the domain of the function 

       

Level Curves

The topographical map shown below is of the Rubicon Trail.  It represents the function that maps a longitude and latitude to an altitude.  

       

Each curve represents a path where the z-coordinate (altitude) is a constant.  Crossing many topo lines in a short distance represents a path that is very steep.

Now  lets make our own contour map of the function.

        f(x,y) =  y - x2  

by setting constant values for z:

z Equation

1 y = x2 + 1

2 y = x2 + 2

We see that each topo line is a parabola and that the y-intercept gives the height.  Below is a contour diagram of this function.

       

Names for the curves drawn are level curves, isotherms (for temperature), isobars (for pressure), and equipotential lines (for electric potential fields) depending on what the two variable function represents.

 

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Limits

Topology Terminology

Let P be a point in the plane then a neighborhood (ball) of P is the set of points that are less than units away from P.  If R is a region, then a point P is called an interior point of R if there is a neighborhood totally contained in R.  If every point of R is an interior point of R then R is called open. A point P is called a boundary point of R if every neighborhood of R contains both points in R and not in R.  R is called closed if it contains all of its boundary points.

        

 

The Definition of a Limit

                          Definition

Let f(x,y) be a function defined near the point P, then 

       

if there is a such that f(x,y) is close to L for all points (except possibly P) in the neighborhood of P.

Equivalently, the limit is L if for all paths that lead to P, the function also tends towards P.  (Recall that for the one variable case we needed to check only the path from the left and from the right.) To show that a limit does not exist at a point we need only find two paths that both lead to P such that f(x,y) tends towards different values.

Techniques For Finding Limits

Example

Show that      

       

Does not exist

 

Solution

 First select the path along the x-axis.  On this path 

        y = 0 

so the function becomes:

                          0        f(x,0)  =              =  0                         x2   

Now choose the path along the y = x line:

                          x2            1        f(0,y)  =              =                                    2x2           2 

Hence the function tends towards two different values for different paths.  We can conclude that the limit does not exist.  The graph is pictured below.

Example

Find

       

We could try the paths from the last example, but both paths give a value of 0 for the limit.  Hence we suspect that the limit exists.  We convert to polar coordinates and take the limit as r approaches 0:

We have 

                          r3cos3 + r3sin3

        f(r,)  =                                    =  rcos3 + rsin3                                      r2  

as r approaches 0 the function also approaches 0 no matter what is.  Hence the limit is 0.

Below is the graph of this function.

Exercises:  Find the limit if it exists

A.

B.

Continuity

We make the following definition for continuity.

                              Definition

A function of several variables is continuous at a point P if the limit exists at P and the function defined at P is equal to this limit.

As with functions of one variable, polynomials are continuous, sums, products, and compositions of continuous functions are continuous.  Quotients of continuous functions are continuous.  A function is continuous if it is continuous at every point.

 

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Partial Derivatives

Definition of a Partial Derivative

Let f(x,y) be a function of two variables.  Then we define the partial derivatives  as

    Definition of the Partial Derivative 

       

if these limits exist.  

Algebraically, we can think of the partial derivative of a function with respect to x as the derivative of the function with y held constant.  Geometrically, the derivative with respect to x at a point P represents the slope of the curve that passes through P whose projection onto the xy plane is a horizontal line.  (If you travel due East, how steep are you climbing?)

Example

Let 

        f(x,y) = 2x + 3y 

then

       

We also use the notation fx  and fy for the partial derivatives with respect to x and y respectively.

Exercise:  

Find fy for the function from the example above.

Finding Partial Derivatives the Easy Way

Since a partial derivative with respect to x is a derivative with the rest of the variables held constant, we can find the partial derivative by taking the regular derivative considering the rest of the variables as constants.

Example   Let 

        f(x,y)  =  3xy2 - 2x2y

then 

        fx  =  3y2 - 4xy

and

        fy  =  6xy - 2x2

Exercises

 Find both partial derivatives for

A. f(x,y) = xy sin x

B.                 x + yf(x,y) =                                     x - y

Higher Order Partials

Just as with function of one variable, we can define second derivatives for functions of two variables.  For functions of two variables, we have four types:

        fxx,     fxy     fyx     and     fyy

Example

Let 

        f(x,y)  =  y ex  

then 

        fx  =  yex

and

        fy  =  ex

Now taking the partials of each of these we get:

        fxx = y ex        fxy = ex        fyx = ex       and       fyy = 0

Notice that    

        fxy  =   fyx  

                                     Theorem  

Let f(x,y) be a function with continuous second order derivatives, then   

           fxy  =   fyx  

Functions of More Than Two Variables

Suppose that 

        f(x,y,z)  =  xy - 2yz 

is a function of three variables, then we can define the partial derivatives in much the same way as we defined the partial derivatives for three variables.  

We have

        fx = y         fy = x - 2z      and       fz = -2y

Application:     The Heat Equation

Suppose that a building has a door open during a snowy day.  It can be shown that the equation

        Ht  =  c2Hxx     

models this situation where H is the heat of the room at the point x feet away from the door at time t.  Show that 

        H = e-t cos(x/c) 

satisfies this differential equation.

Solution

We have

        Ht  =  -e-t cos(x/c) 

        Hx  =  -1/c e-t sin(x/c)

        Hxx  =  -1/c2 e-t cos(x/c)

So that 

        c2Hxx  =  -e-t cos(x/c)

And the result follows.

 

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The Gradient

Directional Derivatives

Suppose you are given a topographical map and want to see how steep it is from a point that is neither due West or due North.  Recall that the slopes due north and due west are the two partial derivatives.  The slopes in other directions will be called the directional derivatives.  Formally, we define

          Definition  

Let f(x,y) be a differentiable function and let u be a unit vector then the directional derivative of f in the direction of u is

      

Note that if u is i then the directional derivative is just fx and if u is j the it is fy.

Just as there is a difficult and an easy way to compute partial derivatives, there is a difficult way and an easy way to compute directional derivatives.

          Theorem

Let f(x,y) be a differentiable function, and u be a unit vector with direction , then

( , ) , cos ,sinu x yD f x y f f

Example:  

Let 

        f(x,y)  =  2x + 3y2 - xy 

and 

        v  =  <3,2> 

Find 

        Dv f(x,y)

Solution

We have 

        fx  =  2 - y 

and 

        fy  =  6y - x 

and 

       

Hence

        Dv f(x,y)  =  <2 - y, 6y - x> . <3/ , 2/ > 

                  2                         3              =                (2 - y)  +            (6y - x)

                                   

 

Exercise

Let 

       

Find  Dv f(x,y)

The Gradient

We define 

  grad f  =  <fx, fy>

Notice that         

     Du f(x,y) = (grad f) . u

The gradient has a special place among directional derivatives.  The theorem below states this relationship.

               Theorem

1. If grad f(x,y) = 0 then for all u,  Du f(x,y) = 02. The direction of grad f(x,y) is the direction with

maximal directional derivative.

3. The direction of -grad f(x,y) is the direction with the minimal directional derivative.

Proof:  

1. If 

        gradf(x,y)  =  0 

then   

        Du f(x,y)  =  grad f . u  =  0 . u  =  02. Du f(x,y)  =  grad f . u  =  ||grad f || cos

This is a maximum when = 0 and a minimum when = .  If = 0 then grad f and u point in the same direction.  If = then u and grad f point in opposite directions.  This proves 2 and 3.

Example:   

Suppose that a hill has altitude 

        w(x,y)  =  x2 - y

Find the direction that is the steepest uphill and the steepest downhill at the point (2,3).

Solution

We find

        grad w  =  <2x, -y>  =  <4, -3>

Hence the steepest uphill is in the direction 

        <4,-3>

while the steepest downhill is in the direction 

        -<4,-3>  =  <-4,3>

The Gradient and Level Curves

If f is differentiable at (a,b) and grad f is nonzero at (a,b) then grad f is perpendicular to the level curve through (a,b).

       

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Tangent Planes and Normal Lines

Tangent Planes

Let z = f(x,y) be a function of two variables.  We can define a new function F(x,y,z) of three variables by subtracting z.  This has the condition

        F(x,y,z)  =  0

Now consider any curve defined parametrically by

        x  =  x(t),      y  =  y(t)      z  =  z(t)

We can write,

        F(x(t), y(t), z(t))  =  0

Differentiating both sides with respect to t, and using the chain rule gives

        Fx(x, y, z) x' + Fy(x, y, z) y' + Fz(x, y, z) z'  =  0

Notice that this is the dot product of the gradient function and the vector <x',y',z'>,

        GradF . <x', y', z'>  =  0

In particular the gradient vector is orthogonal to the tangent line of any curve on the surface.  This leads to

                    Definition

Let F(x,y,z) define a surface that is differentiable at a point (x0,y0,z0), then the tangent plane to F ( x, y, z ) at ( x0 , y0 , z0 ) is the plane with normal vector

          Grad F(x0,y0,z0)

that passes through the point (x0,y0,z0).  In Particular the equation of the tangent plane is

       Grad F(x0,y0,z0) . < x - x0 , y - y0 , z - z0 >  =  0

Example

Find the equation of the tangent plane to

        z  =  3x2 - xy

at the point (1,2,1)

Solution

We let

        F(x,y,z)  =  3x2 - xy - z

then

        Grad F  =  <6x - y, -x, -1>

At the point (1,2,1), the normal vector is

        Grad F(1,2,1) = <4, -1, -1>

Now use the point normal formula for a plan

        <4, -1, -1> . <x - 1, y - 2, z - 1>  =  0

or

        4(x - 1) - (y - 2) - (z - 1)  =  0

Finally we get

        4x - y - z  =  1

Normal Lines

Given a vector and a point, there is a unique line parallel to that vector that passes through the point.  In the context of surfaces, we have the gradient vector of the surface at a given point.  This leads to the following definition.

                        DefinitionLet F(x,y,z) define a surface that is differentiable at a point (x0,y0,z0), then the normal line to F(x,y,z) at ( x0 , y0 , z0 ) is the line with normal vector

          GradF(x0,y0,z0)

that passes through the point (x0,y0,z0).  In Particular the equation of the normal line is

          x(t) = x0 + Fx(x0,y0,z0) t  

          y(t) = y0 + Fy(x0,y0,z0) t  

          z(t) = z0 + Fz(x0,y0,z0) t

Example

Find the parametric equations for the normal line to

        x2yz - y + z - 7  = 0

at the point (1,2,3).

Solution

We compute the  gradient

        Grad F  =  <2xyz, x2z - 1, x2y + 1>  =  <12, 2, 3>

Now use the formula to find

        x(t) = 1 + 12t      y(t) = 2 + 2t      z(t) = 3 + 3t

The diagram below displays the surface and the normal line.

       

Angle of Inclination

Given a plane with normal vector n the angle of inclination, is defined by

                 |n . k|    cos =                                  ||n|| 

More generally, if 

        F(x,y,z)  =  0

is a surface, than the angle of inclination at the point  (x0, y0, z0) is defined by the angle of inclination of the tangent plane at the point.

                    |Grad F(x0, y0, z0) . k|     cos =                                                                  ||Grad F(x0, y0, z0)||

Example

Find the angle of inclination of

         x2             y2          z2

                 +            +             =  1          4            4            8

at the point (1,1,2).

Solution

First compute

        Grad F  =  <x / 2, y / 2, z / 4>

Now plug in to get

        Grad F(1,1,2)  =  <1/2, 1/2, 1/2>

We have

        |<1/2, 1/2, 1/2> . k|  =  1/2

Also,

        ||<1/2, 1/2, 1/2>||  =  / 2

Hence

        cos = (1/2)/[( )/2] = 1/

So the angle of inclination is

          =  arccos(1/ )  .955 radians

The Tangent Line to a Curve

Example

Find the tangent line to the curve of intersection of the sphere

        x2 + y 2 + z2  =  30

and the paraboloid

        z  =  x2 + y2

at the point (1,2,5).

Solution

We find the Grad of the two surfaces at the point

        Grad (x2 + y 2 + z2) = <2x, 2y, 2z> = <2, 4,10>

and

        Grad (x2 + y 2 - z)  =  <2x, 2y, -1> = <2, 4, -1>

These two vectors will both be perpendicular to the tangent line to the curve at the point, hence their cross product will be parallel to this tangent line.  We compute

       

Hence the equation of the tangent line is

        x(t) = 1 - 44t       y(t) = 2 + 22t       z(t) = 5

 

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MATH 202 PRACTICE MIDTERM 1

 

Please work out five of the given six problems and indicate which problem

you are omitting. Credit will be based on the steps that you show towards

the final answer. Show your work.

    Printable Key

PROBLEM 1 Please answer the following true or false. If false, explain why or provide

a counter example. If true explain why

 

A)   (12 Points) If r(t) is parameterized by arclength, then a and N are parallel.

Solution

B)    (13 Points) If r(t) is a differentiable vector valued function then

       

Solution

 

PROBLEM 2 (25 Points)

 

Let

        r(t)  =  2t i - 4t2 j

 

A. Find T(-1).

Solution

B. Find N(-1).

Solution

C. Find the equation of the circle of curvature for r(t) at t  =  -1.  

Solution

 

 

PROBLEM 3 (25 Points) Jason Elam (the football kicker for the Denver Broncos) can

kick a football with an initial velocity of 60 feet per second. At what angle should the

ball be kicked to maximize the horizontal distance that the ball travels before it lands on

the ground? (Use vectors please).

  Solution

 

PROBLEM 4 (25 Points) Prove the following theorem:

Let r(t) be a differentiable vector valued function, then

        |(r x v) . a|  =  ||r'|| ||aN|| |r

. (T x N)|

  Solution

PROBLEM 5 (25 Points)

Find the parametric equations of the tangent line to the curve that is formed by

intersecting the sphere  x2 + y2 + z2  =  2 and the plane x + y - z  =  2 at the point (1,1,0).

  Solution

PROBLEM 6 (25 Points)

If 

        a(t)  =  t i + j - k

find r(5) if 

        r(0)  =  i - k     and         r(1)  =  j + k

  Solution

Extra Credit: Write down one thing that your instructor can do to make the class better

and one thing that is going well.

(Any constructive remark will be worth full credit)

 

 

Iterated Integrals and Area

 

Definition of an Iterated Integral

Just as we can take partial derivative by considering only one of the variables a true variable and holding the rest of the variables constant, we can take a "partial integral".  We indicate which is the true variable by writing "dx", "dy", etc.  Also as with partial derivatives, we can take two "partial integrals" taking one variable at a time.  In practice, we will either take x first then y or y first then x.  We call this an iterated integral or a double integral.

 

Definition of a Double Integral

Let f(x,y) be a function of two variables defined on a region R bounded below and above by 

                y  =  g1(x)          and          y  =  g2(x)

and to the left and right by

                x  =  a          and          x  =  b

then the double integral (or iterated integral) of f(x,y) over R is defined by

 

Example

Find the double integral of  f(x,y)  =   6x2 + 2y  over R where R is the region between y  =  x2 and y  =  4. 

 

Solution

First we have that the inside limits of integration are x2 and 4.  The region is bounded from the left by x  =  -2 and from the right by x  =  2 as indicated by the picture below.

        

We now integrate

       

Changing the Order of Integration

If a region is bounded from the left by  x  =  h1(y) and the right by  x  =  h2(y) and below and above by y  =  c and y  =  d, then we can find the double integral of "dxdy" by first integrating with respect to x then with respect to y.  Sometimes there is a choice to make as to whether to integrate first with respect to x and then with respect to y.  We do whatever is easier.  

 

Example

Find the double integral of  f(x,y)  =  3y over the triangle with vertices (-1,1), (0,0), and (1,1).

       

Solution

If we try to integrate with respect y first, we will have to cut the region into two pieces and perform two iterated integrals.  Instead we integrate with respect to x first.  The region is bounded on the left and the right by x  =  -y  and x  =  y.  The lowest the region gets is y  =  0  and the highest is  y  =  1.  The integral is 

       

 

Example

Evaluate the integral

       

 

Solution

Try as you may, you will not find an antiderivative of and we don't want to get into

power series expansions.  We have another choice.  The picture below shows the region.

       

We can switch the order of integration.  The region is bounded above and below by y  =  1/3 x  and  y  =  0.  The double integral with respect to y first and then with respect to x is 

       

The integrand is just a constant with respect to y so we get

       

This integral can be performed with simple u-substitution. 

        u  =  x2        du  =  2x dx

and the integral becomes

       

Area

Recall from first year calculus, if a region R is bounded below by  y  =  g1(x)  and above by  y  =  g2(x), and  a  <  x  <  b, the area is given by 

       

There is another way of achieving this expression.  If we let the integrand by 1 then the double integral over the region R is 

       

This gives us another way of finding area.  

Theorem:  Area and Double Integrals

 If a region R is bounded below by  y  =  g1(x)  and above by  y  =  g2(x), and  a  <  x  <  b, then the area is given by

              

Remark:  If the region if bounded on the left by x  =  h1(y) and the right by h2(y) with c <  y  <  d, then the double integral of 1 dxdy can also be used to find the area. 

 

Example 

Set up the double integral that gives the area between  y  =  x2  and  y  =  x3.   Then use a computer or calculator to evaluate this integral.

 

Solution

The picture below shows the region

       

We set up the integral 

       

A computer gives the answer of 1/12.

An animation of the "Little Man"

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Double Integrals and Volume

Double Riemann Sums

In first year calculus, the definite integral was defined as a Riemann sum that gave the area under a curve.  There is a similar definition for the volume of a region below a function of two variables.  Let f(x,y) be a positive function of two variables and consider the solid that is bounded below by f(x,y) and above a region R in the xy-plane.  

               

For a two dimensional region, we approximated the area by adding up the areas of many approximating rectangles.  For the volume of a three dimensional solid, we take a similar approach.  Instead of rectangles, we use rectangular solids for the approximation.  We cut the region R into rectangles by drawing vertical and horizontal lines in the xy-plane.  Rectangles will be formed.  We let the rectangles be the base of the solid, while the height is the z-coordinate of the lower left vertex.  One such rectangular solid is shown in the figure.  

           

Taking the limit as the rectangle size approaches zero (and the number of rectangles approaches infinity) will give the volume of the solid.  If we fix a value of x and look at the rectangular solids that contain this x, the union of the solids will be a solid with constant width x.  The face will be approximately equal to the area in the yz-plane of the (one variable since x is held constant) function  z  =  f(x,y).  

           

This area is equal to 

       

If we add up all these slices and take the limit as x approaches zero, we get

       

Which is just the double integral defined in the last section

Instead of fixing the variable x we could have held y constant.  The picture below illustrates the resulting wedge.

By a similar argument, the volume of the wedge is 

       

Adding up all the wedge areas gives the total volume

       

This shows that the volume is equal to the iterated integral no matter which we integrate first.  This is called Fubini's Theorem.  Technically the volume is defined as the double Riemann sum of f(x,y) where we sum over the partition of R in the xy-plane.  We state it below.

Fubini's Theorem

Let f, g1, g2, h1, and h2 be defined and continuous on a region R.  Then the double integral equals

          

 

Notice that all the typical properties of the double integral hold.  For example, constants can be pulled out and the double integral of the sum of two functions is the sum of the double integrals of each function.

Finding Volume

 

 

Example

Set up the integral to find the volume of the solid that lies below the cone

        

and above the xy-plane.  

 

Solution

The cone is sketched below

           

We can see that the region R is the blue circle in the xy-plane.  We can find the equation by setting z  =  0.

       

Solving for y (by moving the square root to the left hand side,  squaring both sides, etc)

gives

       

The "-" gives the lower limit and the "+" gives the upper limit.  For the outer limits, we can see that 

        -4  <  x  <  4

Putting this all together gives

       

Either by hand or by machine we can obtain the result 

        Volume  =  64 /3

Notice that this agrees with the formula 

        Volume  =  r2h/3

 

Exercise

Set up the double integral for this problem with dxdy instead of dydx.  Then show that the two integrals give the same result.

 

Example

Set up the double integral that gives the volume of the solid that lies below the sphere

        x2 + y2 + z2  =  6

and above the paraboloid 

        z  =  x2 + y2 

 

Solution

           

The picture below indicated that the region is the disk that lies inside that circle of intersection of the two surfaces.  We substitute

        x2 + y2 + (x2 + y2)2  =  6

or

        x2 + y2 + (x2 + y2)2 - 6  =  0

Now factor with x2 + y2 as the variable to get

        (x2 + y2 - 2)(x2 + y2 + 3)  =  0

The second factor has no solution, while the first is 

        x2 + y2  =  2

Solving for y gives

       

and

        -   <  x  < 

Just as we did in one variable calculus, the volume between two surfaces is the double integral of the top surface minus the bottom surface.  We have

       

Again we can perform this integral by hand or by machine and get

        Volume  =  7.74

 

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Double Integration in Polar Coordinates

Polar Double Integration Formula

Many of the double integrals that we have encountered so far have involved circles or at least expressions with x2 + y2.  When we see these expressions a bell should ring and we should shout, "Can't we use polar coordinates."  The answer is, "Yes" but only with care.  Recall that when we changed variables in single variable integration such as u  =  2x, we needed to work out the stretching factor du  =  2dx.  The idea is similar with two variable integration.  When we change to polar coordinates, there will also be a stretching factor.  This is evident since the area of the "polar rectangle" is not just as one may expect .  The picture is shown below.

           

Even if r and are very small the area is not the product (r)().  This comes from the definition of radians.  An arc that extends radians a distance r out from the origin has length r.  If both r and are very small then the polar rectangle has area

        Area  =  r r  

This leads us to the following theorem

Theorem:  Double Integration in Polar Coordinates

Let f(x,y) be a continuous function defined over a region R bounded in polar coordinates by 

               r1()  <  r  <  r2()                    1  <    <  2

Then 

              

Notice the extra "r" in the theorem

 

Using Polar Coordinates

 

Example

Find the volume to the part of the paraboloid 

        z  =  9 - x2 - y2

that lies inside the cylinder

        x2 + y2  =  4

 

Solution  

The surfaces are shown below.

           

This is definitely a case for polar coordinates.  The region R is the part of the xy-plane that is inside the cylinder.  In polar coordinates, the cylinder has equation

        r2  =  4

Taking square roots and recalling that r is positive gives

        r  =  2

The inside of the cylinder is thus the polar rectangle 

        0  <  r  <  2        0  <    <  2

The equation of the parabola becomes 

        z  =  9 - r2 

We find the integral 

       

This integral is a matter of routine.  It evaluates to 28.

 

Example

Find the volume of the part of the sphere of radius 3 that is left after drilling a cylindrical hole of radius 2 through the center.

 

Solution

The picture is shown below

       

The region this time is the annulus (washer) between the circles r  =  2 and r  =  3 as shown below.

           

The sphere has equation

        x2 + y2 + z2  =  9

In polar coordinates this reduces to 

        r2 + z2  =  9

Solving for z by subtracting r2 and taking a square root we get top and bottom surfaces of 

       

We get the double integral 

       

This integral can be solved by letting

        u  =  9 - r2           du  =  -2rdr

After substituting we get

       

 

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Center of Mass and Moment of Inertia

Mass

We saw before that the double integral over a region of the constant function 1 measures the area of the region.  If the region has uniform density 1, then the mass is the density times the area which equals the area.  What if the density is not constant.  Suppose that the density is given by the continuous function 

        Density  =  (x,y)

In this case we can cut the region into tiny rectangles where the density is approximately constant.  The area of mass rectangle is given by 

        Mass  =  (Density)(Area)  =  (x,y) x y 

You probably know where this is going.  If we add all to masses together and take the limit as the rectangle size goes to zero, we get a double integral.

Mass

Let (x,y) be the density of a lamina (flat sheet) R at the point (x,y).  Then the total mass of the lamina is the double integral

              

 

Example

A rectangular metal sheet with 2  <  x  < 5  and  0  <  y  <  3 has density function 

        (x,y)  =  x + y

Set up the double integral that gives the mass of the metal sheet.

 

Solution

We just have the integral

       

 

Moments and Center of Mass

We have seen in first year calculus that the moments about an axis are defined by the product of the mass times the distance from the axis.  

        Mx  =  (Mass)(y)        My  =  (Mass)(x)

If we have a region R with density function (x,y), then we do the usual thing.  We cut the region into small rectangles for which the density is constant and add up the moments of each of these rectangles.  Then take the limit as the rectangle size approaches zero.  This will give us the total moment.  

Definition of Moments of Mass and Center of Mass

Suppose that (x,y) is a continuous density function on a lamina R.  Then the moments of mass are 

                             

and if M is the mass of the lamina, then the center of mass is 

                   

 

Example

Set up the integrals that give the center of mass of the rectangle with vertices (0,0), (1,0), (1,1), and (0,1) and density function proportional to the square of the distance from the origin.  Use a calculator or computer to evaluate these integrals.

 

Solution

The mass is given by 

       

The moments are given by 

       

These evaluate to 

        Mx  =  5k/12        and       My  =  5k/12

It should not be a surprise that the moments are equal since there is complete symmetry with respect to x and y.  Finally, we divide to get

        (x,y)  =  (5/8,5/8)

This tells us that the metal plate will balance perfectly if we place a pin at (5/8,5/8)

 

Moments of Inertia

We often call Mx and My the first moments.  They have first powers of y and x in their definitions and help find the center of mass.  We define the moments of inertia (or second moments) by introducing squares of y and x in their definitions.  The moments of inertia help us find the kinetic energy in rotational motion.  Below is the definition

Moments of Inertia

Suppose that (x,y) is a continuous density function on a lamina R.  Then the moments of inertia are 

                             

 

Exercise 

Find the moments of inertia for the previous square metal plate.

 

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Surface Area

 

Definition of Surface Area

In first year calculus we have seen how to find the surface area of revolution.  Now that we how the power of double integration, we are ready to take on the surface area for more general surfaces.  We can think of a smooth surface as a quilt flapping in the wind.  It consists of many rectangles patched together.  More generally and more accurately, let z  =  f(x,y) be a surface in R3 defined over a region R in the xy-plane.  cut the xy-plane into rectangles.  Each rectangle will project vertically to a piece of the surface as shown in the figure below.  Although the area of the rectangle in R is 

        Area  =  yx

The area of the corresponding piece of the surface will not be yx since it is not a rectangle.  Even if we cut finely, we will still not produce a rectangle, but rather will approximately produce a parallelogram.  With a little geometry we can see that the two adjacent sides of  the parallelogram are (in vector form) 

        u  =  x i + fx(x,y)x k 

and

        v  =  fy(x,y)y i + y k 

We can see this by realizing that the partial derivatives are the slopes in each direction.  If we run x  in the i direction, then we will rise  fx(x,y)x in the k direction so that

        rise/run  =   fx(x,y)

Which agrees with the slope idea of the partial derivative.  A similar argument will confirm the equation for the vector v.  Now that we know the adjacent vectors we recall that the area of a parallelogram is the magnitude of the cross product of the two adjacent vectors.  We have

       

This is the area of one of the patches of the quilt.  To find the total area of the surface, we add up all the areas and take the limit as the rectangle size approaches zero.  This results in a double Riemann sum, that is a double integral.  We state the definition below.

Definition of Surface Area

Let z  =  f(x,y) be a differentiable surface defined over a region R.  Then its surface area is given by

                

 

Examples

Example

Find the surface area of the part of the plane 

        z  =  8x + 4y

that lies inside the cylinder

        x2 + y2  =  16

Solution

We calculate partial derivatives

        fx(x,y)  =  8            fy(x,y)  =  4

so that 

        1 + fx2(x,y)  + fy

2(x,y)  =  1 + 64 + 16  =  81

Taking a square root and integrating, we get 

       

We could work this integral out, but there is a much easier way.  The integral of a

constant is just the constant times the area of the region.  Since the region is a circle, we

get

   

     Surface Area  =  9(16)  =  144 

 

In reality, since there is a square root in the formula, most surface area calculations

require intensive integration skills or the use of a machine.  The prior example and the

next example are not meant to deceive, but rather to show how the essence of surface

area problems work without the integration difficulty clouding your understanding.

 

Example

Find the surface area of the part of the paraboloid 

   

     z  =  25 - x

2

- y

2

 

that lies above the xy-plane.

                   

Solution

We calculate partial derivatives

        fx(x,y)  =  -2x            fy(x,y)  =  -2y

so that 

        1 + fx2(x,y)  + fy

2(x,y)  =  1 + 4x2 + 4y2 

At this point if we listen closely, we should hear a little voice pleading "Polar Coordinates".  We listen to its call and realize that the region is just the circle 

        r  =  5

Now convert the integrand to polar coordinates to get

       

2 52

0 0

1 4 r rdrd

Now let

        u  =  1 + 4r2         du  =  8rdr

and substitute

       

1012 101 21/2 3/2

0 1 0 1

1 1 530.958 12

u dud u

 

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Triple Integrals

Definition of the Triple Integral

We have seen that the geometry of a double integral involves cutting the two dimensional region into tiny rectangles, multiplying the areas of the rectangles by the value of the function there, adding the areas up, and taking a limit as the size of the rectangles approaches zero.  We have also seen that this is equivalent to finding the double iterated iterated integral.  

We will now take this idea to the next dimension.  Instead of a region in the xy-plane, we will consider a solid in xyz-space.  Instead of cutting up the region into rectangles, we will cut up the solid into rectangular solids.  And instead of multiplying the function value by the area of the rectangle, we will multiply the function value by the volume of the rectangular solid.  

We can define the triple integral as the limit of the sum of the product of the function times the volume of the rectangular solids. 

Instead of the double integral being equivalent to the double iterated integral, the triple integral is equivalent to the triple iterated integral.

Definition of the Triple Integral

Let f(x,y,z) be a continuous function of three variables defined over a solid Q.  Then the triple integral over Q is defined as

         

where the sum is taken over the rectangular solids included in the solid Q and lim is taken to mean the limit as the side lengths of the rectangular solid.  

 

This definition is only practical for estimating the triple integral when a data set is given.  When we have a symbolically defined function, we use an extension of the fundamental theorem of calculus which is just Fubini's theorem for triple integrals.

 

Theorem for Evaluating Triple Integrals

Let f(x,y,z) be a continuous function over a solid Q defined by

     a  <  x  <  b          h1(x)  <  y  <  h2(x)          g1(x,y)  <  z  <  g2(x,y)

Then the triple integral is equal to the triple iterated integral.

        

 

Remark:  As with double integrals the order of integration can be changed with care.

Examples

 

Example

Evaluate 

       

Where 

        f(x,y,z)  =  1 - x

and Q is the solid that lies in the first octant and below the plain 

        3x + 2y + z  =  6

 

Solution

The picture of the region 

           

The challenge here is to find the limits.  We work on the innermost limit first which corresponds with the variable "z".  Think of standing vertically.  Your feet will rest on the lower limit and your head will touch the higher limit.  The lower limit is the xy-plane or

        z  =  0

The upper limit is the given plane.  Solving for z, we get

        z  =  6 - 3x - 2y

Now we work on the middle limits that correspond to the variable "y".  We look at the projection of the surface in the xy-plane.  It is shown below.

           

Now we find the limits just as we found the limits of double integrals.  The lower limit is just

        y  =  0

If we set z  = 0 and solve for y, we get for the upper limit

        y  =  3 - 3/2 x

Next we find the outer limits, corresponding to the variable "x".  The lowest x gets is 0 and highest x gets is 2.  Hence 

        0  <  x  <  2

The integral is thus

           

 

Example

Switch the order of integration from the previous example so that dydxdz appears.

 

Solution

This time we work on the "y" variable first.  The lower limit for the y-variable is 0.  For the upper limit, we solve for y in the plane to get

        y  =  3 - 3/2 x - 1/2 z

To find the "x" limits, we project onto the xz-plane as shown below

       

The lower limit for x is 0.  To find the upper limit we set y = 0 and solve for x to get

        x  =  2 - 1/3 z

Finally, to get the limits for z, we see that the smallest z will get is 0 and the largest z will get is 6.  We get

                0  <  z  <  6

We can write

       

Mass, Center of Mass, and Moments of Inertia

For a three dimensional solid with constant density, the mass is the density times the volume.  If the density is not constant but rather a continuous function of x, y, and x, then we can cut the solid into very small rectangular solids so that on each rectangular solid the density is approximately constant.  The volume of the rectangle is 

        Mass  =  (Density)(Volume)  =  f(x,y,z) xyz

Now do the usual thing.  We add up all the small masses and take the limit as the rectangular solids get small.  This will give us the triple integral

       

We are often interested in the center of mass of a solid.  For example when the NEAR satellite orbited around the asteroid Eros, NASA scientists needed to compute the center of mass of the asteroid.  Kepler told us that a stable orbit will always orbit in an elliptical orbit with the center of mass as one of the foci.

 

The NEAR satellite orbiting around Eros

We find the center of mass of a solid just as we found the center of mass of a lamina.  Since we are in three dimensions, instead of the moments about the axes, we find the moments about the coordinate planes.  We state the definitions from physics below.

Definition:    Moments and Center of Mass

Let (x,y,z) be the density of a solid Q.  Then the first moments about the coordinate planes are

              

              

              

and the center of mass is given by

              

Notice that letting the density function being identically equal to 1 gives the volume

 

       

Exercise

Find the center of mass of the solid that lies below the paraboloid 

        z  =  4 - x2 - y2 

that lies above the xy-plane if the density of the region is given by 

        (x,y,z)  =  x2 + 2y2 + z

You may use your calculator or computer to evaluate the integrals.

 

Just as with lamina, there are formulas for moments of inertial about the three axes.  They involve multiplying the density function by the square of the distance from the axes.  We have

 

Definition:    Moments of Inertia

Let (x,y,z) be the density of a solid Q.  Then the first moments of inertia about the coordinate axes are

              

              

              

 

Click here for an application to probability

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Triple Integrals in Cylindrical and Spherical Coordinates

 

Cylindrical Coordinates

When we were working with double integrals, we saw that it was often easier to convert to polar coordinates.  For triple integrals we have been introduced to three coordinate systems.  The rectangular coordinate system (x,y,z) is the system that we are used to.  The other two systems, cylindrical coordinates (r,,z) and spherical coordinates (,,) are the topic of this discussion.  

Recall that cylindrical coordinates are most appropriate when the expression 

        x2 + y2 

occurs.  The construction is just an extension of polar coordinates.  

        x  =  r cos         y  =  r sin         z  =  z

Since triple integration can be looked at as iterated integration we have

       

This leads us the the following theorem

 

Theorem:  Integration With Cylindrical Coordinates

Let f(x,y,z) be a continuous function on a solid Q.  Then 

         

 

Example

Find the moment of inertia about the z-axis of the solid that lies below the paraboloid 

        z  =  25 - x2 - y2 

inside the cylinder 

        x2 + y2  =  4

above the xy-plane, and has density function

        (x,y,z)  =  x2 + y2 + 6z

           

Solution    

By the moment of inertia formula, we have

       

The region, being inside of a cylinder is ripe for cylindrical coordinates.  We get

       

Spherical Coordinates

Another coordinate system that often comes into use is the spherical coordinate system.  To review, the transformations are 

        x  =  cos sin        y  =  sin sin        z  =  cos  

In the next section we will show that 

        dzdydx  =  sin ddd  

This leads us to 

 

Theorem:  Integration With Spherical Coordinates

Let f(x,y,z) be a continuous function on a solid Q.  Then 

         

 

Example

Find the volume of solid that lies inside the sphere 

        x2 + y2 + z2  =  2

and outside of the cone

        z2  =  x2 + y2 

           

 

Solution

We convert to spherical coordinates.  The sphere becomes 

          = 

To convert the cone, we add z2 to both sides of the equation

        2z2  =  x2 + y2 +z2 

Now convert to

        22cos  =  2 

Canceling the 2 and solving for we get

          =  cos-1(1/ )  =  /4  or 7/4  

In spherical coordinates (since the coordinates are periodic)

        7/4  =  3/4

To find the volume we compute

       

       

32 24

2

0 04

sin V d d d

 

Evaluating this integral should be routine at this point and is equal to 

                  8     V  =                                      3

 

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Jacobians

Review of the Idea of Substitution

Consider the integral

       

To evaluate this integral we use the u-substitution

        u  =  x2 

This substitution sends the interval [0,2] onto the interval [0,4].  We can see that there is stretching of the interval.  The stretching is not uniform.  In fact, the first part [0,0.5] is actually contracted.  This is the reason why we need to find du.

          du                                 dx             1                   =  2x        or                 =                               dx                                 du             2x

   This is the factor that needs to be multiplied in when we perform the substitution.  Notice for small positive values of x, this factor is greater than 1 and for large values of x, the factor is smaller than 1.  This is how the stretching and contracting is accounted for.       

Jacobians

We have seen that when we convert to polar coordinates, we use

        dydx  =  rdrd

With a geometrical argument, we showed why the "extra r" is included.  Taking the analogy from the one variable case, the transformation to polar coordinates produces stretching and contracting.  The "extra r" takes care of this stretching and contracting.  The goal for this section is to be able to find the "extra factor" for a more general transformation.  We call this "extra factor" the Jacobian of the transformation. We can find it by taking the determinant of the two by two matrix of partial derivatives.

 

Definition of the Jacobian

Let 

               x = g(u,v)       and        y  =  h(u,v)

be a transformation of the plane.  Then the Jacobian of this transformation is 

              

 

Example 

Find the Jacobian of the polar coordinates transformation 

        x(r,)  =  r cos             y(r,)  =  r sin

 

Solution

We have

       

This is comforting since it agrees with the extra factor in integration.

 

Double Integration and the Jacobian

 

Theorem:  Integration and Coordinate Transformations

Let 

                    

given by

               x  =  g(u,v),     y  =  h(u,v)

be a transformation on the plane that is one to one from a region S to a region R.  If g and h have continuous partial derivatives such that the Jacobian is never zero, then 

              

Remark:  A useful fact is that the Jacobian of the inverse transformation is the reciprocal of the Jacobian of the original transformation.

     

This is a consequence of the fact that the determinant of the inverse of a matrix A is the reciprocal of the determinant of A.

 

Idea of the Proof

As usual, we cut S up into tiny rectangles so that the image under T of each rectangle is a parallelogram.  

We need to find the area of the parallelogram.  Considering differentials, we have

        T(u + u,v)    T(u,v) + (xuu,yuu)

        T(u,v + v)    T(u,v) + (xvv,yvv)

Thus the two vectors that make the parallelogram are 

        P  =  guu i + huu j

        Q  =  gvv i + hvv j

To find the area of this parallelogram we just cross the two vectors.

       

and the extra factor is revealed.

 

Example

Use an appropriate change of variables to find the volume of the region below 

        z  =  (x - y)2 

above the x-axis, over the parallelogram with vertices (0,0), (1,1), (2,0), and (1,-1)

           

Solution

We find the equations of the four lines that make the parallelogram to be

        y  =  x        y  =  x - 2        y  =  -x        y  =  -x + 2

or

        x - y  =  0        x - y  =  2        x + y  =  0        x + y  =  2

The region is given by 

        0  <  x - y  <  2        and        0  <  x + y  < 2

This leads us to the inverse transformation

        u(x,y)  =  x - y        v(x,y)  =  x + y

The Jacobian of the inverse transformation is    

       

  Since the Jacobian is the reciprocal of the inverse Jacobian we get

       

The region is given by 

        0  <  u  <  2        and        0  <  v  < 2

and the function is given by 

        z  =  u2 

Putting this all together, we get the double integral

                 

Jacobians and Triple Integrals

For transformations from R3 to R3, we define the Jacobian in a similar way

       

 

Example

Find the Jacobian for the spherical coordinate transformation

        x  =  cos sin        y  =  sin sin        z  =  cos     

Solution

We take partial derivatives and compute

 

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MATH 202 PRACTICE MIDTERM 2

 

Please work out each of the given problems. Credit will be based on the steps that

you show towards the final answer. Show your work.

Printable Key

PROBLEM 1 Please answer the following true or false. If false, explain why or provide

a counter example. If true explain why.

A)   If all six limits of integration of an integral written in spherical coordinates are

constants, then the region of integration is a sphere.

Solution

B)   

  Solution

PROBLEM 2 Set up integrals to evaluate the following. Use the coordinate system that

will most effectively solve the integral.

A.     The mass of the tetrahedron that lies in the first octant and below the plane x + y +

2z  =  2 such that the density function is f(x,y,z)  =  3yz .

  Solution

B.     The surface area of the part of the paraboloid  z  =  9 - x2 - y2  that lies above the

plane z = 5 .

  Solution

C.     The moment of inertia about the z-axis of the solid between the cylinders  x2 + y2  = 

25 and x2 + z2  =  25 that has density function 1.

  Solution

PROBLEM 3

Switch the order of integration.

       

Solution

PROBLEM 4

A master dart thrower has determined that her probability density function is inversely

proportional to one more than the fourth power of the distance in centimeters from the

center of the dartboard.

A. Find the constant of proportionality. (Hint:   )  

Solution

B. Find the probability of her hitting the bull’s-eye which is one centimeter in radius.

  Solution

PROBLEM 5

Consider the solid described by

        (x + 2y)2 + (2y - 2z)2 + (x + z)2  < 25

with density function 

        f(x,y)  =  x + 2y

Show that the mass of the solid is equal to 

Hint: Recall that  

Solution

 

Vector Fields

Definition and Examples of Vector Fields

We have now seen many types of functions.  They are characterized by the domain and the range.  Below is a list of some of the functions that we have encountered so far.

Domain Range Name

R R One variable Function

R R2 Parametric Equations

R2 R Function of 2 Variables

R Vectors Vector Valued Function

By letting the domain be Rn and the range be n-dimensional vectors, we get a new type of functions called a vector field.

 

Definition of a Vector Field

A vector field is be a function where the domain is Rn and the range is n-dimensional vectors.

 

Example

An important vector field that we have already encountered is the gradient vector field.  Let f(x,y) be a differentiable function then the function that take a point (x0,y0) to gradf(x0,y0) is a vector field since the gradient of a function at a point is a vector.  For example, if

        f(x,y)  =  0.1xy - 0.2y

then 

        gradf(x,y)  =  0.1yi + (0.1x - 0.2)j

The sketch of the gradient is pictured below.

           

The best way to sketch a vector field is to use the help of a computer, however it is important to understand how they are sketched.  For this example, we pick a point, say (1,2) and plug it into the vector field

        gradf(1,2)  =  .2i - .1j

Next, sketch the vector that begins at (1,2) and ends at (1 + .2, .2 - .1).  Notice that when we sketch vector fields, we use the definition that involves two points rather than the definition that assumes all vectors emanate from the origin.

 

Example

In physics, many vector fields satisfy the inverse square law.  A vector field F satisfying the inverse square law has the property that if 

        r  =  xi + yj + zj 

and u is the unit vector in the direction of r (u  =  r/||r||), then 

       

Examples of force fields that satisfy the inverse square law are gravitational force fields and electric force fields.  Below is a plot of a vector field that satisfies the inverse square law.

           

Notice that as the distance from the origin gets small, the vector become small quickly.  For gravity, this tells us that as we fly away from the earth, we experience less gravity, until it seems like weightlessness.

Other examples of vector fields that occur in nature include velocity fields.  For example, the currents in the ocean ensure movement of the water.  The if we sketch the velocity vector of at each point of the ocean, we get a vector field.  A similar velocity field is produced from wind in the atmosphere.  

Conservative Vector Fields

Our first and most important example was the gradient vector field.  Can we go backwards?  That is, given a vector field F, can we construct a function f with the property that

        F  =  gradf

The answer to this question is only for some very special vector fields, which we call conservative.

Definition of a Conservative Vector Field

Let F be a vector field.  Then F is called conservative if there is a differentiable

function f such that 

               gradf  =  F

f is called the potential function for F.

Fortunately, we do not need to flounder about to determine if a vector field F is conservative.  We use the following argument.  If 

        F  =  <M,N>

 is conservative, then

        <M,Q>  =  gradf  =  <fx,fy>

So that 

        M  =  fx        and        N  =  fy

Now take partial derivatives with respect to y of the first equation and with respect to x of the second equation to get

        My  =  fxy        and        Nx  =  fyx 

Since mixed partial derivative are the same (for nice functions) we get

        My  =  Nx  

This turns out to be both a necessary and sufficient condition for a vector field to be conservative.  

Theorem:  Testing for Conservativeness

If M and N have continuous first order partial derivatives, then the vector field 

          F  =  Mi + Nj 

is conservative if and only if

          My  =  Nx  

        

Example 

Determine which of the two vector fields are conservative

A.  F  =  3xyi - x2j

B.  G  =  (1 + 2xy)i + (x2 - 2)j

 

Solution

For part A. we find

        My  =  3x        Nx  =  -2x

Since they are not equal the vector field is not conservative.

For part B.  we find 

         My  =  2x        Nx  =  2x

They are equal, so the vector field is conservative.

 

Once we know that a vector field is conservative, how do we find the potential function?  We investigate the relations

        M  =  fx        and      N  =  fy 

 

Example

Find the potential function for the conservative vector field in the last example

        G  =  (1 + 2xy)i + (x2 - 2)j

 

Solution

We have

        M  =  1 + 2xy  =  fx

Integrating both sides with respect to x we get

        x + x2y + c(y)  =  f(x,y)

Notice that the constant of integration may involve y terms since we are treating y as a constant.  Now differentiate with respect to y to get

        x2 + c'(y)  =  fy  =  N  =  x2 - 2

Thus 

        c'(y)  =  -2

Integrating with respect to y, we get

        c(y)  =  -2y

We do not need a constant of integration here since we just want "a" potential function not the general potential function.  Putting it all together, we get the potential function

        f(x,y)  =  x + x2y - 2y

The Curl of a Vector Field

If F is a vector field then we want to look at what operation we can do to it.  One such operation is called the Curl of F.  We define it as follows.

Curl of a Vector Field

If F is a differentiable vector field with 

               F  =  Mi + Nj + Pk 

then

         

 

Notice that for a two dimensional vector field, where there is only a k component for a cross product, that if the curl is zero then the vector field field is conservative.  Actually more is true.  Remembering that all second order partial derivatives are independent of order for nice functions, we have

Theorem

Let F be a three dimensional differentiable vector field with continuous partial derivatives.  Then 

               Curl F  =  0

if and only if F is conservative

 

Example

Determine if the vector field 

        F  =  yz2i + (xz2 + 2)j + (2xyz - 1)k

is conservative.  If it is, find a potential function.

 

Solution

We calculate

       

The vector field is conservative.  Now equate

        fx  =  yz2 

and integrate to get

        f  =  xyz2 + c(y,z)

Notice the constant now depends on both y and z, since they were held constant when we integrated.  Now take a partial with respect to y to get

        xz2 + cy(y,z)  =  fy  =  xz2 + 2

Hence 

        cy(y,z)  =  2

Integrate with respect to y to get

        c(y,z)  =  2y + c(z)

Notice that z is the only "constant" remaining.  We get

        f  =  xyz2 + 2y + c(z)

Now take a derivative with respect to z to get

        2xyz + c'(z)  =  fz  =  2xyz - 1

so that 

        c'(z)  =  -1

finally integrate to get

        c(z)  =  -z

The potential function is

        f(x,y,z)  =  xyz2 + 2y - z

The Divergence of a Vector Field

The curl of a vector field was defined as the cross product of the "dell" operator with the vector field.  A curious student may try to take a dot product instead and see where it leads.  

Definition of the Divergence of a Vector Field

If F is a differentiable vector field with 

               F  =  Mi + Nj + Pk 

then

              

 

Notice that the curl of a vector field is a vector field, while the divergence of a vector field is a real valued function.

 

Example

Find the divergence of the vector field

   F  =  (x-3y)i + (x2z2 + cos(z3))j + (xyz2)k

 

Solution

We just take the three partial derivatives and add them up

        (1) + (0) + (2xyz)  =  1 + 2xyz

 

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Line Integrals

Definition of a Line Integral

By this time you should be used to the construction of an integral.  We break a geometrical figure into tiny pieces, multiply the size of the piece by the function value on that piece and add up all the products.  For one variable integration the geometrical figure is a line segment, for double integration the figure is a region, and for triple integration the figure is a solid. 

The geometrical figure of the day will be a curve.  If we have a function defined on a curve we can break up the curve into tiny line segments, multiply the length of the line segments by the function value on the segment and add up all the products.  As always, we will take a limit as the length of the line segments approaches zero.  This new quantity is called the line integral and can be defined in two, three, or higher dimensions.

Suppose that a wire has as density f(x,y,z) at the point (x,y,z) on the wire.  Then the line integral will equal the total mass of the wire.  Below is the definition in symbols.

Definition of the Line Integral

Let f be a function defined on a curve C of finite length.  Then the line integral of f along C is 

                  (for two dimensions)

            (for three dimensions)

 

Evaluating Line Integrals

This definition is not very useful by itself for finding exact line integrals.  If data is provided, then we can use it as a guide for an approximate answer.  Fortunately, there is an easier way to find the line integral when the curve is given parametrically or as a vector valued function.  We will explain how this is done for curves in R2.  The case for R3 is similar.  

Let 

        r(t)  =  x(t)i + y(t)j

be a differentiable vector valued function.  Then

         

We are now ready to state the theorem that shows us how to compute a line integral.

 

Theorem:  Line Integrals Over Vector Valued

Functions

Let 

               r(t)  =  x(t)i + y(t)j          a < t < b

be a differentiable vector valued function that defines a smooth curve C.  Then

              

and for three dimensions, if

                r(t)  =  x(t)i + y(t)j + z(t)k          a < t < b

then

              

 

Example

Find the line integral

       

where C is the ellipse 

        r(t)  =  (2cos t)i + (3sin t)j          0  <  t  <  2

You may use a calculator or computer to evaluate the final integral.

Solution

We find

       

We have the integral

       

With the help of a machine, we get  

        15.87

 

Work

The main application of line integrals is finding the work done on an object in a force field.  If an object is moving along a curve through a force field F, then we can calculate the total work done by the force field by cutting the curve up into tiny pieces.  The work done W along each piece will be approximately equal to 

        dW  =  F . Tds

Now recall that 

                    r'(t)        T  =                                 ||r'(t)||

and that 

        ds  =  ||r'(t)||dt

Hence 

        dW  =  F . r'(t) dt

As usual, we add up all the small pieces of work and take the limit as the pieces get small to end up with an integral.

Definition of Work

Let F be a vector field and C be a curve defined by the vector valued function r.  Then the work done by F on an object moving along C is given by 

         

 

Example

Find the work done by the vector field

        F(x,y,z)  =  xi + 3xyj - (x + z)k

on a particle moving along the line segment that goes from (1,4,2) to (0,5,1)

 

Solution

We first have to parameterize the curve.  We have 

        r(t)  =  <1,4,2> + [<0,5,1> - <1,4,2>]t  =  <1 - t, 4 + t, 2 - t>

and 

        r'(t)  =  -i + j - k

Taking the dot product, we get

        F . r'(t)  =  -x + 3xy + x + z  =  3xy + z

        =  3(1 - t)(4 + t) + (2 - t)  =  -3t2 -10t + 14

Now we just integrate

       

Notice that work done by a force field on an object moving along a curve depends on the direction that the object goes.  In fact the opposite direction will produce the negative of the work done in the original direction.  This is clear from the fact that everything is the same except the order which we write a and b.

Line Integrals in Differential Form

We can rewrite r'(t)dt as

          dr                 dx          dy          dz                 dt  =  (          i+          j +          k ) dt            dt                 dt           dt           dt

              =  dx i + dyj + dzk 

So that if 

        F  =  Mi + Nj + Pk

then 

        F . r'(t)dt  =  Mdx + Ndy + Pdz

This is called the differential form.

 

Example

Find

       

where C is the part of the helix

        r(t)  =  sin t i + cos t j + t k        0  <  t  <  2

 

Solution

We have 

        r'(t)  =  cos t i - sin t j + k  

so that 

        ydx + zdy  =  (cos2t - t sin t)dt

This leads us to the integral

       

with a little bit of effort (using integration by parts) we get

        3  

 

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Conservative Vector Fields and Independence of Path

The Fundamental Theorem of Line Integrals

Consider the force field representing the wind shown below

       

You are a pilot attempting to minimize the work your engines need to do.  Does it matter which path you take?  Clearly the red path goes with the wind and the green path goes against the wind.  With this vector field, work is dependent on the path that is taken.  

Next consider the vector field 

        F(x,y)  =  yi + xj

shown below

       

It turns out that going from point A to point B, every path leads to the same amount of work done.  What is special about this vector field?  

The key here, as you can quickly check, is that the vector field F is conservative. (My  =  Nx ).  Since for a conservative vector field, all paths produce the same amount of work, we seek a formula that gives this work quantity.

The theorem below shows us how to find this quantity.  Notice the strong resemblance to the fundamental theorem of calculus.

 

The Fundamental Theorem of Line Integrals

Let F be a conservative vector field with potential function f, and C be any smooth curve starting at the point A and ending at the point B.  Then

              

 

The next example demonstrates the power of this theorem.

 

Example

Find the work done by the vector field 

        F(x,y)  =  (2x -3y)i + (3y2 - 3x)j        

along the curve indicated in the graph below

       

Solution

First notice that 

        My  =  -3  =  Nx

We can use the fundamental theorem of line integrals to solve this.  There are two approaches.  

Approach 1

We find the potential function.  We have 

        fx  =  2x - 3y

Integrating we get

        f(x,y)  =  x2 - 3xy + c(y)

Now take the derivative with respect to y to get

        fy  =  -3x + c'(y)  =  3y2 - 3x 

Hence 

        c'(y)  =  3y2

and 

        c(y)  =  y3 

The potential function is 

        f(x,y)  =  x2 - 3xy + y3 

Now use the fundamental theorem of line integrals to get

        f(B) - f(A)  =  f(1,0) - f(0,0)  =  1

 

Approach 2

We since the vector field is conservative any path from point A to point B will produce the same work.  Hence the work over the easier line segment from (0,0) to (1,0) will also give the correct answer.  We parameterize by 

        r(t)  =  ti          0  <  t  <  1

we have

        ri(t)  =  i       

so that

        F . dr  =  ((2x -3y)i + (3y2 - 3x)j) . i  =  2x - 3y  =  2t

Now just integrate

       

        

Proof of the Fundamental Theorem of Line Integrals

To prove the fundamental theorem of line integrals we will use the following outcome of the chain rule:

If 

        r(t)  =  x(t)i + y(t)j 

is a vector valued function, then 

        d/dt f(r(t))  =  fx x'(t) + fy y'(t)

We are now ready to prove the theorem.  We have

       

 

Independence of Path and Closed Curves

Example

Find the work done by the vector field 

         F(x,y)  =  (cos x + y)i + (x+ esin y)j + (sin(cos z))k

along the closed curve shown below

       

 

Solution

First we check that F is conservative.  We have

       

Since the vector field is conservative, we can use the fundamental theorem of line integrals.  Notice that the curve begins and ends at the same place.  We do not even need to find the potential function, since whatever it is, say f, we have

        f(A) - f(A)  =  0

 

In general, the work done by a conservative vector field is zero along any closed curve.  The converse is also true, which we state without proof.

Theorem:  Conservative Vector Fields and Closed Curves

Let F be a vector field with components that have continuous first order partial derivatives and let C be a piecewise smooth curve.  Then the following three statements are equivalent

1.  F is conservative.

2.    is independent of path.

3.    for all closed curves C.

 

 

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Green's Theorem

 

A Little Topology

Before stating the big theorem of the day, we first need to present a few topological ideas.  Consider a closed curve C in R2 defined by 

        r(t)  =  x(t)i + y(t)j        a  <  t  < b

We say C is simple if it does not intersect itself.  A curve intersects itself if 

        r(u)  =  r(v) 

for two distinct values u and v.  A circle is a simple curve while a figure eight is not simple.  

       

A region is called simply connected if it boundary is a single simple closed curve.  Another way of thinking about simply connected regions is that their complement (the space minus the region) consists of only one piece.  Below are examples of simply connected and non-simply connected regions.

       

Our final topological definition is orientation.  We have seen that if we traverse a curve in the opposite direction, then the line integral will be the negative of the original.  We want to have a way to define a positive orientation.  We define it as follows.

Let R be a simply connected region with boundary curve C.  Then C is called positively oriented if facing the direction that the curve is sketched, the region lies to the left of the curve.  Otherwise the curve is said to be negatively oriented.

One way to remember this is to recall that in the standard unit circle angles are measures counterclockwise, that is traveling around the circle you will see the center on your left.

       

Green's Theorem

We have seen that if a vector field 

        F  =  Mi + Nj

has the property that 

        Nx - My  =  0

 then the line integral over any smooth closed curve is zero.  What can we do if the above quantity is nonzero.  Green's theorem states that the line integral is equal to the double integral of this quantity over the enclosed region.  Precisely, we have

 

Green's Theorem

Let R be a simply connected region with smooth boundary C, oriented positively and let M and N have continuous partial derivatives in an open region containing R, then 

              

 

Sketch of the Proof

First we can assume that the region is both vertically and horizontally simple.  Otherwise we can carefully cut the region into parts so that each of the parts are both vertically simple and horizontally simple.  Below is an example of such a cut.  Notice that the line where the regions is cut is drawn once upwards and once downwards.  Thus the two line integrals over this line will cancel each other out.

       

We can assume that the region is as in the figure below

       

We will show that 

       

The proof for the M part is similar.  We will compute both sides and show they are the same.  First we break the curve into its left and right half.  Call the left half C1 and the right half C2.   We have

       

Now we show that the double integral leads to the same expression.  We have 

       

2

1

2

1

( )

( )

( )

2 1( )

( , ) ( ( ), ) ( ( ), )

g yb

x xaR g y

g yb b

a ag y

N dxdy N dxdy

N x y dy N g y y N g y y dy

 

And the two expressions are equal.

 

Using Green's Theorem

Example

Determine the work done by the force field 

                F  =  (x - xy) i + y2 j 

when a particle moves counterclockwise along the rectangle with vertices (0,0), (4,0), (4,6), and (0,6).

 

Solution

We could do this with a line integral, but this would involve four parameterizations (one for each side of the rectangle).  Instead, we use Green's Theorem.  We find

        Nx - My  =  0 - (-x)  =  x

The region is just a rectangle, so the limits are the constants.  We have

       

 

Example 

Calculate the line integral

       

Where C is the union of the unit circle centered at the origin oriented negatively and the circle of radius 2 centered at the origin oriented positively.

 

Solution

We cannot use Green's Theorem directly, since the region is not simply connected.  However, if we think of the region as being the union its left and right half, then we see that the extra cuts cancel each other out.  

       

In this light we can use Green's Theorem on each piece.  We have 

        Nx - My  =  1 - 0  =  1

Hence the line integral is just the double integral of 1, which is the area of the region.  This area is 

        (22 - 12)  =  4

 

Green's Theorem and Area

The example above showed that if 

        Nx - My   =  1

then the line integral gives the area of the enclosed region.  There are three special vector fields, among many, where this equation holds.  We state the following theorem which you should be easily able to prove using Green's Theorem.

 

Theorem:  Using Green's Theorem to Find Area

Let R be a simply connected region with positively oriented smooth boundary C.  Then the area of R is given by each of the following line integrals.

1.      

                   2.                      3.   

 

 

Example

Use the third part of the area formula to find the area of the ellipse 

          x2           y2                    +              =  1           4            9

 

Solution

To compute the line integral, we parameterize the curve

        r(t)  =  2 cos t i + 3 sin t j

        r'(t)  =  -2 sin t i + 3 cos t j

We have 

       

 

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Parametric Surfaces

Definition of a Parametric Surface

We have now seen many kinds of functions.  When we talked about parametric curves, we defined them as functions from R to R2 (plane curves) or R to R3 (space curves).  Because each of these has its domain R, they are one dimensional (you can only go forward or backward).  In this section, we investigate how to parameterize two dimensional surfaces.  Below is the definition.

Definition of Parametric Surfaces

A parametric surface is a function with domain R2 and range R3.

 

Remark:  We typically use the variables u and v for the domain and x, y, and z for the range.  We often use vector notation to exhibit parametric surfaces.

 

Example 

A sphere of radius 7 can be parameterized by 

        r(u,v)  =  7cos u sin v i + 7sin u sin v j + 7 cos v k

Notice that we have just used spherical coordinates with the radius held at 7.  

 

We can use a computer to graph a parametric surface.  Below is the graph of the surface

         r(u,v)  =  sin u i + cos v j + exp(2u1/3 + 2v1/3) k

       

 

Example 

Represent the surface 

        z  =  ex cos(x - y)

parametrically

 

Solution

The idea is similar to parametric curves.  We just let x  =  u and y  =  v, to get

         r(u,v)  =  u i + v j + eu cos(u - v) k

       

Example

A surface is created by revolving the curve 

        y  =  cos x

about the x-axis.  Find parametric equations for this surface.

 

Solution

For a fixed value of x, we get a circle of radius cos x.  Now use polar coordinates (in the yz-plane) to get

         r(u,v)  =  u i + r cos v j + r sin v k

Since u = x and r = cos x, we can substitute cos u for r in the above equation to get

         r(u,v)  =  u i + cos u cos v j + cos u sin v k

       

 

Normal Vectors and Tangent Planes

We have already learned how to find a normal vector of a surface that is presented as a function of tow variables, namely find the gradient vector.  To find the normal vector to a surface r(t) that is defined parametrically, we proceed as follows. 

The partial derivatives 

        ru(u0,v0)         and        rv(u0,v0) 

will lie on the tangent plane to the surface at the point (u0,v0).  This is true, because fixing one variable constant and letting the other vary, produced a curve on the surface through

(u0,v0). ru(u0,v0) will be tangent to this curve.  The tangent plane contains all vectors tangent to curves passing through the point.  

To find a normal vector, we just cross the two tangent vectors.

 

Example

Find the equation of the tangent plane to the surface 

        r(u,v)  =  (u2 - v2) i + (u + v) j + (uv) k

at the point (1,2).

 

Solution

We have 

        ru(u,v)  =  (2u) i +  j + v k

        rv(u,v)  =  (-2v) i +  j + u k

so that 

        ru(1,2)  =  2 i +  j + 2 k

        rv(1,2)  =  -4 i +  j + k

        r(1,2)  =  -3 i + 3 j + 3 k

Now cross these vectors together to get

       

We now have the normal vector and a point (-3,3,2).  We use the normal vector-point equation for a plane

        -1(x + 3) - 10(y - 3) + 6(z - 2)  =  0

        -x - 10y + 6z  =  -15        or        x + 10y - 6z  =  15

Surface Area

To find the surface area of a parametrically defined surface, we proceed in a similar way as in the case as a surface defined by a function.  Instead of projecting down to the region in the xy-plane, we project back to a region in the uv-plane.  We cut the region into small rectangles which map approximately to small parallelograms with adjacent defining vectors ru and rv.  The area of these parallelograms will equal the magnitude of the cross product of ru and rv.  Finally add the areas up and take the limit as the rectangles get small.  This will produce a double integral.

 

Area of a Parametric Surface

Let S be a smooth surface defined parametrically by 

                 r(u,v)  =  x(u,v) i + y(u,v) j + z(u,v) k   

where u and v are contained in a region R.  Then the surface area of S is given by 

              

 

Since the magnitude of a cross product involves a square root, the integral in the surface area formula is usually impossible or nearly impossible to evaluate without power series or by approximation techniques.  

 

Example

Find the surface area of the surface given by 

        r(u,v)  =  (v2) i + (u - v) j + (u2) k        0  <  u  <  2        1  <  v  <  4

 

Solution

We calculate

        ru(u,v)  =   j + 2u k

        rv(u,v)  =  (2v) i -  j

The cross product is 

       

The surface area formula gives

       

This integral is probably impossible to compute exactly.  Instead, a calculator can be used

to obtain

a surface area of 70.9.

 

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Surface Integrals

Surface Integrals for Parametric Surfaces

In the last section, we learned how to find the surface area for parametric surfaces.  We cut the region in the uv-plane into tiny rectangles and added up the area of the corresponding tiny parallelograms in the xy-plane.  The area of these parallelograms was 

       

If we think of the surface as having varying density f(x,y,z), then the mass of this parallelogram will be 

       

and adding up all these masses and taking the limit as the rectangle sizes approach zero, gives the definition of the surface integral.

 

Definition of the Surface Integral

Let S be a smooth surface given by the vector valued function

               r(u,v)  =  x(u,v)i + y(u,v)j + z(u,v)k

and f(x,y,z) be a continuous function.  Then the surface integral of f over S is 

              

 

As with finding the surface area the integral typically results in an impossible integral.

 

Example

Find 

       

where S is the surface 

        r(u,v)  =  ui + u2j + (u+ v)k        0  <  u  <  2        1  <  v  <  4

and

        f(x,y,z)  =  x + 2z

 

Solution

We find

        ru  =  i + (2u)j + k

        rv  =  k

and take the cross product

       

We have 

        f(x(u,v),y(u,v),z(u,v))  =  x(u,v) +2z(u,v)  =  u +2(u + v)  =  3u + v

We find 

       

Although this integral is possible, its solution is quite involved.  You can verify that the surface integral evaluates to approximately 525.27.

Surface Integrals for Surfaces that are Functions of Two Variables

We have seen before that if

        z  =  g(x,y) 

is a surface such that g has continuous first order partial derivatives, then the parameterization

        r(u,v)  =  ui + vj + g(u,v)k

has the property that 

       

This leads to the formula for surface integrals.

 

Theorem: The Surface Integral for Surfaces of the Form   z  =  g(x,y)

Let S be a surface given by

               z  =  g(x,y)

over a region R such that both first order partial derivatives of g are continuous and let f(x,y,z) be a continuous function.  Then the surface integral of f over S is 

              

 

Example

Find 

       

where S is the part of the paraboloid 

   

     z  =  x

2

+ y

2

 

that lies inside the cylinder

   

      x

2

+ y

2

  =  1

and

   

     f(x,y,z)  =  z

 

Solution

We have

   

    

and 

        f(x,y,z)  =  z  = 

x

2

+ y

2

  

At this point, you should be thinking, "This looks like a job for polar coordinates."  And

we get

   

    

Let

   

     u  =  1 + 4r

2   

     du  =  8r dr        r

2

  =  1/4 u - 1/4

and the substitution gives us 

   

    

Oriented Surfaces and Flux

We have seen how a region R with boundary curve C can be oriented.  Traveling along

C, we look to see if the region is on the right or left.  Unfortunately, this definition does

not work will for surfaces in three dimensions.  The idea of right and left is not well

defined.  In fact not all surfaces can be oriented. 

We say that a surface is orientable if a unit normal vector can be defined on the surface

such that it varies continuously over the surface.

Below is an example of a non-orientable surface (called the Mobius Strip)

   

    

You can see that there is no front or back of this surface.  

Recall that a unit normal vector to a surface can be given by 

   

    

There is another choice for the normal vector to the surface, namely the vector in the opposite direction, -N.  

By this point, you may have noticed the similarity between the formulas for the unit

normal vector and the surface integral.  This idea leads us to the definition of the Flux

Integral

Consider a fluid flowing through a surface S.  The Flux of the fluid across S measures the

amount of fluid passing through the surface per unit time.  If the fluid flow is represented

by the vector field F, then for a small piece with area S of the surface the flux will equal

to 

   

     Flux  =  F

.

N S 

Adding up all these together and taking  a limit, we get

 

Definition of the Flux Integral

Let F be a differentiable vector field on a surface S oriented by a unit normal vector N.  The flux integral of F across N is given by 

              

   

     

Notice that the denominator of N and the formula for dS both involve ||r

u

x r

v

||. 

Canceling, we get

   

     NdS  =  ru

x rv

dvdu

for a surface that is defined by the function z  =  g(x,y), we get the nice formula

        NdS  =  

-gx(x,y)i - gy(x,y)j + k     (oriented upward)

or

        NdS  =  

gx(x,y)i + gy(x,y)j - k     (oriented downward)

 

Example

Find the flux of 

         F(x,y,z)  =  xi + 2yj + zk

across the part of the surface

        z  =  x + y2 

with upward pointing normal that lies within the box

        0  <  x  <  3        2  <  y  <  5

 

Solution

We compute

        NdS  =  -i - 2yj + k  dydx

and 

        F . N dS  =  -x - 4y2 + x + y2  =  -3y2 

The flux integral is 

       

 

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Divergence Theorem

The Divergence Theorem

When we looked at Greens Theorem, we saw that there was a relationship between a region and the curve that encloses it.  This gave us the relationship between the line integral and the double integral.  Moving to three dimensions, the divergence theorem provides us with a relationship between a triple integral over a solid and the surface integral over the surface that encloses the solid.  

 

The Divergence Theorem

Let Q be a solid region bounded by a closed surface oriented with outward pointing unit normal vector N, and let F be a differentiable vector field (components have continuous partial derivatives).  Then

              

 

Example

Find 

       

Where

         F(x,y,z)  =  y2i + ex(1-cos(x2 + z2))j + (x + z)k

and S is the unit sphere centered at the point (1,4,6) with outwardly pointing normal vector.

 

Solution

This seemingly difficult problem turns out to be quite easy once we have the divergence theorem.  We have

        divF  =  0 + 0 + 1  =  1

Now recall that a triple integral of the function 1 is the volume of the solid.  Since the solid is a sphere of radius 1 we get .

Part of the Proof of the Divergence Theorem

As usual, we will make some simplifying remarks and then prove part of the divergence theorem.

We assume that the solid is bounded below by 

        z  =  g1(x,y) 

and above by 

        z  =  g2(x,y) 

                                           

Notice that the outward pointing normal vector is upward on the top surface and downward for the bottom region.  We also note that the divergence theorem can be written as 

       

We will show that 

       

We have on the top surface

        Pk . N dS  =  Pk . ( (-g2)x i - (g2)y j + k)  =  P(x,y,g2(x,y))

On the bottom surface, we get

        Pk . N dS  =  Pk . ( (g1)x i + (g1)y j - k)  =  -P(x,y,g1(x,y))

Putting these together we get 

       

For the triple integral, the fundamental theorem of calculus tell us that 

       

An Interpretation Of Divergence

We have seen that the flux is the amount fluid flow per unit time through a surface.  If the surface is closed, then the total flux will equal the flow out of the solid minus the flow in.  Often in the solid there is a source (such as a star when the flow is electromagnetic radiation) or a sink (such as the earth collecting solar radiation)  If we have a small solid S(P) containing a point P, then the divergence of the vector field is approximately constant, which leads to the approximation

       

The divergence theorem expresses the approximation 

   

     Flux through S(P)    divF(P) (Volume)

Dividing by the volume, we get that the divergence of F at P is the Flux per unit volume.  

If the divergence is positive, then the P is a source.  

If the divergence is negative, then P is a sink.

 

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Stokes' Theorem

Stokes' Theorem

The divergence theorem is used to find a surface integral over a closed surface and Green's theorem is use to find a line integral that encloses a surface (region) in the xy-plane.  The theorem of the day, Stokes' theorem relates the surface integral to a line integral.

Since we will be working in three dimensions, we need to discus what it means for a curve to be oriented positively.  

Let S be a oriented surface with unit normal vector N and let C be the boundary of S.  Then C is positively oriented if its orientation follows the right hand rule, that is if you right hand curls around N in the direction of C's orientation, then your thumb will be pointing in the direction of N.  

                       

Now we are ready to state Stokes' Theorem.  The proof will be left for a more advanced course.

 

Stokes' Theorem

Let S be an oriented surface with unit normal vector N and C be the positively oriented boundary of S.  If F is a vector field with continuous first order partial derivatives then

              

 

Example

Let S be the part of the plane 

        z  =  4 - x - 2y

with upwardly pointing unit normal vector.  Use Stokes' theorem to find

       

Where 

        F  =  yi + zj - xyk

Solution

First notice that without Stokes' theorem, we would have to parameterize three different line segments.  Instead we can find this with just one double integral.

We have

       

and 

        N dS  =  i + 2j + k

So that 

        Curl F . N dS  =  1 + x + 2y - 1  =  x  2y

We integrate

       

Curl and Circulation

Just as the divergence theorem assisted us in understanding the divergence of a function at a point, Stokes' theorem helps us understand what the Curl of a vector field is.  Let P be a point on the surface and Ce be a tiny circle around P on the surface.  The 

       

measures the amount of circulation around P.  You can see this by noticing that if F flows in the direction of the tangent vector, then F . dr will be positive.  If it flows in the opposite direction, then it will be negative.  The stronger the force field in the direction of the tangent vector, the greater the circulation.

Since the region enclosed by Ce is tiny, the surface integral can be approximated by 

       

or

        Curl F . N  =  Circulation per unit area

So the curl tell us how much the force field rotates around the point.  

           

We can see that if this is a small piece of the surface containing  P, then 

        Curl F . N  >  0

 

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Math 202 Practice Midterm 3

Please work out each of the given problems.  Credit will be based on the steps that you show towards the final answer.  Show your work.

Printable Key

Problem 1  Please answer the following true or false.  If false, explain why or provide a counter example.  If true explain why.  

A.  Let f(x,y,z) be a function with continuous second order partial derivatives and let F(x,y,z) be the gradient of f(x,y,z).  If S is the ellipsoid 

            x2                      +  y2 + z2  =  1            4

oriented outward, then 

       

Solution

B.  Let F(x,y) be a conservative vector field, then

   

    

Solution

Problem 2

Show that 

for any closed surface S.  

Solution

Problem 3

A fish starting at the origin swims in a straight path to the point (0.2,0.1,0.3), then changes direction and swims along the circular path through the point (0.5,0.2,0.6) and the point (0.7,0.5,0.8), and finally changes directions heading straight to the point (1.5,1,2).  The current can by represented by the vector field

        F(x,y,z)  =  (2x + 2z)i + (1 - 3z)j + (2x - 3y + 5)k

Find the total work done by the current.

Solution

Problem 4

Evaluate

where

F is the vector field 

       

and S is the rectangular solid with vertices (0,0,0), (1,0,0), (1,2,0), (0,2,0), (0,0,3), (1,0,3), (1,2,3), (0,2,3).

Solution

Problem 5

Find the work done by sailing a ship from the point (2,3) the the point (-1,2) against the wind with velocity field

        F(x,y)  =  yi + (3x + 2y)j

Solution

Problem 6

Find the flux of F through the surface S where 

        F(x,y,z)  =  3zi - 4j + yk

and S is the part of the plane

        x + y + z  =  1

in the first octant with upwardly pointing unit normal.

Solution

 

Math 202 Practice Final

Please work out each of the given problems.  Credit will be based on the steps that you show towards the final answer.  Show your Work

Printable key

Problem 1  Please answer the following true or false.  If false, explain why or provide a counter example.  If true, explain why.

A.  Let Q be a three dimensional solid and let 

        F(x,y,z)  =  (x2y + sin z)i+ (cos x - xy2)j + (3xy + z)k

and let S be the boundary of Q with outwardly pointing normal.  The the volume so Q is given by

       

Solution

B.  Let F  =  3xy i + cosx j and let C1 and C2 be as shown below.  Then

       

       

Solution

 

C.  A new particle, the fluxon, has been discovered to be emitted from the sun.  The particle emits a force field 

        F(x,y,z)  =  (y2 - z) i + (x2 - z) j + (x2 + y2) k

where the origin represents the center of the sun.  If the total flux through the earth's northern hemisphere has been calculated as 10,000, then the total flux through the earth's southern hemisphere must also be 10,000.

Solution

Problem 2  You are the captain of the spaceship Potential that you have programmed to follow the vector-valued function 

        r(t) =  (t2 + 5) i + (t - 3) j + t3 k 

where t is measured in hours.  However, at time t  =  2, your engines fail and your ship begins drifting in deep space.  There is a deep space station located at (6,2,38).  

A.  Find the vector-valued function that describes the Potential's flight after the engines failed. Use t  =  2 to represent the time at which your engines first shut down. (Hint:  This should be a linear vector valued function.)

Solution

B.  Will your ship make it to the station, or will you float helplessly for eternity?

Solution

 

Problem 3  Show that the helix

        r(t)  =  (R cos t) i + (R sin t) j + t k 

where R is a positive constant, has the property that N(t) . r(t) is a constant.  Find this

constant.

Solution

Problem 4  The probability density function for an event is given by

       

where R is the square with vertices (4,0), (6,2), (4,4), and (2,2).

A.  Use the appropriate change of variables (Jacobians) to find k that is solve 

       

Solution

B.  Find the probability that  0  <  x - y  <  1

Solution

Problem 5  Switch the order of integration and write as one double integral

       

Solution

Problem 6  Set up the integrals that give the following.  Use the most appropriate coordinate system.

A.  The mass of the solid that lies inside the sphere

        x2 + y2 + z2  =  9

and outside the cone

        z2  =  x2 + y2 

that has density function

       

Solution

B.  The surface area of the part of the paraboloid 

        z  =  x2 + y2

that lies inside the cylinder

        x2 + y2  =  4

Solution

Problem 7  Find the work done by the force field 

        F(x,y)  =  (3x2 - y) i + (x2 - y3) j

as a particle moves counterclockwise around the rectangle with vertices (2,1), (5,1), (5,5), and (2,5).

Solution

Problem 8

Verify Stokes Theorem where 

        F(x,y,z)  =  (2x) i + (2y) j + (z sin z3) k

and S is part of the paraboloid 

        z  =  9 - x2 - y2 

that lies above the plane z  =  8 oriented upward.

Solution

 

 Math 202 Practice Midterm 3

Please work out each of the given problems.  Credit will be based on the steps that you show towards the final answer.  Show your work.

Problem 1  Please answer the following true or false.  If false, explain why or provide a counter example.  If true explain why.  

A.  Let f(x,y,z) be a function with continuous second order partial derivatives and let F(x,y,z) be the gradient of f(x,y,z).  If S is the ellipsoid 

            x2                      +  y2 + z2  =  1            4

oriented outward, then 

       

Solution

False

Using the divergence theorem, we have

   

    

If gradf(x,y,z)  =  F, then 

   

     divF  =  f

xx

+ f

yy

+ f

zz

 

There theorem would be true if the function was harmonic, however if it not harmonic. 

All bets are off.  For example, if

   

     f(x,y,z)  =  1/6 (x

2

+ y

2

+ z

2

)

and 

   

     divF  =  1/6 (2 + 2 + 2)  =  1

Hence the integral represents the volume of the ellipse which is certainly not zero.  

 

B.  Let F(x,y) be a conservative vector field, then

   

    

Solution

True,  

The first line integral traces out the line segment from (0,1) to (1,0) and the second traces out the quarter-circle from (0,1) to (1,0).  Notice that in the first integral

        r1(t)  =  (1 - t)i + tj         r1'(t)  =  -i + j  

and in the second integral 

        r2()  =  (cos )i + (sin )j         r1'(t)  =  (-sin )i +(cos ) j  

By the fundamental theorem of line integral, the integral is independence of path, hence the two integrals are equal.

 

Problem 2

Show that 

for any closed surface S.  

 

Solution

We use Stokes' Theorem.  We have

       

Where C is the boundary of the surface S.  But since S is a closed surface, it has not boundary.  Hence C is a curve of zero length and the right hand integral is zero.

 

Problem 3

A fish starting at the origin swims in a straight path to the point (0.2,0.1,0.3), then changes direction and swims along the circular path through the point (0.5,0.2,0.6) and the point (0.7,0.5,0.8), and finally changes directions heading straight to the point (1.5,1,2).  The current can by represented by the vector field

        F(x,y,z)  =  (2x + 2z)i + (1 - 3z)j + (2x - 3y + 5)k

Find the total work done by the current.

 

Solution

The important thing to not here is that 

       

Since F is conservative, we can use the fundamental theorem of line integrals.  We seek a potential function f.  We have

        fx  =  2x + 2z

Integrating with respect to x gives

        f  =  x2 + 2xz + C(y,z)

Now taking the derivative with respect to y gives

        fy  =  Cy(y,z)  =  1 - 3z

Integrating with respect to y gives

        C(y,z)  =  y - 3yz + C(z)

so that

        f  =  x2 + 2xz + y - 3yz + C(z)

Now we take the derivative with respect to z to get

        2z - 3y + C'(z)  =  2x - 3y + 5

so that 

        C'(z)  =  5

Integrate with respect to z to get

        C(z)  =  5z

Hence 

        f(x,y,z)  =   x2 + 2xz + y - 3yz + 5z

The fundamental theorem of line integrals gives that the integral is

        f(1.5,1,2) - f(0.2,0.1,0.3)  =  

        [(1.5)2 + 2(1.5)(2) + 1 - 3(1)(2) + 5(2)] - [(0.2)2 + 2(0.2)(0.3) + 0.1 - 3(0.1)(0.3) + 5(0.3)]

        =  11.58

 

Problem 4

Evaluate

where

F is the vector field 

       

and S is the rectangular solid with vertices (0,0,0), (1,0,0), (1,2,0), (0,2,0), (0,0,3), (1,0,3), (1,2,3), (0,2,3).

 

Solution

We use the divergence theorem.  We have

        divF  =  1 - 1 + 2  =  2

We have 

       

Since the integrand of the right hand side is just a constant, its value is equal to the constant times the volume of the solid.  Since the solid is a rectangular solid with side lengths 1, 2, and 3, we have

        2(Volume E)  =  2(1)(2)(3)  =  12

 

Problem 5

Find the work done by sailing a ship from the point (2,3) the the point (-1,2) against the wind with velocity field

        F(x,y)  =  yi + (3x + 2y)j

 

Solution

First notice that F is not a conservative vector field.  We need to parameterize the curve and perform the line integral.  The curve can be parameterized by

        r(t)  =  (2 + (-1 - 2)t)i + (3 + (2 - 3))j  =  (2 - 3t)i + (3 - t)j

We have

        dr  =  -3i - j        

        F(t)  =  (3 - t)i + (3(2 - 3t) + 2(3 - t))j  =  (3 - t)i + (12 - 11t)j 

The integrand becomes

        F . dr  =  -3(3 - t) + (-1)(12 - 11t)  =  -21 + 14t

Now we integrate

       

 

Problem 6

Find the flux of F through the surface S where 

        F(x,y,z)  =  3zi - 4j + yk

and S is the part of the plane

        x + y + z  =  1

in the first octant with upwardly pointing unit normal.

 

Solution

We use Stokes' theorem.  We have 

       

We can write the surface as

   

     z  =  1 - x - y

Using Stokes theorem we get

       

 

Math 202 Practice Final

Please work out each of the given problems.  Credit will be based on the steps that you show towards the final answer.  Show your Work

Problem 1  Please answer the following true or false.  If false, explain why or provide a counter example.  If true, explain why.

A.  Let Q be a three dimensional solid and let 

        F(x,y,z)  =  (x2y + sin z)i+ (cos x - xy2)j + (3xy + z)k

and let S be the boundary of Q with outwardly pointing normal.  The the volume so Q is given by

       

Solution

True,

We have

        divF  =  2xy - 2xy + 1  =  1

Using the divergence theorem, we see that 

       

which is just the volume of the solid.

 

B.  Let F  =  3xy i + cosx j and let C1 and C2 be as shown below.  Then

       

       

Solution

 

 

C.  A new particle, the fluxon, has been discovered to be emitted from the sun.  The particle emits a force field 

        F(x,y,z)  =  (y2 - z) i + (x2 - z) j + (x2 + y2) k

where the origin represents the center of the sun.  If the total flux through the earth's northern hemisphere has been calculated as 10,000, then the total flux through the earth's southern hemisphere must also be 10,000.

Solution

False,  since divF  =  0, the total flux must be zero.  If the flux through the northern hemisphere is 10,000, then the flux through the southern hemisphere must be -10,000.

 

Problem 2  You are the captain of the spaceship Potential that you have programmed to follow the vector-valued function 

        r(t) =  (t2 + 5) i + (t - 3) j + t3 k 

where t is measured in hours.  However, at time t  =  2, your engines fail and your ship begins drifting in deep space.  There is a deep space station located at (6,2,38).  

A.  Find the vector-valued function that describes the Potential's flight after the engines failed. Use t  =  2 to represent the time at which your engines first shut down. (Hint:  This should be a linear vector valued function.)

Solution

The flight will go in a linear path in the direction of the unit tangent vector with speed equal to the speed when the engines fail.  We have

        r'(t)  =  2t i + j + 3t2 k

        r'(2)  =  4i + j + 12k

When t  =  2, the spacecraft is at 

        r(2)  =  9i - j + 8k 

The flight can be described by

        rf(t)  =  r(2) + (t - 2)r'(2)  =  9i - j + 8k + (t - 2)(4i + j + 12k)

        =  (1 + 4t)i + (-3 + t)j + (-16 + 12t)k 

 

B.  Will your ship make it to the station, or will you float helplessly for eternity?

Solution

We are looking for a time t with 

        rf(t)  = 6i + 2j + 38j

Setting the j components equal we get

        -3 + t  =  3        t  =  5

However

        rf(5)  =  9i + 2j + 44k

since the i and k components of rf and the station are different, we can conclude that our spaceship will drift away to eternity.

 

 

Problem 3  Show that the helix

        r(t)  =  (R cos t) i + (R sin t) j + t k 

where R is a positive constant, has the property that N(t) . r(t) is a constant.  Find this

constant.

Solution

We first calculate T(t)

        r'(t)  =  (-R sin t) i + (R cos t) j + k 

        || r'(t)||  =  (R2 sin2 t + R2 cos2 t + 1)1/2  =  (R2 + 1)1/2

Hence 

        T(t)  =  (R2 + 1)-1/2 [(-R sin t) i + (R cos t) j + k]

Next, we have

        T'(t)  =  (R2 + 1)-1/2 [(-R cos t) i + (-R sin t) j]

and

        ||T'(t)||  =  (R2 + 1)-1/2 [R2 cos2 t + R2 sin2 t]1/2  =   (R2 + 1)-1/2 [R] 

Dividing gives

        N(t)  =  -cos t i - sin t j

Finally, we take the dot product

        N(t) . r(t)  =  [(R cos t) i + (R sin t) j + t k]

. [-cos t i - sin t j]

        =  R cos2 t + R sin2 t  =  R

 

Problem 4  The probability density function for an event is given by

       

where R is the square with vertices (4,0), (6,2), (4,4), and (2,2).

A.  Use the appropriate change of variables (Jacobians) to find k that is solve 

       

Solution

We sketch the picture and find the equation of the four lines that border the square.

       

We can also write

        0  <  x - y  < 4        and        4  <  x + y  <  8

We let 

        u  =  x - y        v  =  x + y

Adding the two equations gives

        u + v  =  2x        x  =  1/2 (u + v)

Subtracting the two equations gives

        v - u  =  2y        y  =  1/2 (v - u)

We can compute the Jacobian

       

We have

               

Setting this equal to 1 gives

                    3        k  =                                      896

 

        

B.  Find the probability that  0  <  x - y  <  1

Solution

Since

        u  =  x - 1

we just adjust the limits appropriately

       

Problem 5  Switch the order of integration and write as one double integral

       

Solution

The key to this problem is to sketch the picture which is shown below

       

Now we can realize the region as being bounded from below by y  =  x2 and above by y  =  x + 2.  We have

       

        

 

Problem 6  Set up the integrals that give the following.  Use the most appropriate coordinate system.

A.  The mass of the solid that lies inside the sphere

        x2 + y2 + z2  =  9

and outside the cone

        z2  =  x2 + y2 

that has density function

       

Solution

We use spherical coordinates.  The sphere becomes

          =  3

and to find the equation of the cone, we add z2 to both sides to get

        2z2  =  x2 + y2 + z2 

        22cos2   =  2         cos2   = 1/2          = /4

Now we can write

       

B.  The surface area of the part of the paraboloid 

        z  =  x2 + y2

that lies inside the cylinder

        x2 + y2  =  4

Solution

We calculate the partials

        zx  =  2x        zy  =  2y        (1 + zz + zy)1/2  =  (1 + 4x2 + 4y2)1/2

Since the region is a circle of radius 2, we convert to polar coordinates to get

       

 

Problem 7  Find the work done by the force field 

        F(x,y)  =  (3x2 - y) i + (x2 - y3) j

as a particle moves counterclockwise around the rectangle with vertices (2,1), (5,1), (5,5), and (2,5).

Solution

We use Green's Theorem.  We have

        Nx - My  =  2x + 1

We have

       

 

Problem 8

Verify Stokes Theorem where 

        F(x,y,z)  =  (2x) i + (2y) j + (z sin z3) k

and S is part of the paraboloid 

        z  =  9 - x2 - y2 

that lies above the plane z  =  8 oriented upward.

Solution

We first compute the line integral 

       

We notice that the intersection of the paraboloid and the plane is given by 

        9 - x2 - y2  =  8        x2 + y2  =  1

This is the circle of radius 1 raised 8 units above the xy-plane.  Its parameterization is given by

        r(t)  =  (cos t) i + (sin t) j + 8k 

        r'(t)  =  (-sin t) i + (cos t) j  

so that 

        F .

dr  =  [(2cos t) i + (2sin t) j + (8 sin 83) k] . [(-sin t) i + (cos t) j]

        =  -2 sin t cos t + 2 sin t cos t  =  0

Since the integrand is zero, so is the integral.

Now we use Stokes Theorem.  We have 

       

So that the surface integral is zero.

 

 

 

 

  

MATH 202 PRACTICE MIDTERM 2

 

Please work out each of the given problems. Credit will be based on the steps that

you show towards the final answer. Show your work.

PROBLEM 1 Please answer the following true or false. If false, explain why or provide

a counter example. If true explain why.

A)   If all six limits of integration of an integral written in spherical coordinates are

constants, then the region of integration is a sphere.

Solution

False, it can be a part of a sphere.  For example

       

represents a cone with a rounded top.

B)   

Solution

False,  the integrands work out when you convert, but the

regions are different.  The left side is the part of the

sphere of radius 9 inside the cylinder of radius 3 and the

right side is a cone.

PROBLEM 2 Set up integrals to evaluate the following.

Use the coordinate system that will most effectively solve

the integral.

A.     The mass of the tetrahedron that lies in the first octant and below the plane x + y +

2z  =  2 such that the density function is f(x,y,z)  =  3yz .

  Solution

The tetrahedron is pictured below.  We find the triple integral (using dzdydx)  The

bottom is z  =  0 and the top is 

z  =  2 - .5(x + y)

To find the outer limits we draw the projection of the surface on

the xy-plane as shown to the right.  The bottom curve is y  =  0

and the top curve is  y  =  -x + 2.  The farthest to the left that x

gets is 0 and the farthest to the right is 2.  Putting this together,

we get

       

B.     The surface area of the part of the paraboloid  z  =  9 - x2 - y2  that lies above the

plane z = 5 .  

Solution

The picture is shown to the right. We can use polar coordinates to simplify the region

which is a circle of radius 2 since solving 

        9 - r2  =  5 

gives r  =  2.

 The paraboloid becomes.  We have 

        zx  =  -2x        zy  =  -2y

so that

        1 + zx2 + zy

2  =  1 + 4x2 + 4y2  =  1 + 4r2 

Putting this all together gives

       

        

 

C.     The moment of inertia about the z-axis of the solid between the cylinders  x2 + y2  = 

25 and x2 + z2  =  25 that has density function 1.  

Solution

  Changing to polar coordinates gives

        r  =  5        and      r2cos2 q + z2  =  25

the inner limits are 

       

The outer limits just come from the circle of radius 5.  The integrand is 

        x2 + y2  =  r2 r  =  r3 

Putting this together gives

       

PROBLEM 3

Switch the order of integration.

       

Solution

The key to this problem is to graph the region and notice that to integrate with dxdy we

must break the region into two pieces.  As shown to the right.  We get

       

 

PROBLEM 4

A master dart thrower has determined that her probability density function is inversely

proportional to one more than the fourth power of the distance in centimeters from the

center of the dartboard.

A.  Find the constant of proportionality. (Hint:   )  

Solution

We integrate

       

by transforming to polar coordinates.  We get

       

To integrate this we can use u-substitution with

        u  =  r2      du  =  2r dr

This gives

       

Since the probability must be equal to one, the constant of proportionality must be the reciprocal of this double integral that is

                    2        k  =                                        2

       

B.  Find the probability of her hitting the bull’s-eye which is one centimeter in radius.

  Solution

We work the same problem except that the region of integration is the circle of radius 1

and we include the constant of proportionality.  We have

       

PROBLEM 5

Consider the solid described by

        (x + 2y)2 + (2y - 2z)2 + (x + z)2  < 25

with density function 

        f(x,y)  =  x + 2y

Show that the mass of the solid is equal to 

Hint: Recall that  

 

Solution

We use Jacobians with 

        u  =  x + 2y        v  =  2y+2z        w  =  x + z

We have

       

Thus the Jacobian is 1/6.  The function f becomes u.  The region becomes 

        u2 + v2 + w2  < 25 

which is a sphere of radius 5.  The result follows when we put this region in spherical

coordinates (using (u,v,w) instead of (x,y,z)).

 

   

MATH 202 MIDTERM 1

 

Please work out five of the given six problems and indicate which problem

you are omitting. Credit will be based on the steps that you show towards

the final answer. Show your work.

   

PROBLEM 1 Please answer the following true or false. If false, explain why or provide

a counter example. If true explain why

 

A)   (12 Points) If r(t) is parameterized by arclength, then a and N are parallel.

Solution

True,  since 

        s(t)  =  t

We have 

        s''(t)  =  0

So that 

        a  =  s''(t)T + k(s')2N  =  k(s')2N  

so that a and N are multiples of each other.  Hence they are parallel.

        

B)    (13 Points) If r(t) is a differentiable vector valued function then

       

Solution

False,  for example if

        r(t)  =   i + t j

then

       

and

        ||r'(t)||  =  || j ||  =  1

So they are different.

 

PROBLEM 2 (25 Points)

Let

        r(t)  =  2t i - 4t2 j

A. Find T(-1).

 

Solution

We have

        r'(t)  =  2 i - 8t j 

so that 

       

Now just plug in -1 for t to get

       

B. Find N(-1).

Solution

Use the quotient rule to get

       

Now get rid of the denominator and multiply by the root to get

        (-4j)(1 + 16t

2

) - (i - 4tj)(16t)

        =  4[-j(1 + 16t

2

) - (i - 4tj)(4t)]

Now divide by 4 and plug in -1 for t to get

        -17 j - (i + 4j)(-4)  =  4i - j

Dividing by the magnitude gives

       

       

C. Find the equation of the circle of curvature for r(t) at t  =  -1.

 Solution

The acceleration vector is 

        a  =  r''(t)  =  -8j 

Dotting with the normal vector gives the component of the acceleration in the direction of

the normal vector.  We get

        aN  = 

and

        (ds/dt)

2

  =  ||r'(t)||

2

  =  1 + 64t

2

 

evaluating at t  =  -1 gives

        65

Using the curvature formula gives

        65K  =  an  

so that 

       

The radius of the circle is just the reciprocal of the curvature.

To find the center, we add the vectors

        Center  =  r(-1) + KN

which gives

       

Call this 

        ai + bj 

Then the equation of the circle is just

        (x - a)2 + (y - b)2  =  1/K2

 

PROBLEM 3 (25 Points) Jason Elam (the football kicker for the Denver Broncos) can

kick a football with an initial velocity of 60 feet per second. At what angle should the

ball be kicked to maximize the horizontal distance that the ball travels before it lands on

the ground? (Use vectors please).

  Solution

The acceleration is

        a(t)  =  -32 j 

Integrating gives

        v(t)  =  vx i + (-32t + vy) j  =  60cos i + (-32t + 60sin) j

Now integrate again to get

        r(t)  =  60t cos i + (-16t2 + 60sin t) j

Notice that these constants are all zero since the ball starts at the origin.  The ball will

reach its maximal horizontal distance when the j component equals 0.  We have

        -16t2 + 60sin t  =  0

        t  =  15/4 sin

Now plug back into the i component and maximize

        h  =  60 (15/4 sin) cos   =  225/2 sin(2)     

Now to maximize, we take the derivative and set it equal to zero

        h'  =  225 cos(2)

This is zero when   =  /4.

  Jason Elam should punt the ball with an initial angle of /4.

PROBLEM 4 (25 Points) Prove the following theorem:

Let r(t) be a differentiable vector valued function, then

        |(r x v) . a|  =  ||r'|| ||aN|| |r

. (T x N)|

Solution

Since 

        v  =  ||r'|| T

we have

         |(r x v) . a|  =  |(r x ||r'||T) . a| =  |(r x ||r'||T) . (aTT + aNN)| 

        =  ||r'|| |(r x T) . aTT + (r x T) . aNN)|  

Since (r x T) is orthogonal to T, the first term is zero.  We get

        =  ||r'|| |(r x T) . aNN)|  =   ||r'|| |aN|  |(r x T) . N)| 

 

PROBLEM 5 (25 Points)

Find the parametric equations of the tangent line to the curve that is formed by

intersecting the sphere  x2 + y2 + z2  =  2 and the plane x + y - z  =  2 at the point (1,1,0).

Solution

  The tangent line to this curve lies on the tangent planes of each of the two surfaces.  We

can conclude that this line is orthogonal to both normal vectors.  The gradient vectors are

        grad F  =  2x i + 2yj + 2z k        and        grad G  =  i + j - k

Evaluating these at the point (1,1,0) gives

        n1  =  2 i + 2 j        and        n1  =   i + j - k

The cross product is

       

Now we use the formula for a line given a point and a parallel vector

   

     r(t)  =  i + j + t(-2i + 2j)

The parametric equations are 

   

     x(t)  =  1 - 2t        y(t)  =  1 + 2t        z(t)  =  0

 

PROBLEM 6 (25 Points)

If 

        a(t)  =  t i + j - k

find r(5) if 

        r(0)  =  i - k     and         r(1)  =  j + k  

  Solution

We integrate to get the velocity function

        v(t)  =  (1/2 t2 + vx )i + (t + vy )j + (-t + vz )k

Integrate again to get 

        r(t)  =  (1/6 t3 + vxt + rx)i + (1/2 t2 + vyt + ry)j + (-1/2t2 + vzt + rz )k

Now plug in the first initial condition to get

        rx i + ry j + rz k  =   i - k

So that 

        rx  =  1        ry  =  0        rz  =  -1

Plugging these in and using the second initial condition gives

        (1/6 + vx + 1)i + (1/2 + vy)j + (-1/2 + vz - 1)k  =   j + k  

So that 

        vx  =  -7/6        vy  =  1/2        vz  =  5/2

Substituting gives

        r(t)  =  (1/6 t3 - 7/6 t + 1)i + (1/2 t2 + 1/2 t)j + (-1/2t2 + 5/2 t - 1)k

Now plug in 5 to get

        r(5)  =  (125/6  - 35/6 + 1)i + (25/2 + 5/2)j + (-25/2 + 25/2 - 1 )k

        =  16 i + 15 j - k

 

Extra Credit: Write down one thing that your instructor can do to make the class better

and one thing that is going well.

(Any constructive remark will be worth full credit)