Math 55 LE2 Notes

U P K E M

M E M B E R S H I P A C A D E M I C D E V E L O P M E N T

Math 55

2nd Long Exam Notes

J.Quintos

T.Delos Santos

Preface

This handout is intended as a reviewer only and

should not be substituted for a complete lecture,

or used as a reference material. The goal of this

reviewer is to refresh the student on the concepts

and techniques in one reading. But this is more

than enough to replace your own notes :)

A Topological Note

Definition (Region). A region is an open, con-

nected, and non-empty set of points in Rn.

In this reviewer notes, well usually denote re-

gions with D,E, or R. Well also be working

mostly with closed regions, in which we include

the boundary points of the region.

1 Triple Integrals

From integration over regions in R2, we progressto regions in R3.

Definition (Triple Integral). Given a region

D R3 and a function f : D R, the tripleintegral of f over D is

D

f = lim0

li=0

mj=0

nk=0

f(xi , yj , zk)Vijk

provided the limit exists. If the limit does exist,

f is said to be integrable over D.

Remark. For a more detailed development of the

definition, refer to the appendix.

Other notations used for a triple integral include:D

f(x, y, z) dV

D

f(x, y, z) dx dy dz

To compute triple integrals, we invoke again Fu-

binis Theorem, which equates a triple integral

into an iterated integral.

1.1 Applications

Ill discuss first the typical applications of the triple

integral because theyll be used again and again

in most examples.

1.1.1 Volume

The volume of a region D in R3 is

1

University of the Philippines Chemical Engineering Society, Inc. (UP KEM)

Math 55 2nd Long Exam Reviewer

V(D) =

D

dV (1)

1.1.2 Mass and Density

You are very familiar with the equation

=M

V

Where ,m, and V are density, mass and volume

respectively. Rearranging and with equation (1),

we have:

M =

D

dV (2)

This is only applicable whenever the density is

uniform/constant throughout the object.

Suppose the density varies in R3 space. Let bea function of position in space. Then:

M =

D

(x, y, z) dV (3)

1.1.3 Center of Mass

The center of mass of an object is the average

position of all the mass in the object.

x =1

M

D

x dV (4)

y =1

M

D

y dV (5)

z =1

M

D

z dV (6)

Where M is given by equation (3).

Now we move on to actually computing triple in-

tegrals.

1.2 Box-like regions

For a simple case of a box region

D = [a, b] [i, j] [m,n]

the triple integral becomes:D

f(x, y, z) dV =

nm

ji

ba

f(x, y, z) dx dy dz

The right-hand side expression can be thought of

as evaluating the integrals from the inside out,

as in: nm

[ ji

( ba

f(x, y, z) dx

)dy

]dz

treating outer variables constant when evaluat-

ing the inner integral with respect to the inner

variable

Example. EvaluateE

xyz dV

where the region E = [1, 2] [3, 4] [5, 6].

Solution. Using Fubinis Theorem we have,

=

65

43

21

xyz dxdy dz

Working from the innermost integral to the out-

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ermost, 21

xyz dx = yz

((2)2 (1)2

2

)

=3

2yz

43

3

2yz dy =

3

2z

((4)2 (3)2

2

)

=21

4z

65

21

4dz =

21

4

((6)2 (5)2

2

)

=231

8

Try to change the order of integration. Verify that

for box-like regions, the order of integration does

not matter.

1.3 General regions

Consider the region in figure 1.3.1a. If we project

the shadow of the blob on the xy-plane, we get a

planar region D in R2 (figure 1.3.1b).

Observe that D can be described, previously

learned in double integrals, as the set of points:

D = {(x, y) : a x b and f(x) y F (x)}

It means that as we move x in the direction from

point a to point b, our y includes the values

from f(x) to F (x). All the points (x, y) that are

touched belong to the region D.

With similar reasoning, we can think of region E

as the set of points

E = {(x, y, z) : (x, y) D and g(x, y) z G(x, y)}

Think of g(x, y) as a floor and G(x, y) as a ceil-

ing of the region. For any point (x, y) that be-

long in region D : z goes from its floor function

g(x, y), including every value in between, up to its

ceiling function G(x, y).

x

y

z

E

(a) a blob

x

y

z

E

D

f(x) F(x)

g(x,y)

G(x,y)

a

b

(b) a blob shadow

Figure 1.3.1: A blob-like region in R3

The triple iterated integral of a function over the

blob region E is then: ba

F (x)f(x)

G(x,y)g(x,y)

(some function) dz dy dz

Remark. I emphasised by typing some func-

tion that the function to be integrated is not

necessarily the same function describing the re-

gion. This has been a common mistake, as noted

by my instructor.

The challenge is cleverly thinking of a convenient

representation of the region that simplifies com-

putation compared to other forms.

Note that the order of integration is not inter-

changeable for a particular set of bounds describ-

ing a general region. Reordering the integral needs

rewriting the bounds.

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Example. Setup the iterated triple integral that

will give the volume of a half-cylindrical region

with r = 2 and l = 1, shown below. Write in the

order (a) dxdy dz and (b) dz dy dx.

1

2

2

x

y

z

Solution.

(a) The outermost variable is z, which takes val-

ues 0 z 2.Next is y. Since z and y are related by

the circle z2 + y2 = 22, we can express

y(z) =

4 z2. As z goes from 0 to 2, ygoes from

4 z2 to

4 z2. So theres

our bounds.

The x variable doesnt depend on the previ-

ous variables, and we have 0 x 1. Theiterated integral is then:

20

4z24z2

10

1 dxdy dz

(b) The outermost variable is x which is just 0

to 1.

For y, observe that it goes from 2 to +2.Now z is a function of y alone (recall the

circle), for each y value, z goes from 0 to4 y2. The integral is:

10

22

4y20

1 dz dy dx

1.4 Cylindrical coordinates

We now look at the analogue of the polar coordin-

ates in R3. In figure (1.4.1), we see a coordinate

O

L

A

z

r

Figure 1.4.1: Cylindrical coordinate system.

system with origin at O, a longitudinal axis L and

a polar axis A.

A point in 3D space can be described by its radial

distance (r) from the origin; its angular rotation

from the A-axis (); and its longitudinal distance

(z) from the origin.

Taking z- and x-axis to be the longitudinal and

polar axes respectively, the coordinate transform-

ation is:

x = r cos

y = r sin

z = z

x2 + y2 = rs

And the differential volume becomes:

dVrect = r dVcyl

Remark. There are cases where z-axis is not

the most convenient choice for the longitudinal

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axis. You should choose carefully which it should

be and apply the corresponding coordinate trans-

formation.

Example. Evaluate the triple integral in the pre-

vious example using cylindrical coordinates.

Solution. We choose the x-axis as our longitud-

inal axis, and the y-axis as our polar axis.

Our function is just the constant 1, so no trans-

formation needed. However we must analyse what

our bounds in terms of the new coordinates will

be.

x : from 0 to 1

r : from 0 to 2

: from 0 to pi

The iterated integral is then: 10

pi0

20

r dr d dx

Which evaluates to 2pi

In this next example, we show how to apply the

transformation

Example. Transform

11

1y20

x2+y2x2+y2

xyz dz dxdy

into cylindrical coordinates.

Solution. Observe the bounds of the region:

1 y 1

0 x

1 y2

x2 + y2 z x2 + y2

The planar shadow of this region projected onto

the xy-plane is a half disc of radius 1 lying on

the 1st and 4th quadrant. So, pi2 pi2 and

0 r 1.

As for the longitudinal component, note that

x2 + y2 = r2, so by algebra, r2 z r. We nowhave the bounds.

Transforming the function:

xyz = (r cos )(r sin )z

= r2z cos sin

The integral is then

pi2

pi2

10

rr2r3z cos sin dz dr d

1.5 Spherical coordinates

x

z(,,)

y

Figure 1.5.1: Spherical coordinates

A point in R3 can also be described by its

radial distance, , from the origin;

polar angle, ; and

azimuthal angle, .

The transformation to spherical coordinates, with

z axis as the zenith direction, is:

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x = sin cos

y = sin sin

z = cos

x2 + y2 + z2 = 2

The corresponding differential volume is

dVrect = 2 sindVsph

Remark. Some instructors use different conven-

tions other than (, , ) . Be careful.

Example. Find the volume of the region

bounded above by the cone = pi4 and below by

the sphere = 4 cos. Refer to the figure below.

(sample problem from [3])

x y

z

4

(a) = pi/4 and = 4 cos

y

z

O

P

()

(b) Projected view on the zy-plane

Figure 1.5.2: Cone and sphere

First we need to setup the bounds. A slice of the

region of interest is the shaded region in figure

1.5.2b.

Observe that our region of interest starts at =

pi/4 (zOP ) and ends with = pi/2 (zOy).

At each , goes from 0 at the origin to the sphere

= 4 cos.

As for , we fully revolve this slice about z to get

our desired region, so 0 2pi

Our iterated integral is then 2pi0

pi2

pi4

4 cos 0

2 sinddd

Which evaluates to8pi

3

1.6 Change of variables* (optional)

Back in Math 54, you were taught the u-

substitution technique:f(g(x))g(x) dx =

f(u) du

Where u = g(x). This can be extended to multiple

integrals. You may have noticed that when we

changed from rectangular to cylindrical or spher-

ical, there were extra terms we considered. We

begin the change in variables for multiple integ-

rals by defining the Jacobian for a particular case.

Definition (Jacobian). Suppose we have the

transformation x = g(u, v) and y = h(u, v). Then

the Jacobian of the transformation is the determ-

inant

JT =

xu xvyu yv

Where the subscripts denote partial differenti-

ation. You can extend this to 3 or more variables,

just take note of the row and column pattern.

Now with the theorem

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Theorem 1 (Change of variables for Double In-

tegrals). Suppose that we want to integrate f(x, y)

over the region D.

Under the transformation x = g(u, v), y = h(u, v)

the region becomes E and the integral becomes,D

f(x, y) dA =

E

f(g(u, v), h(u, v))|JT |dudv

We extend this theorem to the triple integral in

the following example:

Example. Verify with the change of variable

theorem the transformation from rectangular to

spherical coordinates.

Solution. We compute the Jacobian of the trans-

formation

JT =

( sin cos ) ( sin cos ) ( sin cos )

( sin sin ) ( sin sin ) ( sin sin )

( cos) ( cos) ( cos)

=

sin cos sin sin cos cos sin sin sin cos cos sin

cos 0 sin

with some handwavery math magic:

JT = 2 sin

Take the absolute value, and there you go.

2 Vector Calculus

In this section we discuss vector fields, two new

types of integral, and important theorems about

those new types of integral.

2.1 Vector fields

Definition (Vector field). A vector field in two or

three dimensional space is a function ~F that out-

puts a vector at each point in the domain ((x, y)

or (x, y, z)) given by ~F (x, y) or ~F (x, y, z)

Notation for vector fields follow that of vector

function.

Angled bracket notation

~F = P,Q

~G = P,Q,R

Unit vector notation

~F = P i +Q j

~G = P i +Q j +R k

Where in all cases, P,Q,R are scalar functions of

x, y or x, y, z.

Some examples of vector fields IRL are shown in

figure 2.1.1.

(a) Magnetic field lines

(b) Surface wind over the Ph.

Figure 2.1.1: Examples of vector fields in reality

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Visual representations do not include all of the

vectors in a vector field though. Otherwise, youd

just end up covering the whole image and see

nothing. In reality, every point in the plane (or

space) has a vector.

2.2 Divergence and Curl

Definition (Del operator). We let the symbol ~denote the Del operator, such that:

~ =

x,

y,

z

We then introduce two new objects, the diver-

gence and curl.

Definition (Divergence). The divergence of a

vector field

~F = P (x, y, z, ), Q(x, y, z), R(x, y, z)

is a scalar expression given by

div(~F ) = ~ ~F

Equivalently,

div(~F ) =

x,

y,

z

P,Q,R

=P

x+Q

y+R

z

= Px +Qy +Rz

Definition (Curl). The curl of a vector field

~F = P (x, y, z, ), Q(x, y, z), R(x, y, z)

is a vector given by

curl(~F ) = ~ ~F

Equivalently

curl(~F ) =

x,

y,

z

P,Q,R

= (Ry Qz), (Pz Rx), (Qx Py)

= (Ry Qz) i +(Pz Rx) j +(Qx Py) k

When a point in (x, y, z) is specified, the diver-

gence and curl expressions can be evaluated to

give scalar and vector values, respectively.

A simplification for 2-D vector fields can be easily

derived for

~F = P (x, y, z, ), Q(x, y, z), R(x, y, z)

by setting R = 0, and noting that P and Q will

be functions of x and y only. Verify the following:

div(~F ) = Px +Qy

curl(~F ) = 0, 0, Qx Py

2.3 Conservative fields

Definition (Conservative field). A vector field ~F

is said to be conservative if there exists a scalar

function such that:

~ = ~F = P,Q,R

The scalar function is called the potential func-

tion of ~F .

An easier method utilises a theorem, the proof of

which will not be shown here.

Theorem 2. ~F is conservative if and only if

curl(~F ) = 0

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Remark 2.1. For 2D vector fields, this simply

means that

Qx = Py

as shown in the previous subsection.

Upon showing that the vector field is conservative,

we can apply the definition of gradient in the x

term. Considering only 2D:

d

dx= P

d =

P dx

Because of integration, this initial potential func-

tion has a constant of integration which may

be a function of y. To check, differentiate with

respect to y and compare this to Q.

The usefulness of conservative vector fields and

the potential function will be apparent later on.

2.4 Line Integrals

Definition (Line Integral). We define the line in-

tegral of f over a curve C as:C

f(x, y, z) ds = limt0

ni=1

f(xi (t), yi (t), z

i (t))is

where is is the arc length of a segment of C,

and t is the interval between subsequent points

on C.

In a 2D sense, the line integral is the area of

the curtain that connects the points of curve

C plotted in the xy-plane to points in z defined

as z = f(x, y).

2.4.1 Scalar Fields

The evaluation of the line integral is simple when

the curve, the function, and the limits of integ-

ration are given in parametric form, or can be

transformed to parametric form easily. The line

integral can be expressed as:

C

f(x, y, z) ds =

t=bt=a

f(x(t), y(t), z(t))ds

dtdt

where

ds

dt= ||~R(t)|| (7)

~R(t) = x(t), y(t), z(t) (8)

Note that when the function is unity, the line in-

tegral simplifies to the formula of the arc length

of the curve from a to b.

2.4.2 Vector Fields

The result of an evaluated integral is usually a

number. To force this on vector fields such as ~F =

P (x, y, z, ), Q(x, y, z), R(x, y, z), the line integralof vector fields is given by:

C

~F d~R =C

~F T ds

=

C

~F d~Rdtdsdt

ds

=

C

~F d~Rdtdsdt

ds

dtdt

=

t=bt=a

~F d~R

dtdt

By using the dot product, we mean to get the

component of the vector field ~F along the curve,

signified by T. We multiply this component to a

distance dt, then integrate over a curve. Doesnt

this look familiar? This is the concept of work for

any vector field. This is why the Line Integral is

also called the Work Integral.

Again, the evaluation of the line integral is simple

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when the curve, the function, and the limits of

integration are given in parametric form, or can

be transformed to parametric form easily.

2.4.3 Fundamental Theorem for Line Integ-

rals

Theorem 3. Let C be a piecewise, smooth para-

metric curve lying in an open region D having

initial and terminal points (x0, y0) and (x1, y1),

respectively. If ~F (x, y) = M(x, y), N(x, y) is aconservative vector field on D, where M and N

are continuous on D, then for any potential func-

tion of ~F on D:C

~F d~R = (x, y)(x1,y1)(x0,y0)

This theorem provides us another means of cal-

culating the line integral if we can show that the

function ~F is conservative (that is curl(~F ) = 0)

and we can successfully find the potential func-

tion .

2.4.4 Path independence

By a line integral independent of the path, we

mean an integral C

~F d~R

satisfying C

~F d~R =C~F d~R

for any other curve C having the same initial andterminal points as C.

The importance of this concept is its ability to

simplify a complex path without a parametric

form into a simple path whose parametric form

is easily obtainable (e.g. lines). To show that a

line integral is not independent of the path, simply

show that the line integral yields different answers

in different paths.

To show that a line integral is independent of the

path, we utilise the theorem, whose proof will not

be shown here.

Theorem 4. If M(x, y) and N(x, y) are continu-

ous on open-connected region D, then the follow-

ing are equivalent statements:

1. ~F (x, y) = M(x, y), N(x, y) is conservative onD.

2.

C

~F d~R is independent of the path to and fromany point on D.

3.

C

~F d~R = 0 for any piecewise, smooth,closed curve C in D.

2.5 Greens Theorem

As stated above, if the curve is closed and the

vector field conservative, then the line integral is

simply zero. Greens Theorem provides for the

method to calculate the line integral over a closed

curve with a vector field which is not conservative

and in doing so, relates the line integral to a

double integral.

Theorem 5 (Greens Theorem). If

~F (x, y) = P (x, y), Q(x, y)

is a smooth vector field on both the closed curve

C and the region R it encloses, then:C

~F ~dR =R

(Q

x Py

)dA

Notice that when the function is conservative,(Q

x Py

)= 0

and the line integral is zero, as stated in the pre-

ceding subsection.

Notice further that this method requires double

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integration, under which are procedures on chan-

ging the order of integration (e.g. from Dxdy to

Dxdy) or changing coordinate axes (e.g. from

rectangular to polar). Review on these.

Greens theorem has important application in sur-

veying. What this video (watch me!) to be

amazed. See how Greens theorem applies here.

2.6 Surface integrals

Definition (Surface Integral).

f(x, y, z) dS =

limP0

nj=1

mi=1

f(xij(u, v), yij(u, v), z

ij(u, v))iS

where iS is the infinitesimal area element, and

P is the interval between subsequent points.

This doesnt really give us anything substantial

calculation-wise.

2.6.1 Scalar fields

If we express the infinitesimal area element using a

double integral, we see that for a region, function,

and limits parametrized in u and v:

dS = ||~Ru ~Rv||dA

and we can write:

f dS

=

v1v0

u1u0

f(x(u, v), y(u, v), z(u, v))||~Ru ~Rv|| dudv

where:

: ~R(u, v) = x(u, v), y(u, v, ), z(u, v)

In rectangular coordinates:

dS =

1 + [gx]2 + [gy]2 dA

and we can write:

f dS

=

y1y0

x1x0

f(x, y, g(x, y))

1 + [gx]2 + [gy]2 dA

where:

: z = g(x, y)

2.6.2 Vector fields

Similar to the extension of Line Integrals to Vector

Fields, the Surface Integral will be calculated as

follows:

~F N dS

where N is an oriented unit normal to .

In taking the component of the vector field along

the normal, were taking the flux of the vector field

as it passes through the surface. That is why the

Surface Integral is also known as the Flux Integral.

It is important to note that the unit normal may

be defined in two ways for every surface - one way

is the opposite of the other.

In parametric form, the surface is given by~R(u, v) = x(u, v), y(u, v, ), z(u, v), such that:

N =~Ru ~Rv||~Ru ~Rv||

dS = ||~Ru ~Rv||dA

where N is defined as the positive orientation.

The negative orientation is the negative of N.

Plugging in the two equalities to the definition,

using positive orientation:

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~F N dS =

D

~F (~Ru ~Rv) dA

In rectangular coordinates, the surface is given

by z = g(x, y), such that:

N =gx,gy, 11 + [gx]2 + [gy]2

dS =

1 + [gx]2 + [gy]2 dA

where N is defined as the upward unit normal.

The downward unit normal is the negative of N.

Plugging in the two equalities to the definition,

using upward unit normal:

~F N dS =

D

~F gx,gy, 1dA

The Surface Integral is just a complex exercise of

plug-and-play. Be wary of the distinction between

scalar and vector field, parametric and rectangular

coordinates, positive and negative orientation, or

upward and downward unit normal.

References

[1] Wikipedia, the Free Encyclopedia.

[2] Paul Dawkins. Calculus III, 2012.

[3] Joe Erickson. Calculus 3 Notes.

Page 12 of 12

Triple IntegralsApplicationsVolumeMass and DensityCenter of Mass

Box-like regionsGeneral regionsCylindrical coordinatesSpherical coordinatesChange of variables* (optional)

Vector CalculusVector fieldsDivergence and CurlConservative fieldsLine IntegralsScalar FieldsVector FieldsFundamental Theorem for Line IntegralsPath independence

Green's TheoremSurface integralsScalar fieldsVector fields

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