Chapter III · Web viewWe will refer to the definite integral of a continuous function f on as ....

Unit II Multiple Integral

Unit II

Multiple Integrals

In this section we will compute:

- Volume of solid regions.

- Surface areas of the boundaries of some solids.

2.1 Double Integrals

2.1.1 Introduction

We will refer to the definite integral of a continuous function f on as single

integral.Let R be a closed region in the xy-plane and f be a non-negative continuous function on R.

Let D be the solid region bounded by the graph of f, the region R and on the side by the vertical

surface passing through the boundary of R. D is called the solid region between the graph of f

and R. Now we want to define the volume of the solid region D.

Assumption: Let the volume V of a rectangular parallelepiped (box) with base area A

and height h be V = Ah

Now consider the following:

i) Suppose R is a closed rectangular region on and f is a non-negative continuous function on

R. Since f is continuous on R, f has minimum value m and maximum value M on R.

Let V be the volume of the solid D and A be the area of the region R.

Then m A V M A

Now let us partition R into n sub-rectangles , , , ... , . For each integer k between

1 and n let be the minimum value and be the maximum value of f on and let be the

area of . Then

V

Prepared bt Tekleyohannes Negussie

yA

Maximum value of f = M

The graph of f

Minimum value of f

x

z

R

37


If we increase the number of sub-rectangles we get as close to V as we wish.

ii) Let R be any closed region on and let f be a non-negative continuous function on R.

Now let R' be a rectangle containing R. Partition R' into a collection Ρ of rectangles. The

rectangles in Ρ are entirely contained in R or partially contained in R or contain no points of R.

Let , , , ... , be the rectangles in Ρ that are entirely contained in R and let , , , . . ., be those rectangles in Ρ that are partially contained in R and let the

remaining rectangles contain no points of R.

Let be the minimum value of f on for i = 1, 2, 3, …, n and be the maximum value of f on

for i = 1, 2, 3, …, p and let be the area of for i = 1, 2, 3, …, p.

Then .

The sum on the left side is called the lower sum of f for p and is dented by ( Ρ ) and the right side is called the upper sum of f for p and is denoted by ( Ρ ).

Thus ( Ρ ) v ( Ρ ).

Prepared bt Tekleyohannes Negussie 38


Definition 2.1 Let R be a bounded region in the xy-plane and f a function continuous

on R.

a) If there is a unique number I satisfying ( Ρ ) I ( Ρ ) for every partition Ρ of any rectangle containing R, then f is integrable on R.

We denote the unique number by

and call it double integral of f over R.

b) If f is non-negative and integrable on R, then the volume V of the solid

region between the graph of f and R is given by

V =

Note that: Let Ρ be a partition of a rectangle containing R into sub rectangles, number so that

, , , ... , are those entirely contained in R. For each integer k between

1 and n let ( , ) be a point in . Then the sum

is called a Riemann sum for f on R.

Theorem 2.1 Let f be integrable on a bounded region R, and let R' be a rectangle

containing R. For any 0 there is a number 0 such that the following

statement holds.

If Ρ is a partition of R' into sub rectangles whose dimensions are all less

than and if , , , ... , are those rectangles contained in R, then

< ε

where the point ( , ) is arbitrarily chosen in for 1 k n.



The conclusion of theorem 2.1 is usually expressed as:

=

Where denotes the largest of the dimensions of the rectangles in Ρ and it is called the norm

of the partition Ρ .

Example 1 Let R be the triangular region bounded by the lines y = 2x, x = 0 and y = 4 and let R′

be the rectangle whose sides are the lines x = 0, x = 2, y = 0 and y = 4. Suppose the partition

Ρ of R′ consists of the squares whose sides are 1 unit long.

If f (x, y) = x + y for (x, y) in R, find

i) ( Ρ ) ii) ( Ρ )

iii) the Riemann sum of f that uses the midpoints of the rectangles that are contained in R.

Solution Let R′ and the partition Ρ = { , , , ... , } be as shown below.

and = 5 , = 4 , = 3 , = , = 6 , = and = 3 , ( Ρ ) = 27.

Since only the rectangles and in the partition are contained in R, the required Riemann sum is:

= 7.

2.1.2 Vertically and Horizontally Simple Regions

Definition 2.2 a) Vertically Simple Regions

A plane region R is vertically simple if there are two continuous functions

and on an interval a, b such that (x) (x) for a x b and such

that R is the region between the graphs of and on a, b. In this case

we say that R is the vertically simple region between the graphs of and

on a, b.


x

y y = 2xNow

Out of these eight sub rectangles only and

are entirely contained in R and only , , ... ,

contain points of R. Thus n = 2 and p = 7.

Since = 3 and = 2, ( Ρ ) = 5

R1 R5

R2 R6

R3 R7

R4 R8

x

40


b) Horizontally Simple Regions

A plane region R is horizontally simple if there are two continuous functions

and on an interval c, d such that (y) (y) for c y d and such

that R is the region between the graphs of ad on c, d. In this case we say

that R is the horizontally simple region between the graphs of and on c, d.

c) Simple Region

A plane region R is simple if it is both vertically simple and horizontally simple.

Remark: Any vertical line x = c where a < x < b (horizontal line y = k where

m < k < n) intersects the boundary of a vertically simple region R on a, b

horizontally simple region R on m, n at most twice.

Vertically simple Horizontally simple simple Neither

Example 2 Let R be the region between the graphs of y = and y = x + 6. Show that R is simple.

Solution y = and y = x + 6 − x − 6 = 0 (x + 2) (x − 3) = 0 x = − 2 or x = 3.

Thus the intersection points for the graphs of y = and y = x + 6 are (− 2, 4) and (3, 6).

Now we need to show that R is simple.

i) Vertically simple ii) Horizontally simple

Let g1 (x) = and g2 (x) = x + 6 Let

Then x + 6 x [− 2, 3] and and

and are continuous on [− 2, 3]. Then (y) (y) y [0, 9] and and

Thus R is vertically simple. are continuous on [0, 9].


y yy y

x

x x

x

41


Therefore, R is a simple region. Thus R is horizontally simple.

2.1.3 Evaluation of Double Integrals

Let f be a non-negative continuous function on a vertically simple region R between the graphs of

and on [a, b]. Let D be the solid region between the graphs of f and R. The volume V of D is

given by:

V =

The volume V of the solid D is:

V =

But the cross sectional area A (x) is given by:

A (x) = .

Hence the volume V of the solid D is given by:

V = .

Similarly if R is a horizontally simple region between the graphs of and on [c, d] and f is a non-

negative continuous function on R, then the volume V of the solid D between the graph of f

and R is given by:

= .

The integrals and are called iterated integrals.

Theorem 2.2 Let f be continuous on R in the xy –plane.


b

a y

z

x

z = f (x, y)

A (x)

g2 (x)g1 (x)

Rx

42


a) If R is the vertically simple region between the graphs of and on [a, b],

then f is integrable on R and

=

b) If R is the horizontally simple region between the graphs of and on [c, d],

then f is integrable on R and

=

Note that: If R is simple, then

=

.

Example 3 Evaluate the double integral , if R is the region containing points

(x, y) for which − 1 x 2 and 1 y 3.

Solution Now R is a simple region.

=

= =

= = − 24.

Therefore, = − 24.

Example 4 Let R be the region between the graphs of y = and y = x + 6. Evaluate .

Solution Now R is simple.

= =

= = = .



Therefore, = .

Example 5 Find the volume V of the solid D bounded above by the paraboloid z = 4 − and

below by the xy-plane.

Solution On the xy-plane z = 0.

Hence = 4 is a simple region and y = where − 2 x 2.

Thus V =

= =

Now let x = 2 sin t, dx = 2 cos t dt and = 8 .

Hence V = = .

= = 8π.

Therefore, V = 8π cubic units.

Example 6 Find the volume of the solid above the xy-plane bounded by the elliptic paraboloid

z = and the cylinder = 4.

Solution Let f (x, y) = .

Now = 4 y = where − 2 x 2.

Hence V =

= = 2

=

Now let x = 2 sin t, dx = 2 cos t dt and = 2 cos t.



Hence V = = .

= =

= 4π.

Therefore, V = 4π cubic units.

2.1.4 Area

The area A of a plane region R is defined by:

By theorem 2.2 we get: A =

Now let R be a region between the graphs of the two continuous functions g1 and g2 on [a, b] such

that g1(x) g2 (x) on x [a, b].

=

= which is consistent to our previous finding.

Example 7 Find the area of the region bounded by the ellipse ,

where a > 0 and b > 0.

Solution , where − a x a.

Hence area = where α = = 2

Now let x = a sin t, dx = a cos t dt and = a cos t.

Thus area = =



= = ab .

Therefore, the area of the region is ab square units.

2.1.5 Reversing the Order of Integration

Remark: If R is a simple region, then can be evaluated as either

or .

Which of these iterated integrals we use depends on the nature of the integrand,

the limits of integration, and our convenience. The change from one iterated

integral to the other is called reversing the order of integration.

Example 8 Reversing the order of integration evaluate

.

Solution x = h1 (y) = y = ; x 0 and x = (y), thus x = 3 where 0 y 9.

Then =

= = = .

Therefore, = .

2.1.6 Double Integration over More General Regions

If R is composed of two or more vertically or horizontally simple sub-regions , , , ... ,

with the property that any two sub-regions have only boundaries in common, then any function f

that is continuous on R is integrable on R, and

=

Example 9 Let R be the region between the graph of y = x and y = . Evaluate .

Solution y = x and y = x = − 1 or x = 0 or x = 1.



Hence the boundaries of R intersect at (− 1, − 1), (0, 0) and (1, 1).

Thus R is composed of the two vertically simple regions and .

Hence = + =

.

= .

=

= .

Therefore, = .

2.2 Double Integral in Polar Coordinates

2.2.1 Polar Coordinates

Let P (0, 0) be a point in the xy-plane and let r be the distance of P from the origin O and be the

angle between the positive x-axis and . The ordered pair (r, ) is called the polar coordinates for

the point p. If r > 0 and 0 2π, then for any point p (0, 0) there corresponds a unique polar

coordinates (r, ) and vice versa.

2.2.2 Conversion between Cartesian and Polar Coordinates

Let P (0, 0) be a point in the xy-plane with r = OP and , 0 2π the angle between the positive x-axis and . Then x = r cos and

P (x, y) y = r sin tan =

y = r sin where x 0.

x = r cos

Note that: If x = 0 and y > 0, then = and if x = 0 and y < 0, then = .

Example 10 Find the Cartesian coordinates of the point p having polar coordinates (3, ).

Solution Now r = 3 and = .

Hence x = 3 cos ( ) and y = 3 sin ( ).



= − 3 cos = − 3 cos

= =

Therefore, the coordinates of P are ( , ).

Example 11 Find the set of all polar coordinates for the point p having Cartesian coordinates

(− 2, ). Solution and r > 0 r = 4. Now tan = − = tan − 1 (− ).

But x < 0 and y > 0, and hence is a second quadrant angle.

Thus = 2n π + where n Z.

Therefore, the polar coordinates of P are (4, 2n π + ), where n Z.

Note that: The polar equation of a line passing through the origin and making an angle with

positive x – axis is = .

Let P be a point on the line on the first quadrant and let OP = r, then

x = r cos and y = r sin .

Hence = tan − 1 (tan ) = .

Therefore, the equation of the line is = .

2.2.3 Polar Equations and Graphs

Example 12 Find the polar equation of:i) ; a > 0. ii) ; a > 0.

iii) ; a > 0.

Solutions i) From we get . Since a > 0, r = a.

Therefore, r = a is the polar equation of the circle with center at (0, 0) and radius a.

ii) Since x = r cos and y = r sin , we get:

= ar cos r = a cos Therefore, r = a cos is the required polar equation.

iii) Similarly we get:

r = a sin .

Example 13 Graphs of Cardioids r = 1 + sin and r = 1+ cos .



Example 14 The graph of the Limacon r = .

Since cos (− ) = cos , whenever (r, ) satisfies the equation (r, − ) does. Hence the graph is

symmetric with respect to the x-axis. Hence sketch the graph for 0 π with respect tothe x-

axis.

If (r, ) satisfies the equation, then so does (r, − ) or (− r, π − ) with respect to the y-axis.

If (r, ) satisfies the equation, then so does (− r, − ) or (r, π − ) with respect to the origin. If

(r, ) satisfies the equation, then so does (− r, ) or (r, π + ).

Example 15 The graph of the three-leaved rose r = cos 3.


x

yy

x

The graph of The graph of r = 1 + sin r = 1 + cos

The graph of Lie-me-sohn

x

y

The graph of r = cos 3

49


2.2.4 Double Integral in Polar Coordinates

In evaluating a multiple integral in the xy-plane over a region R it is often convenient to transform

the coordinate system to another coordinate system.

If we let (u, v) be a point in the uv coordinate system corresponding to (x, y), then there will be a set

of transformation equation.

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

In such a case the region R of the xy coordinate plane is mapped to the region S of the uv plane.

Then we get:

= .

where G (u, v) = F ( f (u, v), g (u, v)) and

= is the Jacobian of x and y with respect to u and v.

Now let f be a function that is continuous on a closed region R in the xy - plane. Then x = r cos and y = r sin

are the transformation equations that maps the region R into the region S of the polar coordinate

plane and hence

= =

where = = = r.

Therefore, = .

Now suppose that and are continuous on an interval [, ], where 0 − 2π and

( ) () for .

Let R be the region in the xy plane bounded by the lines = and by the polar graphs r = ( ) and

r = ( ). Then we say R is the region between the polar graphs h1 and h2 on [, ].

If f is a continuous function on R, then f is integrable on R.

Theorem 2.3 Suppose that and are continuous on [, ], where 0 − 2π

and that ( ) () for . Let R be the region between the

graphs of r = () and r = ( ) for . If f is continuous on R,

then



=

Remark: If f is non-negative on R, then:

i) the volume V of the solid between the graph of f and R is given by

V = .

ii) the area A of R is given by

A = .

Example 15 Let R be the region bounded by the circles r = 1 and r = 2 and the lines = and

= , where 0 − 2π, express as an iterated integrals

in polar coordinates and evaluate the iterated integral for

i) = 0 and = ii) = 0 and = iii) = 0 and = 2 .

Solutions i) = 0 and = .

Now let x = r cos and y = r sin .

Then =

=

=

= = 7 + .

Therefore, = 7 + .

ii) = 0 and =

Similarly = = 15 π.

Therefore, = 15 π.

iii) = 0 and = 2 .



Similarly = = 30 π.

Therefore, = 30 π.

Example 16 Suppose D is the solid region bounded on the sides by the cylinder r = cos , above the

cone z = and below by the xy plane. Determine the volume V of D.

Solution Let f (x, y) = .

Now R is the region between the polar graphs r = 0 and r = cos for .

Hence V = =

= =

= = .

Therefore, V = cubic units.

Example 17 Let D be the solid region bounded above by the paraboloid and

below by the xy plane. Find the volume V of D.

Solution On the xy plane z = 0 x 2 + y 2 = 4.

Hence r = 2 and

and V = = = = 8 π.

Therefore, V = 8 π cubic units.

Example 18 Let R be the region between the polar graphs r = and r = 2 for .

Evaluate .

Solution = = =

= = .

Therefore, = .



Example 19 Find the area A of the region in the xy plane of one leaf of the three leaved rose

bounded by the graph of r = 2 sin 3.

Solution Now take .

A = = =

= = = .

Therefore, A = square units.

Example 20 Determine the area of the region inside the circle r = 2 cos and out side the

circle r = 1.

Solution First determine the intersection points of the two circles.

2 cos = 1 .

Hence A = =

= =

= = 3

+

Therefore, A = 3

+ square units.

Example 21 Find the area of the region enclosed by the cardiod r = 1 + cos .

Solution

A = = =

= = = .

Therefore, A = square units.

2.3 Triple Integrals

Let R be a vertically or horizontally simple region in the xy plane, and let and be continuous

on R satisfying



(x, y) (x, y) for (x, y) in R. Let D denote the solid region consisting of all points (x, y, z) such that (x, y) is in R and

(x, y) Z (x, y).We refer to D as the solid region between the graphs of and on R.

Let f be a non-negative continuous function defined on D and D ' be a rectangular parallelepiped

containing D. Let p be a partition of D ' into smaller rectangular parallelepiped.

Let D1, D2, D3, ..., Dn be the rectangular parallelepiped in p that are entirely contained in D and let

Dn + 1, Dn + 2, Dn + 3, ..., Dp be the rectangular parallelepiped in p that are partially contained in D.

Let mk be the minimum value of f on Dk for 1 k n and Mk the maximum value of f on Dk for

1 k p. If denotes the volume of Dk and Lf (p) = and Uf (p) =

, then there is exactly one number I such that Lf (p) I Uf (p)

for every partition p of any parallelepiped D ' containing D.

Defn 2.3 Let D be the solid region between the graphs of two continuous functions F1

and F2 on a vertically or horizontally simple region R in the xy plane. If f is

continuous on D, we write

for the unique number that lies between Lf (p) and Uf (p) for every

partition p of any parallelepiped containing D. The number

is called the triple integral of f on D.

Now let (xk, yk, zk) be an arbitrary point in Dk for 1 x n. Then the sum

is a Riemann sum for f on D which approximates .

Theorem 2.4

Let f be continuous on the solid region D between the graphs of two continuous

functions and let D ' be a rectangular parallelepiped containing D. For any 0

there is a number 0 such that the following statement holds. If p is a partition

of D ' into sub rectangular parallelepipeds whose dimensions are all less than

and if D1, D2, D3, ..., Dn are the sub rectangular parallelepipeds in D ' entirely



contained in D, then

where is an arbitrary point in Dk for 1 k n.

The conclusion of theorem 2.4 is usually expressed as:

=

Where is the largest of the dimensions of the sub rectangular parallelepipeds in p and is called

the norm of the partition p.

2.3.1 Evaluation of Triple Integrals

In general, lower sum, upper sum and Riemann sums are not very effective in evaluating a triple

integral. Once again, iterated integrals provide a method of evaluating triple integrals.

Theorem 3.5

Let D be the solid region between the graphs of two continuous functions F1

and F2 on a vertically or horizontally simple region R in the xy plane, and let

f be continuous on D. Then

= .

Remark

i) If R is a vertically simple region between the graphs of g1 and g2 on a, b, then

= .

(1)

ii) If R is a Horizontally simple region between the graphs of h1 and h2 on c, d, then



=

(2)

The integrals on the right side of (1) and (2) are called iterated integrals and are

usually written without brackets.

Example 1. Let R be the triangular region in the xy plane between the graphs of y = 0 and y = x

for 0 x 1, and let D be the solid between the graphs of the surfaces z = y2 and

z = x2 for (x, y) in R. Evaluate .

Solution. = =

= =

= =

= = cubic units.

Example 2. Let D be the solid region bounded by the portions of the two circular paraboloids

z = 3 x2 y2 and z = 5 + x2 + y2 for which x 0 and y 0. Evaluate

Solution. The two paraboloids intersect on the plane z = 1, where x2 + y2 = 4. Hence the

corresponding region R in the xy plane is the horizontally simple region in the first

quadrant that lies inside the circle x2 + y2 = 4 , between the graphs of x = 0 and

x = for 0 y 2.

Since 3 x2 y2 5 + x2 + y2 for (x, y) in R.

= =

= =



= = .

Therefore, = .

2.3.2 Volume by Triple Integrals

Let D be the solid region between the graphs of the two continuous functions F1 and F2 on R in

the xy plane. The volume of the solid D is defined by:

v = and hence v = .

Example 1 Find the volume of the solid bounded by the cylinder x2 + y2 = 25, the plane

x + y + z = 8 and the xy plane.

Solution x + y + z = 8 z = 8 x y and x2 + y2 = 25 y = for 5 x 5.

Now let F1 (x, y) = 0 and F2 (x, y) = 8 x y for (x, y) (x, y) : x2 + y2 25.

Thus v = = =

= =

=

Now let x = 5 sin t, then dx = 5 cos t dt.

Thus = =

= = 200 .

and = = 0.

Therefore, v = 200 .

Example 2 Suppose R is a vertically simple region between the graphs of the two non-negative



continuous functions f and g on a, b with g (x) f (x) for x a, b. Let D be the solid

region generated by revolving R about the x axis. Show that the volume v of D is given by:

v = .

Solution Let v1 be the volume of the solid region D1 generated by revolving R1 bounded by the

graph of g and the x axis on a, b about the x axis and let v2 be the volume of the solid

region D2 generated by revolving R2 bounded by the graph of f and the x axis on a, b

about the x axis.

Thus v = v2 v1 and R1 and R2 are given by:

z2 + y2 = and z2 + y2 = for x a, b respectively.

Hence v1 = = =

Now let y = g (x) sin t, then dy = g (x) cos t dt.

Thus v1 = =

Similarly v2 = .

Therefore, v = .

Remark: The volume of the solid region generated by revolving the region R

bounded by the graph of a non-negative continuous function f and

i) the y axis on c, d about the y axis given by:

v =

ii) the z axis on m, n about the z axis given by:

v =



Remark: Sometimes it is advantageous to regard the solid D as a solid region

between two functions of x and z or two functions of y and z.

Example 3 For the region shown in the figure below, evaluate .

Solution. Now let R be the region in the y = 1

plane consisting points (x, y): x2 + z2 1.

Now x2 + y2 + z2 = 1 y2 = 1 x2 z2

and x2 + z2 = 1 z = for x 1.

Thus =

=

=

=

= = = .

Therefore, = cubic units.

2.3.3 Triple Integrals in Cylindrical Coordinates

Just as certain double integrals are easier to evaluate by means of polar coordinates than by

rectangular coordinates, certain triple integrals are easier to evaluate by coordinates other than

rectangular coordinates.

Cylindrical Coordinates


x

z

y

x2 + z2 = 1

59


Let p be a point in space with rectangular coordinates (x, y, z). If (r, ) is a set of coordinates

for the point (x, y) then we call (r, , z) a set of cylindrical coordinates for p.

Q has polar coordinates (r, )

Relation between cylindrical and polar

coordinates

i) x2 + y2 = r2 and tan =

where x 0.

ii) x = r cos and y = r sin .

Equations in cylindrical coordinates

surfaces rectangular

cylindrical

cylinder r = a

sphere double circular cone r = a or r = cot

circular paraboloid

Double circular cone.

Theorem 2.6 Let D be the solid region between the graphs of F1 and F2 on R, where

R is the plane region between the polar graphs of h1 and h2 on ,

with 0 2 and 0 h1 ( ) h2 ( ) for . If f is

continuous on D, then


x

y

z

p (x, y, z)

lower cone

upper cone

y

x

z

Q (r, )

60


Example 1 Find the volume v of the solid region D that the cylinder r = a cos cuts out of the

sphere of radius a centered at the origin.

Solution The plane region R is bounded by the polar graphs r = 0 and r = a cos for

and D is the solid region determined by the sphere between the graphs of z =

and z = for (r, ) in R.

Since D is symmetric with respect to the planes z = 0 and y = 0,

v = = = 4

= 4 =

= a3 = a3

= a3 = .

Therefore, v = cubic units.

Example 2 Let D be the solid region between the sphere r2 + z2 = a2 and the nappe of the cone

z = r cot that makes an angle with the positive z axis. Show that the volume v of D is

given by:

v = .

Solution There are three cases to consider.

Case 1 0 .

r2 + z2 = a2 and z = r cot r2 csc 2 = a2 r = a sin .

Thus D is the solid region between the upper nappe of the cone and the sphere on the plane region R

bounded by the circle r = a sin .



Hence v = = =

=

=

= = .

Case 2 .

v = (volume of the portion of the sphere inside the lower nappe of the cone)

=

=


Example 3 Evaluate

Solution From the limits of integrations on the first and the second integrals we get that these two

integrals are taken over the region bounded by the circle x2 + y2 = 4 or r = 2.

Thus =

=

=

=

=

=



= 14

= 28.

Therefore = 28.

2.3.4 Triple Integrals in Spherical Coordinates

Spherical coordinates

Let P be a point in space, P (0, 0, 0). Let (x, y, z) and (r, , z) be the set of rectangular and

cylindrical coordinates of p with r 0.

Let = OP. If 0, then = the angle between and the positive z axis with 0 .

If = 0, then may be chosen arbitrarily.

Let be the angle between and the positive x axis where Q is a point in the xy plane such that

is parallel to the z axis.

The triple (, , ) (in some books (,, ) ) is called a set of spherical coordinates of the point p.

Exercise Plot the following points whose spherical coordinates are:

i) A(1, , ) ii) B(2, , ) ii) C(1, , ).

Relation between Spherical and Cartesian Coordinates

Example 1 Find the rectangular coordinates of the point P with spherical coordinates (2, , ).

Solution = 2, = and = .

Hence x = 2 sin ( ) cos ( ) = , y = 2sin ( ) sin( ) = and z = 2cos ( ) = 1.

Therefore ( , , 1) is the rectangular coordinates for P.

Example 2 Find the cylindrical and spherical coordinates of the point P with rectangular


From the figure we get: r = sin and z = cos i) x = r cos and y = r cos ii) From i) and ii) we get: x = sin cos y = sin sin and z = cos

P(x, y, z)

Q(r, )

63


coordinates (4, 2, 4).

Solution 2 = 42 + 22 + 42 = 36 = 6, = tan 1 ( ) = tan 1 ( )

and = cos 1 ( ) = cos 1 ( ).

Thus cos = cos 2 = and hence sin 2 = and

sin = and r = 2 .

Therefore, (2 , tan 1 ( ), 4) and (6, cos 1 ( ), tan 1 ( )) are the required solutions.

Equations in Spherical Coordinates

Surface

1. Sphere

2. Cone

3. Vertical half

plane

Rectangular

x2 + y2 + z2 = a2

x2 + y2 = a2 z2

y = c x

Cylindrical

r2 + z2 = a2

r = a z

= const.

Spherical

= a

= const.

= const.

Example 1 Find the equation in Cartesian coordinates for the following equations in spherical

coordinates.

a) cos = b b) sin = a.

Solution a) cos = b z = b.

Therefore the Cartesian equation is z = b.

b) sin = a r = a r2 = a2 x2 + y2 = a2.

Therefore the Cartesian equation is x2 + y2 = a2.

Note that: If there are transformation equations

x = sin cos , y = sin sin and z = cos , then

.



Thus we get the following theorem.

Theorem 2.7

Let and be real numbers with + 2. Let h1, h2, F1 and F2 be

continuous functions with 0 h1 h2 and 0 F1 F2. Let D be the

solid region consisting of all points in space whose spherical coordinates

(, , ) satisfy

, h1 () h2 () and F1 (, ) F2 (, ).

If f is continuous on D, then

Note that: i) 0 ii) 0 ii) 0 .

Example 1 Let D be the solid region between the spheres = 1 and = 2. Find .

Solution 0 2 , 0 , 1 2 and z2 = 2 cos 2 .

= .

= .

= .

= = = .

Therefore, = .

Example 2 Find the volume v of the solid region D between the spheres x2 + y2 + z2 = 1 and

x2 + y2 + z2 = 9 and above by the upper nappe of the cone z2 = 3(x2 + y2).

Solution x2 + y2 + z2 = 1 = 1 and x2 + y2 + z2 = 9 = 3.

z2 = 3(x2 + y2) cos 2 = 3 r2 cos 2 = 3 sin 2 .



Since D lies above the upper nappe of the cone 0.

Hence tan2 = and 0 = .

Thus 0 .

Therefore, v = =

= = .


Some Applications of the Double Integrals

Surface Area

Defn 2.4 Let R be a vertically or horizontally simple region, and let f have

continuous partial derivatives on R. If is the graph of f on R, then the

surface area S of is defined by:

S = (1)

Remark: It is usually difficult to compute surface area using formula (1).

Example 1 Find the surface area of the portion of a parabolic sheet z = x2 directly above the triangle

with vertices (0, 0, 0), (1, 0, 0) and (1, 1, 0).

Solution Let f (x, y) = x2 , and R be the triangular region bounded by the lines y = 0 and y = x for

0 x 1.

Thus S = =

Now let t = , then dt = 8x dx.

Hence S = = = .

Therefore, S = square units.

Example 2 Find the surface area S of the portion of the paraboloid z = 4 x2 y2 above the xy



plane.

Solution R is bounded by x2 + y2 = 4.

Let f (x, y) = 4 x2 y2 , then fx (x, y) = 2x and fy (x, y) = 2y.

Hence S = =

= = .

Therefore, S = square units.

Example 3 Let a 0. Find the surface area S of the frustum of the cone z = a with

minimum and maximum radii r1 and r2 respectively.

Solution R is bounded by the annulus

r1 2 r2

2.

If f (x, y) = a , then

and .

Hence S =

=

= ( area of R)

= .

Therefore S = square units.


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