This document provides an overview of key concepts in multivariable calculus covered in MAT 10B at Williams College in Fall 2011, including: 1) Definitions and computations of double integrals, including over rectangles and more general regions, and Fubini's theorem on changing the order of integration. 2) Triple integrals as a generalization of double integrals to functions of three variables over boxes in R3. 3) Techniques for changing the order of integration in iterated integrals when one order may be more convenient than another.

1. MAT 10B: Highlights
Fall 2011 - Williams
Latest Update: November 22, 2011
Section 5.1. Introduction: Areas and Volumes
Definition (Standard rectangle in R2
). Let a, b, c, d be real numbers such that a < b and
c < d. These numbers determine a closed rectangle
R = {(x, y) | a  x  b , c  y  d} ⇢ R2
.
We also denote this rectangle as
R = [a, b] ⇥ [c, d] .
Computational Tool: Iterated integrals
Let R = [a, b] ⇥ [c, d] be a rectangle. Let f : R ! [0, 1) be a continuous function. Then the
integral Z d
c
Z b
a
f(x, y) dx dy
is interpreted as the iterated integral
Z d
c
✓Z b
a
f(x, y) dx
◆
dy .
When computing the “inner integral”
Z b
a
f(x, y) dx ,
treat y as a constant.
Similarly, Z b
a
Z d
c
f(x, y) dy dx =
Z b
a
✓Z d
c
f(x, y) dy
◆
dx .
1

2. Example.
Z 2
0
Z 3
1
x2
+ y dy dx =
Z 2
0
✓Z 3
1
x2
+ y dy
◆
dx
=
Z 2
0
h
x2
y +
y2
2
iy=3
y=1
dx
=
Z 2
0
✓
3x2
+
9
2
◆ ✓
x2
+
1
2
◆
dx
=
Z 2
0
2x2
+ 4 dx
=
40
3
.
We compute the associated integral in which the order of integration is switched.
Z 3
1
Z 2
0
x2
+ y dx dy =
Z 3
1
✓Z 2
0
x2
+ y dx
◆
dy
=
Z 3
1
hx3
3
+ xy
ix=2
x=0
dy
=
Z 3
1
8
3
+ 2y dy
=
40
3
.
We happen to get the same answer of 40
3
. This is no coincidence since the integral represents
the same geometric quantity; also see Fubini’s Theorem in the next section.
Proposition. Let R = [a, b] ⇥ [c, d] be a rectangle.
Let f : R ! [0, 1) be a continuous function.
If V denotes the volume of the region between the graph of f and the xy–plane, then
V =
Z b
a
Z d
c
f(x, y) dy dx =
Z d
c
Z b
a
f(x, y) dx dy .
A simplified notation for the above iterated integral is
ZZ
R
f dA .
Example (Volume of a box). Suppose that a box has dimensions l, w, h. Draw a picture in
R3
. Then the volume of the box is
V =
Z l
0
Z w
0
h dy dx =
Z l
0
wh dx = lwh .
2

3. Example (Volume of a complicated region). Find the volume of the region bounded by the
graph of
f(x, y) = 9 x2
y2
over the rectangle R = [1, 2] ⇥ [0, 1].
The volume is
V =
Z 2
1
Z 1
0
9 x2
y2
dy dx
=
Z 2
1
✓
26
3
x2
◆
dx
=
19
3
.
Section 5.2. Double Integrals
Goals: Define the double integral ZZ
R
f dA
1. for a more general function f : R ! R (not necessarily nonnegative nor continuous)
2. over a more general region R ⇢ R (not necessarily a rectangle).
Part 1: Integral of a function over a rectangle
Definition (Partition of order n). Let R = [a, b] ⇥ [c, d] be a rectangle, and let n be a
positive integer. Consider partitions
a = x0 < x1 < x2 < · · · < xn = b for [a, b], and
c = y0 < y1 < y2 < · · · < yn = d for [c, d].
A partition of order n (of R) is the collection of subrectangles
Rij = [xi 1, xi] ⇥ [yj 1, yj]
for i, j = 1, . . . , n. Visualize the partition as a tiling of R.
Given Rij, the base is xi = xi xi 1 and the height is yj = yj yj 1.
The area of Rij is
Aij = xi · yj .
Furthermore, the partition is called regular if xi = b a
n
and yj = d c
n
for all i, j =
1, . . . , n.
3

4. Definition (Riemann Sum). Let R = [a, b] ⇥ [c, d] be a rectangle.
Choose a partition {Rij} for R.
For each i, j = 1, . . . , n, choose a sample point cij 2 Rij.
Let f : R ! R be a function.
The quantity
S =
nX
i,j=1
f(cij) Aij ,
is called a Riemann sum.
Now we give the definition of the double integral.
Definition (Double Integral). Let R = [a, b] ⇥ [c, d] be a rectangle.
Let f : R ! R be a function.
The double integral of f over R is defined to be
ZZ
R
f dA = lim
n!1
nX
i,j=1
f(cij) Aij
over all regular partitions Rij of R.
Remark. The above definition of
RR
R
f dA is equivalent to the one in the textbook. If the
integral
RR
R
f dA exists, we say that f is integrable on R.
Theorem. If R is a rectangle and f : R ! R is continuous, then
RR
R
f dA exists.
Theorem (Fubini’s Theorem (Simple Form)). Let R = [a, b] ⇥ [c, d] be a rectangle. If
f : R ! R is continuous, then
ZZ
R
f dA =
Z b
a
Z d
c
f(x, y) dy dx =
Z d
c
Z b
a
f(x, y) dx dy .
Example.
Z 2
0
Z 3
1
x2
y dy dx =
Z 2
0
✓Z 3
1
x2
y dy
◆
dx
=
Z 2
0
h
x2
y
y2
2
iy=3
y=1
dx
=
Z 2
0
✓
3x2 9
2
◆ ✓
x2 1
2
◆
dx
=
Z 2
0
2x2
4 dx
=
8
3
.
4

5. Proposition (Properties of the integral). Suppose that f and g are both integrable on the
standard rectangle R. Then
1. f + g is also integrable on R and
ZZ
R
f + g dA =
ZZ
R
f dA +
ZZ
R
g dA .
2. cf is also integrable on R for any c 2 R and
ZZ
R
cf dA = c
ZZ
R
f dA .
3. If f(x, y)  g(x, y) for all (x, y) 2 R, then
ZZ
R
f dA 
ZZ
R
g dA .
4. |f| is also integrable on R and
ZZ
R
f dA 
ZZ
R
|f| dA .
Examples (Properties of the integral). Let R ⇢ R2
be a standard rectangle.
1. ZZ
R
xy + y2
dA =
ZZ
R
xy dA +
ZZ
R
y2
dA .
2. ZZ
R
5xy dA = 5
ZZ
R
xy dA .
3. Suppose that R = [0, 2] ⇥ [0, 5]. Let’s prove the inequality
ZZ
R
x2
y dA  200 .
Since 0  x  2 and 0  y  5, we conclude that x2
y  22
· 5 = 20 for all (x, y) 2 R.
By Property 3, we have
ZZ
R
x2
y dA 
ZZ
R
20 dA = 20 · 2 · 5 = 200 .
4. Suppose that R = [ 1, 2] ⇥ [ 3, 5]. Let’s prove the inequality
ZZ
R
xy dA  240 .
Since 1  x  2 and 3  y  5, we conclude that |xy| = |x| · |y|  2 · 5 = 10 for all
(x, y) 2 R. By Properties 3 and 4, we have
ZZ
R
xy dA 
ZZ
R
|xy| dA 
ZZ
R
10 dA = 10 · 3 · 8 = 240 .
5

6. Part 2: Integral over a general region
Definition (Elementary Region). A subset D ⇢ R2
is called an elementary region if it
admits a description of one (or both) of the following types.
1. D = {(x, y) | (x)  y  (x) , a  x  b} for continuous functions (x) and (x), or
2. D = {(x, y) | (y)  x  (y) , c  y  d} for continuous functions (y) and (y).
Theorem. Let D ⇢ R2
be an elementary region. Suppose that f : D ! R is a continuous
function. Then f is integrable on D. Furthermore
• If D is of type 1, then
ZZ
D
f dA =
Z b
a
Z (x)
(x)
f(x, y) dy dx .
• If D is of type 2, then
ZZ
D
f dA =
Z d
c
Z (y)
(y)
f(x, y) dx dy .
Example. Sometimes a region can be represented as both types. Let D be the region
bounded by the curves y = x and y = x2
. As a type 1 region, express D as
D = {(x, y) | x2
 y  x , 0  x  1} .
Then
ZZ
D
xy dA =
Z 1
0
Z x
x2
xy dy dx
=
Z 1
0
✓
xy2
2
iy=x
y=x2
◆
dx
=
Z 1
0
✓
x3
2
x5
2
◆
dx
=
1
24
.
As a type 2 region, express D as
D = {(x, y) | y  x 
p
y , 0  y  1} .
6

7. Then
ZZ
D
xy dA =
Z 1
0
Z p
y
y
xy dx dy
=
Z 1
0
✓
x2
y
2
ix=
p
y
x=y
◆
dy
=
Z 1
0
✓
y2
2
y3
3
◆
dy
=
1
24
.
Double integrals can also be used to calculate areas of elementary regions.
Theorem (Area of an elementary region in R2
). Suppose that D is an elementary region in
R2
. Then
Area(D) =
ZZ
D
1 dA =
ZZ
D
dA .
Putting elementary regions together
The main idea is that we can integrate over a union of elementary regions as long as any
two of the elementary pieces do not overlap too much. More precisely, we have
Theorem. Let D1 and D2 be elementary regions in R2
. Let D = D1 [ D2. If D1 D2 has
zero area, (e.g. D1 D2 is a line segment), then
ZZ
D
f(x, y) dA =
ZZ
D1
f(x, y) dA +
ZZ
D2
f(x, y) dA .
See Example 6 from pages 302-303 in the textbook. We will later describe the annulus and
any integrals over it in a more appropriate way (using polar coordinates).
Section 5.3. Changing the Order of Integration
Main Idea. Given an iterated integral, it is sometimes helpful to change the order of inte-
gration when
• the given integral is impossible to evaluate by hand, or
• the given integral is more convenient to evaluate (by hand or computer) in the other
order.
7

8. Example. Consider the iterated integral
Z 1
0
Z 2
2y
ex2
dx dy .
There is no simple technique to calculate an antiderivative of ex2
, so we want to integrate
ex2
with respect to y. The given integral tells us that the (relevant) domain of f(x, y) = ex2
is
R = {(x, y) | 2y  x  2, 0  y  1} .
Draw the region R in the plane. It is easy to see that R is a triangle with vertices (0, 0), (2, 0)
and (2, 1). Changing perspective, re-interpret R as
R = {(x, y) | 0  y  x/2, 0  x  2} .
Therefore
Z 1
0
Z 2
2y
ex2
dx dy =
Z 2
0
Z x/2
0
ex2
dy dx
=
Z 2
0
✓
yex2
iy=x/2
y=0
◆
dx
=
Z 2
0
x
2
ex2
dx
= (e4
1)/4 .
Section 5.4. Triple Integrals
Main Idea. Generalize the double integral for a function of three variables.
Computational tool: iterated integrals
Triple iterated integrals just generalize double iterated integrals.
Example.
Z 1
0
Z 2
1
Z 3
1
3yz2
dx dy dz =
Z 1
0
Z 2
1
h
3yz2
x
ix=3
x=1
dy dz
=
Z 1
0
Z 2
1
6yz2
dy dz
=
Z 1
0
h
3y2
z2
iy=2
y= 1
dz
=
Z 1
0
9z2
dz
= 3 .
8

9. Definition (Standard box in R3
). Let a, b, c, d, p, q be real numbers such that a < b, c < d,
and p < q. These numbers determine a (closed) box
B = {(x, y, z) | a  x  b , c  y  d , p  z  q} ⇢ R3
.
We also denote this box as
B = [a, b] ⇥ [c, d] ⇥ [p, q] .
Goals: Define the triple integral ZZZ
B
f dV
1. for a function f : B ! R where B is a standard box.
2. over a more general region B ⇢ R3
(not necessarily a standard box).
Part 1: Integral of a function over a box
As with double integrals, we
• partition the box B = [a, b] ⇥ [c, d] ⇥ [p, q]
• compute f(c) for various sample points c = cijk 2 B
• form the Riemann sum
S =
nX
i,j,k=1
f(cijk) Vijk
• define the integral ZZZ
B
f dV = lim
n!1
S ,
the limit of Riemann sums.
Remark. See the textbook for the precise definition of the triple integral. If the integralRRR
B
f dV exists, we say that f is integrable on B.
Theorem. If B is a standard box and f : B ! R is continuous, then
RRR
B
f dV exists.
9

10. Theorem (Fubini’s Theorem (Simple Form)). Let B = [a, b] ⇥ [c, d] ⇥ [p, q] be a box. If
f : B ! R is continuous, then
ZZZ
B
f dV
=
Z b
a
Z d
c
Z q
p
f(x, y, z) dz dy dx =
Z b
a
Z q
p
Z d
c
f(x, y, z) dy dz dx
=
Z d
c
Z b
a
Z q
p
f(x, y, z) dz dx dy =
Z d
c
Z q
p
Z b
a
f(x, y, z) dx dz dy
=
Z q
p
Z b
a
Z d
c
f(x, y, z) dy dx dz =
Z q
p
Z d
c
Z b
a
f(x, y, z) dx dy dz .
Example. The integral Z 1
0
Z 2
1
Z 3
1
3yz2
dx dy dz
can also be calculated as
Z 3
1
Z 2
1
Z 1
0
3yz2
dz dy dx =
Z 3
1
Z 2
1
h
yz3
iz=1
z=0
dy dx
=
Z 3
1
Z 2
1
y dy dx
=
Z 3
1
3
2
dx
= 3 .
Proposition (Properties of the integral). Suppose that f and g are both integrable on the
closed box B. Then
1. f + g is also integrable on B and
ZZZ
B
f + g dV =
ZZZ
B
f dV +
ZZZ
B
g dV .
2. cf is also integrable on B for any c 2 R and
ZZZ
B
cf dV = c
ZZZ
B
f dV .
3. If f(x, y, z)  g(x, y, z) for all (x, y, z) 2 B, then
ZZZ
B
f dV 
ZZZ
B
g dV .
4. |f| is also integrable on B and
ZZZ
B
f dV 
ZZZ
B
|f| dV .
10

11. Part 2: Integral over a general region
Definition (Elementary Region). A subset W ⇢ R3
is called an elementary region if it
admits a description of at least one of the following types.
1. Type 1 Regions [The most fundamental of regions]
(a) W = {(x, y, z) | '(x, y)  z  (x, y) , (x)  y  (x) , a  x  b} , or
(b) W = {(x, y, z) | '(x, y)  z  (x, y) , ↵(y)  x  (y) , c  y  d} .
Simply put, the region W is the region between the graphs of z = '(x, y) and z =
(x, y) over an elementary region in the xy–plane.
2. Type 2 Regions
(a) W = {(x, y, z) | ↵(y, z)  x  (y, z) , (z)  y  (z) , p  z  q} , or
(b) W = {(x, y, z) | ↵(y, z)  x  (y, z) , '(y)  z  (y) , c  y  d} .
Simply put, the region W is the region between the graphs of x = ↵(y, z) and x =
(y, z) over an elementary region in the yz–plane.
3. Type 3 Regions
(a) W = {(x, y, z) | (x, z)  y  (x, z) , ↵(z)  x  (z) , p  z  q} , or
(b) W = {(x, y, z) | (x, y)  y  (x, y) , '(x)  z  (x) , a  x  b} .
Simply put, the region W is the region between the graphs of y = (x, z) and y = (x, z)
over an elementary region in the xz–plane.
Theorem. Let W be an elementary region in R3
and let f be a function on W.
1. If W is a type 1 region, then
ZZZ
W
f dV =
Z b
a
Z (x)
(x)
Z (x,y)
'(x,y)
f(x, y, z) dz dy dx ,
or ZZZ
W
f dV =
Z d
c
Z (y)
↵(y)
Z (x,y)
'(x,y)
f(x, y, z) dz dx dy .
2. If W is a type 2 region, then
ZZZ
W
f dV =
Z q
p
Z (z)
(z)
Z (y,z)
↵(y,z)
f(x, y, z) dx dy dz ,
or ZZZ
W
f dV =
Z d
c
Z (y)
'(y)
Z (y,z)
↵(y,z)
f(x, y, z) dx dz dy .
11

12. 3. If W is a type 3 region, then
ZZZ
W
f dV =
Z q
p
Z (z)
↵(z)
Z (x,z)
(x,z)
f(x, y, z) dy dx dz ,
or ZZZ
W
f dV =
Z b
a
Z (x)
'(x)
Z (x,z)
(x,z)
f(x, y, z) dy dz dx .
Triple integrals can also be used to compute volume of elementary regions in R3
.
Theorem (Volume of an elementary region in R3
). Suppose that W is an elementary region
in R3
. Then
Volume(W) =
ZZZ
W
1 dV =
ZZZ
W
dV .
Let’s take a look at some examples of each type.
Examples of type 1 regions
Example. Let W ⇢ R3
be the solid containing (0, 0, 0) in the first octant bounded by
z = 2 x2
y2
, z = 0 and x + y = 1. Compute the integral
ZZZ
W
x2
y dV .
Solution: The shadow of W in the xy–plane is the triangle in the first quadrant bounded by
the line x + y = 1. Hence the shadow can be expressed as
Wshadow = {(x, y) | 0  y  1 x , 0  x  1} .
It follows that
W = {(x, y, z) | 0  z  2 x2
y2
, 0  y  1 x , 0  x  1} .
Therefore, the iterated integral that computes
RRR
W
x2
y dV is
Z 1
0
Z 1 x
0
Z 2 x2 y2
0
x2
y dz dy dx .
12

13. Now we evaluate the integral:
Z 1
0
Z 1 x
0
Z 2 x2 y2
0
x2
y dz dy dx =
Z 1
0
Z 1 x
0
h
x2
yz
iz=2 x2 y2
z=0
dy dx
=
Z 1
0
Z 1 x
0
x2
y(2 x2
y2
) dy dx
=
Z 1
0
h
x2
y2 x4
y2
2
x2
y4
4
iy=1 x
y=0
dx
=
Z 1
0
✓
x2
(1 x)2 x4
(1 x)2
2
x2
(1 x)4
4
◆
dx
=
11
420
.
Since the shadow of W can be expressed as
Wshadow = {(x, y) | 0  x  1 y , 0  y  1} ,
the integral can also be set-up as
Z 1
0
Z 1 y
0
Z 2 x2 y2
0
x2
y dz dx dy .
Example. Let W ⇢ R3
be the solid bounded by z = 3x2
+ 2y2
and z = 16 x2
2y2
.
Compute the integrals
(a) ZZZ
W
xz dV
(b) ZZZ
W
dV =
ZZZ
W
1 dV
Solution: Find the shadow of W in the xy–plane by solving for the intersection of the surfaces:
z = 3x2
+ 2y2
meets z = 16 x2
2y2
=) 3x2
+ 2y2
= 16 x2
2y2
=) 4x2
+ 4y2
= 16
=) x2
+ y2
= 4 .
So the shadow of W in the xy–plane is
Wshadow = {(x, y) | x2
+ y2
 4} = {(x, y) |
p
4 x2  y 
p
4 x2 , 2  x  2} .
13

14. (a) An iterated integral that computes
RRR
W
xz dV is
Z 2
2
Z p
4 x2
p
4 x2
Z 16 x2 2y2
3x2+2y2
xz dz dy dx .
An easy way to compute this integral is to use the symmetry of the solid W across the
plane x = 0. The function f(x, y, z) = xz satisfies f( x, y, z) = f( x, y, z). In a
sense, the function f is odd with respect to the yz–plane. Therefore, the integral has
the value 0.
(b) An iterated integral that computes
RRR
W
dV is
Z 2
2
Z p
4 x2
p
4 x2
Z 16 x2 2y2
3x2+2y2
dz dy dx .
Of course the integral
RRR
W
dV represents the volume of W, so we expect the integral
to be positive.
Z 2
2
Z p
4 x2
p
4 x2
Z 16 x2 2y2
3x2+2y2
dz dy dx =
Z 2
2
Z p
4 x2
p
4 x2
⇥
z
⇤z=16 x2 2y2
z=3x2+2y2 dy dx
=
Z 2
2
Z p
4 x2
p
4 x2
(16 x2
2y2
) (3x2
+ 2y2
) dy dx
=
Z 2
2
Z p
4 x2
p
4 x2
16 4(x2
+ y2
) dy dx .
To finish the calculation, we should use polar coordinates as in Section 5.5 (to be
discussed). It turns out that this transformation will turn our integral into a simpler
double integral:
Z 2⇡
0
Z 2
0
(16 4r2
)r dr d✓ = 32⇡ .
Examples of type 2 and type 3 regions
Example. Let W be the solid in the first octant bounded by the plane x + 2y + 3z = 6. Let
f(x, y, z) be a continuous function. Set-up an iterated integral that computes
ZZZ
W
f dV
by considering W as
14

15. (a) a type 1 region
(b) a type 2 region
(c) a type 3 region
Solution: First, we should draw the plane x + 2y + 3z = 6 in the first octant. This surface
is the triangle with vertices (6, 0, 0), (0, 3, 0) and (0, 0, 2).
(a) As a type 1 region, W is the region between z = 0 and z = 6 x 2y
3
; the shadow of W
in the xy–plane is the triangle with vertices (0, 0), (6, 0) and (0, 3). It follows that
W = {(x, y, z) | 0  z 
6 x 2y
3
, 0  y 
6 x
2
, 0  x  6} .
Therefore ZZZ
W
f dV =
Z 6
0
Z 6 x
2
0
Z 6 x 2y
3
0
f(x, y, z) dz dy dx .
(b) As a type 2 region, W is the region between x = 0 and x = 6 2y 3z; the shadow of
W in the yz–plane is the triangle with vertices (0, 0), (3, 0) and (0, 2). It follows that
W = {(x, y, z) | 0  x  6 2y 3z , 0  y 
6 3z
2
, 0  z  2} .
Therefore ZZZ
W
f dV =
Z 2
0
Z 6 3z
2
0
Z 6 2y 3z
0
f(x, y, z) dx dy dz .
(c) As a type 3 region, W is the region between y = 0 and y = 6 x 3z
2
; the shadow of W
in the xz–plane is the triangle with vertices (0, 0), (6, 0) and (0, 2). It follows that
W = {(x, y, z) | 0  y 
6 x 3z
2
, 0  x  6 3z , 0  z  2} .
Therefore ZZZ
W
f dV =
Z 2
0
Z 6 3z
0
Z 6 x 3z
2
0
f(x, y, z) dy dx dz .
Putting elementary regions together
The main idea is that we can integrate over a union of elementary regions as long as any
two of the elementary pieces do not overlap too much. More precisely, we have
Theorem. Let W1 and W2 be elementary regions in R3
. Let W = W1 [ W2. If W1 W2 has
zero volume, (e.g. W1 W2 is a surface), then
ZZ
W
f(x, y, z) dV =
ZZ
W1
f(x, y, z) dV +
ZZ
W2
f(x, y, z) dV .
15

16. Section 5.5. Change of Variables: Double Integrals
Some basic linear algebra
Definition (Matrix). A 2 by 2 matrix is a square array of real numbers a, b, c, d 2 R

a b
c d
.
Definition (Determinant). The determinant of a matrix is a real number given by the
formula
det

a b
c d
= ad bc .
Definition (Linear transformation). A linear transformation is a function of the form
T : R2
uv ! R2
xy , T(u, v) = (x, y)
given by the equations
x = au + bv
y = cu + dv
where a, b, c, d 2 R. In this case, T admits a matrix form

x
y
=

a b
c d

u
v
.
It is convenient to write
T ⌘

a b
c d
and to define the determinant of T to be the determinant of its matrix.
Remark. The determinant of T tells us how T distorts (or rescales) area. See Proposition
5.1 in the textbook.
Definition (Invertible Linear Transformation). A linear transformation T is invertible
(a.k.a. nonsingular) if ↵ = det T 6= 0. The inverse of T is denoted by T 1
, and the matrix
of T 1
is given by the formula
T 1
⌘

d/↵ b/↵
c/↵ a/↵
.
In this case, the determinants of T and T 1
are reciprocals of each other:
det(T 1
) = 1 det(T) .
16

17. Linear change of variables for integrals
Theorem (Linear change of variables). Suppose that T : R2
uv ! R2
xy is an invertible
linear transformation. Suppose that D⇤
⇢ R2
uv and D ⇢ R2
xy are elementary regions
such that T(D⇤
) = D. If f : D ! R is continuous, then
ZZ
D
f(x, y) dx dy =
ZZ
D⇤
f(x(u, v), y(u, v)) | det T| du dv .
Example (Problem 9). Evaluate the integral
Z 2
0
Z (x/2)+1
x/2
x5
(2y x)e(2y x)2
dy dx
by making the substitution u = x and v = 2y x.
Solution:
The substitution u = x and v = 2y x gives a linear transformation
T 1
: R2
xy ! R2
uv
whose matrix is 
1 0
1 2
.
So det T 1
= 2. Therefore, the linear transformation
T : R2
uv ! R2
xy
has det T = 1/2.
A quick sketch of the region D reveals that D is the parallelogram spanned by the vectors
(2, 1) and (0, 1). We would like to find D⇤
so that T(D⇤
) = D.
To do this, we notice that
(x, y) = (2, 1) =) (u, v) = (2, 0) and (x, y) = (0, 1) =) (u, v) = (0, 2) .
It follows that D⇤
is the parallelogram spanned by the vectors (2, 0) and (0, 2). Hence
D⇤
= {(u, v) | 0  u  2 , 0  v  2} = [0, 2] ⇥ [0, 2] .
17

18. Therefore
ZZ
D⇤
f(x(u, v), y(u, v)) | det T| du dv =
Z 2
0
Z 2
0
u5
vev2 1
2
du dv
=
1
2
✓Z 2
0
u5
du
◆ ✓Z 2
0
vev2
dv
◆
=
1
2
✓
64
6
◆ ✓
e4
1
2
◆
=
8
3
(e4
1) .
Example (Problem 10). Determine
ZZ
D
r
x + y
x 2y
dA
where D is the region enclosed by the lines y = x/2, y = 0, and x + y = 1.
Solution: Let’s try the substitution u = x + y and v = x 2y. This gives a linear transfor-
mation
T 1
: R2
xy ! R2
uv
whose matrix is 
1 1
1 2
.
So det T 1
= 2 1 = 3 . Therefore, the linear transformation
T : R2
uv ! R2
xy
has det T = 1/3.
A quick sketch of D reveals that it is the triangle with verticies (0, 0), (1, 0) and (2/3, 1/3).
We would like to find D⇤
so that T(D⇤
) = D.
To do this, we notice that
(x, y) = (1, 0) =) (u, v) = (1, 1) and (x, y) = (2/3, 1/3) =) (u, v) = (1, 0) .
So D⇤
is the triangle with verticies (0, 0), (1, 0) and (1, 1):
D⇤
= {(u, v) | 0  u  1 , 0  v  u}
18

19. It follows that
ZZ
D
r
x + y
x 2y
dA =
ZZ
D⇤
p
u
p
v
|det T| du dv
=
Z 1
0
Z u
0
p
u
p
v
(1/3) dv du
= 1/3 .
General change of variables
Recall a couple of definitions about functions. The definitions below generalize to subsets of
Rn
and Rm
.
Definition (C1
function). A function T : Rn
! Rm
is of class C1
if its first-order partial
derivatives exist and are continuous.
Definitions (One-to-one function, Onto function). A function T : Rn
! Rm
is called
• one-to-one if
T(a) = T(b) =) a = b
for every a, b 2 Rn
;
• onto if T(Rn
) = Rm
.
Definition (Jacobian matrix and Jacobian determinant). Let T : R2
uv ! R2
xy be a C1
function. The Jacobian matrix of T is the matrix
DT =
" @x
@u
@x
@v
@y
@u
@y
@v
#
.
The determinant of this matrix is denoted by
@(x, y)
@(u, v)
and is called the Jacobian determinant; in our textbook, this determinant is simply called
“the Jacobian”.
19

20. Theorem (Change of Variables (Double Integrals)). Let D⇤
⇢ R2
uv and D ⇢ R2
xy
be elementary regions. Suppose that
T : R2
uv ! R2
xy
is a C1
function such that T(D⇤
) = D and T is one-to-one on D⇤
. If f : D ! R is
integrable, then
ZZ
D
f(x, y) dx dy =
ZZ
D⇤
f(x(u, v), y(u, v))
@(x, y)
@(u, v)
du dv .
Remark. In some computations, it will be easier to set-up the Jacobian of T 1
. It is a fact
that this Jacobian is the reciprocal of the Jacobian of T:
det(DT 1
) = 1/ det(DT) .
In coordinate notation for T : R2
uv ! R2
xy , this is expressed as
@(u, v)
@(x, y)
= 1
. ✓
@(x, y)
@(u, v)
◆
.
Example (Problem 12). Evaluate
ZZ
D
(2x + y 3)2
(2y x + 6)2
dx dy ,
where D is the square with verticies (0, 0), (2, 1), (3, 1), and (1, 2).
Solution: For this problem, we have to come up with our own substitution formulas for u
and v. A good substitution would be
u = 2x + y 3
v = 2y x + 6 .
This gives the transformation T 1
: R2
xy ! R2
uv. The Jacobian matrix of T 1
is
" @u
@x
@u
@y
@v
@x
@v
@y
#
=

2 1
1 2
.
The Jacobian of T 1
is
@(u, v)
@(x, y)
= 5 .
So the Jacobian of T is the reciprocal:
@(x, y)
@(u, v)
= 1
. ✓
@(u, v)
@(x, y)
◆
= 1/5 .
20

21. Since T 1
is an a ne (linear + translation) transformation, it takes straight-lines to straight-
lines. It follows that D⇤
= T 1
(D) is a parallelogram with vertices T 1
(0, 0) = ( 3, 6),
T 1
(2, 1) = (2, 6), T 1
(3, 1) = (2, 1), and T 1
(1, 2) = ( 3, 1). Therefore
D⇤
= [ 3, 2] ⇥ [1, 6] .
Therefore
ZZ
D
(2x + y 3)2
(2y x + 6)2
dx dy =
ZZ
D⇤
u2
v2
·
1
5
du dv =
Z 2
3
Z 6
1
u2
5v2
dv du =
35
18
.
Example (Simple rescaling). Consider the linear transformation T : R2
uv ! R2
xy defined by
(x, y) = T(u, v) = (2u, 3v) .
It is easy to see that T transforms the unit square
[0, 1] ⇥ [0, 1] ⇢ R2
uv
into the rectangle
[0, 2] ⇥ [0, 3] ⇢ R2
xy .
Also, T transforms the unit disk
D⇤
= {(u, v) | u2
+ v2
 1} ⇢ R2
uv
into the elliptical region
D = {(x, y) | (x/2)2
+ (y/3)2
 1} ⇢ R2
xy .
The determinant of T is
det T = 6 .
This linear transformation allows us to compute
Area(D) =
ZZ
D
1 dx dy =
ZZ
D⇤
|det T| du dv = |det T|
✓ZZ
D⇤
1 du dv
◆
= 6·Area(D⇤
) = 6⇡ .
Review of polar coordinates (See section 1.7)
Polar coordinates are also called angular coordinates. These are the best coordinates for
describing geometric objects that possesses circular symmetry. The coordinates are (r, ✓);
here r represents the radial coordinate and ✓ represents the angular (or circular) coordinate.
Example (Disk). Consider the closed disk centered at the origin with radius 2:
D = {(x, y) | x2
+ y2
 4} .
We can describe this region in polar coordinates as
D⇤
= {(r, ✓) | 0  r  2 , 0  ✓  2⇡} .
21

22. Example (Annulus). Consider the closed annulus centered at the orign with inner radius
1 and outer radius 2:
A = {(x, y) | 1  x2
+ y2
 4} .
We can describe this region in polar coordinates as simply
A⇤
= {(r, ✓) | 1  r  2 , 0  ✓  2⇡} .
Conversions between Rectangular and Polar coordinates
For more complicated examples besides the disk and annulus, we need to a systematic way
to go between the two coordinate systems.
The basic (explicit) conversions from Rectangular to Polar coordinates are
x = r cos ✓ , y = r sin ✓
The basic (implicit) conversions from Polar to Rectangular coordinates are
r2
= x2
+ y2
, tan ✓ = y/x
The equations above can be solved for explicit values of r and ✓; this can be tricky for ✓, so
it is recommended that you use a diagram of the unit circle for help.
Example (Another Disk). Consider the disk D bounded by the circle
(x 1)2
+ y2
= 1 .
You should be able to easily plot this curve in R2
. This disk can be described in rectangular
coordinates as
D = {(x, y) | 0  x  2 ,
p
1 (x 1)2  y 
p
1 (x 1)2} .
Use the conversion formulas to rewrite the equation of the circle
(x 1)2
+ y2
= 1
in terms of r and ✓:
(x 1)2
+ y2
= 1 =) (r cos ✓ 1)2
+ (r sin ✓)2
= 1
=) (r2
cos2
✓ 2r cos ✓ + 1) + (r2
sin2
✓) = 1
=) (r2
cos2
✓ + r2
sin2
✓) 2r cos ✓ = 0
=) r2
= 2r cos ✓
=) r = 2 cos ✓ .
The picture of D, with polar coordinates in mind, allows us to describe D as
D⇤
= {(r, ✓) | 0  r  2 cos ✓ , ⇡/2  ✓  ⇡/2} .
22

23. Integrals in polar coordinates
Consider the polar coordinate transformation
T : R2
r✓ ! R2
xy
defined by the formula
(x, y) = T(r, ✓) = (r cos ✓, r sin ✓) .
The Jacobian matrix for T is
DT =
" @x
@r
@x
@✓
@y
@r
@y
@✓
#
=

cos ✓ r sin ✓
sin ✓ r cos ✓
.
So the Jacobian determinant is
@(x, y)
@(r, ✓)
= r cos2
✓ + r sin2
✓ = r .
When converting to polar coordinates, D⇤
⇢ R2
r✓ is typically an elementary region, but D =
T(D⇤
) ⇢ R2
xy need not be (e.g. D is an annulus). Also, the polar coordinate transformation
is not quite one-to-one. Nevertheless, the formula for change of variables applies:
Change to polar coordinates: The rule for transforming a double integral in
rectangular coordinates over D to a double integral in polar coordinates over D⇤
is
ZZ
D
f(x, y) dx dy =
ZZ
D⇤
f(r cos ✓, r sin ✓) r dr d✓ .
Example. Previously, we encountered the integral
Z 2
2
Z p
4 x2
p
4 x2
16 4(x2
+ y2
) dy dx .
The region D over which we are integrating is the disk centered at the origin with radius 2.
In polar coordinates, this disk is simply expressed as
D⇤
= {(r, ✓) | 0  r  2 , 0  ✓  2⇡} .
We convert to polar coorinates: x = r cos ✓ , y = r sin ✓, r2
= x2
+ y2
. The integral simplies
to
23

24. Z 2
2
Z p
4 x2
p
4 x2
16 4(x2
+ y2
) dy dx =
Z 2⇡
0
Z 2
0
(16 4r2
) r dr d✓
=
Z 2⇡
0
Z 2
0
(16r 4r3
) dr d✓
=
Z 2⇡
0
h
8r2
r4
ir=2
r=0
d✓
=
Z 2⇡
0
16 d✓
= 32⇡ .
Example (Another Disk (revisited)). Consider (again) the disk D bounded by the circle
(x 1)2
+ y2
= 1 .
Recall that, in polar coordinates, D corresponds to
D⇤
= {(r, ✓) | 0  r  2 cos ✓ , ⇡/2  ✓  ⇡/2} .
So given a continuous function f(x, y) on D, we have
ZZ
D
f(x, y) dx dy =
ZZ
D⇤
f(r cos ✓, r sin ✓) r dr d✓ =
Z ⇡/2
⇡/2
Z 2 cos ✓
0
f(r cos ✓, r sin ✓) r dr d✓ .
For example, the area of D can be calculated as
ZZ
D
dx dy =
Z ⇡/2
⇡/2
Z 2 cos ✓
0
r dr d✓
=
Z ⇡/2
⇡/2
hr2
2
ir=2 cos ✓
r=0
d✓
=
Z ⇡/2
⇡/2
(2 cos ✓)2
2
d✓
=
Z ⇡/2
⇡/2
2 cos2
✓ d✓
=
Z ⇡/2
⇡/2
1 + cos 2✓ d✓
=
h
✓ +
sin 2✓
2
i⇡/2
⇡/2
= ⇡ .
24

25. Example (Polar coordinates with a twist). Consider the curve
4x2
+ y2
8x + 4y 8 = 0
in R2
. This curve is an ellipse:
4x2
+ y2
8x + 4y 8 = 0 =) 4(x2
2x) + (y2
+ 4y) = 8
=) 4(x2
2x + 1) + (y2
+ 4y + 4) = 8 + 4 + 4
=) 4(x 1)2
+ (y + 2)2
= 16
=)
✓
x 1
2
◆2
+
✓
y + 2
4
◆2
= 1 .
Suppose that we want to integrate a continuous function f(x, y) over the region D bounded
by this ellipse:
D =
(
(x, y)
✓
x 1
2
◆2
+
✓
y + 2
4
◆2
 1
)
.
We could use an a ne (linear + translation) change of variables
u =
x 1
2
, v =
y + 2
4
and then change to polar coordinates, or we can do everything at once.
Consider the substitution
r cos ✓ =
x 1
2
, r sin ✓ =
y + 2
4
.
This substitution corresponds to the transformation
T : R2
r✓ ! R2
xy
defined by the equations
x = 2r cos ✓ + 1 , y = 4r sin ✓ 2 .
Just like in ordinary polar coordinates, we have D⇤
= [0, 1] ⇥ [0, 2⇡] ⇢ R2
r✓.
The Jacobian matrix for T is
DT =
" @x
@r
@x
@✓
@y
@r
@y
@✓
#
=

2 cos ✓ 2r sin ✓
4 sin ✓ 4r cos ✓
.
So the Jacobian determinant is
@(x, y)
@(r, ✓)
= det DT = 8r cos2
✓ + 8r sin2
✓ = 8r .
For instance, we can now directly compute the area of D :
Area(D) =
ZZ
D
1 dx dy =
ZZ
D⇤
8r dr d✓ =
Z 2⇡
0
Z 1
0
8r dr d✓ = 8⇡ .
25

26. Section 5.5. Change of Variables: Triple Integrals
In our class, we will focus only on changes to cylindrical and spherical coordinates for triple
integrals.
Review of cylindrical coordinates (See section 1.7)
See Figures 1.89 to 1.95 in the textbook.
Cylindrical coordinates (r, ✓, z) arise from simply adding the coordinate z to polar coordi-
nates.
Conversions between Rectangular and Cylindrical Coordinates
The basic conversions between Rectangular and Cylindrical coordinates are
x = r cos ✓ , y = r sin ✓ , z = z
and
r2
= x2
+ y2
, tan ✓ = y/x , z = z
Integrals in cylindrical coordinates
Consider the cylindrical coordinate transformation
T : R3
r✓z ! R3
xyz
defined by the formula
(x, y, z) = T(r, ✓, z) = (r cos ✓, r sin ✓, z) .
Change to cylindrical coordinates: The rule for transforming a triple integral
in rectangular coordinates over W to a triple integral in cylindrical coordinates over
W⇤
is
ZZZ
W
f(x, y, z) dx dy dz =
ZZZ
W⇤
f(r cos ✓, r sin ✓, z) r dr d✓ dz .
26

27. Example (Region between a Plane and a Paraboloid). Let W be the solid in R3
bounded
by the plane z = 9 and the paraboloid z = x2
+ y2
. Use cylindrical coordinates to
(a) compute the volume of W.
(b) integrate the function
f(x, y, z) =
p
x2 + y2 x2
y2
over W.
Solution:
Notice that the paraboloid meets the plane in the circle
{(x, y, 9) | x2
+ y2
= 9} .
The shadow of W in the xy–plane is given as
Wshadow = {(x, y) | x2
+ y2
 9}
which, in polar coordinates, is given as
W⇤
shadow = {(r, ✓) | 0  r  3 , 0  ✓  2⇡} .
Therefore, W can be described, in cylindrical coordinates, as
W⇤
c = {(r, ✓, z) | 0  r  3 , 0  ✓  2⇡ , r2
 z  9} .
To compute integrals, we make the substitutions
x = r cos ✓ , y = r sin ✓ , z = z .
(a) The volume of W is the integral of 1 over W is
ZZZ
W
1 dx dy dz =
ZZZ
W⇤
c
r dr d✓ dz
=
Z 2⇡
0
Z 3
0
Z 9
r2
r dz dr d✓
=
Z 2⇡
0
Z 3
0
h
rz
iz=9
z=r2
dr d✓
=
Z 2⇡
0
Z 3
0
9r r3
dr d✓
= 243⇡ .
27

28. (b) The integral of f(x, y, z) =
p
x2 + y2 x2
y2
over W is
ZZZ
W
⇣p
x2 + y2 x2
y2
⌘
dx dy dz =
ZZZ
W⇤
c
(r r2
)r dr d✓ dz
=
Z 2⇡
0
Z 3
0
Z 9
r2
(r r2
)r dz dr d✓
=
Z 2⇡
0
Z 3
0
Z 9
r2
(r2
r3
) dz dr d✓
=
Z 2⇡
0
Z 3
0
h
(r2
r3
)z
iz=9
z=r2
dr d✓
=
Z 2⇡
0
Z 3
0
(r2
r3
)(9 r2
) dr d✓
=
Z 2⇡
0
Z 3
0
(9r2
9r3
r4
+ r5
) dr d✓
=
29889⇡
140
.
Review of spherical coordinates (See section 1.7)
See Figures 1.96 to 1.102 in the textbook.
Spherical coordinates (⇢, ', ✓) are useful for describing solids or surfaces in R3
which have
spherical symmetry. The (usually nonnegative) coordinate ⇢ is the radial coordinate, and
there are two angular coordinates ' (or ) and the same ✓ coordinate from polar or cylindrical
coordinates. The new angular coordinate ' measures the angle between a line through the
origin and the z–axis. See Figure 1.97.
Conversions between Spherical coordinates and Cylindrical coordinates
The basic conversions between Spherical and Cylindrical coordinates are
r = ⇢ sin ' , ✓ = ✓ , z = ⇢ cos '
and
⇢2
= r2
+ z2
, tan ' = r/z , ✓ = ✓
Conversions between Spherical coordinates and Rectangular coordinates
The basic conversions between Spherical and Rectangular coordinates are
x = ⇢ sin ' cos ✓ , y = ⇢ sin ' sin ✓ , z = ⇢ cos '
and
⇢2
= x2
+ y2
+ z2
, tan ' =
p
x2 + y2/z , tan ✓ = y/x
28

29. Integrals in spherical coordinates
Consider the spherical coordinate transformation
T : R3
⇢'✓ ! R3
xyz
defined by the formula
(x, y, z) = T(⇢, ', ✓) = (⇢ sin ' cos ✓, ⇢ sin ' sin ✓, ⇢ cos ') .
Change to spherical coordinates: The rule for transforming a triple integral
in rectangular coordinates over W to a triple integral in spherical coordinates over
W⇤
is
ZZZ
W
f(x, y, z) dx dy dz =
ZZZ
W⇤
f(⇢ sin ' cos ✓, ⇢ sin ' sin ✓, ⇢ cos ') ⇢2
sin ' d⇢ d' d✓ .
Example (Ice Cream Cone). Let W ⇢ R3
be the solid bounded by the cone z =
p
x2 + y2
and the sphere x2
+ y2
+ z2
= 9. Use spherical coordinates to
(a) compute the volume of W.
(b) set-up the interated integral which computes the integral of f(x, y, z) = 1p
100 x y2+z3
over W.
Solution:
The solid W looks like an “ice cream cone”.
In spherical coordinates, the equation for cone transforms into
⇢ cos ' =
p
(⇢ sin ' cos ✓)2 + (⇢ sin ' sin ✓)2 =) ⇢ cos ' = ⇢ sin '
=) cos ' = sin '
=) ' = ⇡/4 .
The sphere is naturally described by the equation ⇢ = 3.
Therefore W has the following description in spherical coordinates:
W⇤
s = {(⇢, ', ✓) | 0  ⇢  3 , 0  '  ⇡/4 , 0  ✓  2⇡} .
(a) It follows that the volume of W is
29

30. ZZZ
W
dx dy dz =
ZZZ
W⇤
s
⇢2
sin ' d⇢ d' d✓
=
Z 2⇡
0
Z ⇡/4
0
Z 3
0
⇢2
sin ' d⇢ d' d✓
=
Z 2⇡
0
Z ⇡/4
0
h⇢3
3
sin '
i⇢=3
⇢=0
d' d✓
=
Z 2⇡
0
Z ⇡/4
0
9 sin ' d' d✓
=
Z 2⇡
0
Z ⇡/4
0
h
9 cos '
i'=⇡/4
'=0
d✓
=
Z 2⇡
0
✓
9
9
p
2
◆
d✓
= 2⇡
✓
9
9
p
2
◆
.
(b) The integral of f(x, y, z) = 1p
100 x y2+z3
over W is
ZZZ
W
1
p
100 x y2 + z3
dx dy dz
=
ZZZ
W⇤
s
1
p
100 (⇢ sin ' cos ✓) (⇢ sin ' sin ✓)2 + (⇢ cos ')3
⇢2
sin ' d⇢ d' d✓
=
Z 2⇡
0
Z ⇡/4
0
Z 3
0
1
p
100 (⇢ sin ' cos ✓) (⇢ sin ' sin ✓)2 + (⇢ cos ')3
⇢2
sin ' d⇢ d' d✓ .
Example (Ice Cream Cone (revisited)). Let W ⇢ R3
be the solid bounded by the cone
z =
p
x2 + y2 and the sphere x2
+ y2
+ z2
= 9. Use cylindrical coordinates to
(a) compute the volume of W.
(b) set-up the interated integral which computes the integral of f(x, y, z) = 1p
100 x y2+z3
over W.
Solution:
To give a description of W in cylindrical coordinates, first observe that the cone’s equation
is z = r, and the upper hemisphere’s equation is
r2
+ z2
= 9 =) z =
p
9 r2 .
30

31. So the intersection of the cone and the sphere satisfies the equation
r =
p
9 r2 =) r =
3
p
2
.
Therefore W has the following description in cylindrical coordinates:
W⇤
c = {(r, ✓, z) | 0  r 
3
p
2
, 0  ✓  2⇡ , r  z 
p
9 r2} .
(a) The volume of W can be calculated as
ZZZ
W
dx dy dz =
ZZZ
W⇤
c
r dr d✓ dz
=
Z 2⇡
0
Z 3p
2
0
Z p
9 r2
r
r dz dr d✓
=
Z 2⇡
0
Z 3p
2
0
h
rz
iz=
p
9 r2
z=r
dr d✓
=
Z 2⇡
0
Z 3p
2
0
⇣
r
p
9 r2 r
⌘
dr d✓
=
Z 2⇡
0
✓
9
9
p
2
◆
d✓
= 2⇡
✓
9
9
p
2
◆
.
(b) The integral of f(x, y, z) = 1p
100 x y2+z3
over W is
ZZZ
W
1
p
100 x y2 + z3
dx dy dz
=
ZZZ
W⇤
c
1
p
100 (r cos ✓) (r sin ✓)2 + z3
r dr d✓ dz
=
Z 2⇡
0
Z 3p
2
0
Z p
9 r2
r
1
p
100 (r cos ✓) (r sin ✓)2 + z3
r dz dr d✓ .
Example (Region between a Plane and a Sphere). Let W be the region between the plane
z = 2 and the (upper) hemisphere
x2
+ y2
+ z2
= 16 =) ⇢ = 4 .
Compute the volume of W.
31

32. Solution:
The plane meets the sphere when
z = ⇢ cos ' =) 2 = 4 cos ' =) 1/2 = cos ' =) ' = arccos(1/2) = ⇡/3 .
So far, we have shown that the bounds for ' and ✓ in W must be
0  '  ⇡/3 and 0  ✓  2⇡ .
The bounds for ⇢ will require a little more work. Consider the equation z = ⇢ cos '. On the
plane z = 2, the equation simplifies to ⇢ = 2 sec '. So the bounds for ⇢ (depending on ')
are
2 sec '  ⇢  4 .
Therefore W has the following description in spherical coordinates:
W⇤
= {(⇢, ', ✓) | 2 sec '  ⇢  4 , 0  '  ⇡/3 , 0  ✓  2⇡} .
The volume of W is
ZZZ
W
dx dy dz =
ZZ
W⇤
⇢2
sin ' d⇢ d' d✓
=
Z 2⇡
0
Z ⇡/3
0
Z 4
2 sec '
⇢2
sin ' d⇢ d' d✓
=
Z 2⇡
0
Z ⇡/3
0
h⇢3
3
sin '
i⇢=4
⇢=2 sec '
d' d✓
=
Z 2⇡
0
Z ⇡/3
0
✓
64
3
sin '
8 sec3
'
3
sin '
◆
d' d✓
=
Z 2⇡
0
Z ⇡/3
0
✓
64
3
sin '
8
3 cos3 '
sin '
◆
d' d✓
=
Z 2⇡
0
h 64
3
cos '
4
3 cos2 '
i'=⇡/3
'=0
d✓
=
Z 2⇡
0
✓
32
3
16
3
+
64
3
+
4
3
◆
d✓
=
Z 2⇡
0
20
3
d✓
=
40⇡
3
.
32

33. Section 6.1. Scaler and vector line integrals
It will be useful for you to review parts of Chapter 3.
All paths (curves) under consideration will be piecewise C1
, unless otherwise specified. Also,
all scaler functions f : Rn
! R and vector fields F : Rn
! Rn
will be assumed to be
continuous unless otherwise specified.
The following is a useful way to parametrize a straight line connecting two points P and Q.
Straight-line parametrizations: A convenient way to get a parametrization of
a line segment in Rn
beginning at a point P and ending at a point Q is to use the
formula
x(t) = (1 t)P + tQ .
This gives a straight-line path
x : [0, 1] ! Rn
with x(0) = P and x(1) = Q.
Scaler line integrals
Let x : [a, b] ! Rn
be a path. Let f : Rn
! R be given. Then the scaler line integral of
f along x is
Z
x
f ds =
Z b
a
f(x(t)) ||x0
(t)|| dt .
Notice that ||x0
(t)|| is the speed. In particular, the length of the path is given by integrating
the speed:
length(x) =
Z
x
1 ds =
Z b
a
||x0
(t)|| dt .
Example (Problem 4). Let f(x, y, z) = 3x + xy + z3
. Consider the parametrization
x(t) = (cos 4t, sin 4t, 3t)
for 0  t  2⇡. Compute
R
x
f ds.
Solution:
f(x(t)) = 3 cos 4t + (cos 4t)(sin 4t) + (3t)3
33

34. and
||x0
(t)|| = ||( 4 sin 4t, 4 cos 4t, 3)||
=
p
( 4 sin 4t)2 + (4 cos 4t)2 + 32
=
p
16 sin2
4t + 16 cos2 4t + 9
=
p
16 + 9
= 5 .
Therefore
Z
x
f ds =
Z 2⇡
0
f(x(t)) ||x0
(t)|| dt
=
Z 2⇡
0
3 cos 4t + (cos 4t)(sin 4t) + (3t)3
5 dt
=
Z 2⇡
0
✓
3 cos 4t +
sin 8t
2
+ (3t)3
◆
5 dt
= 180⇡4
.
Example (Problem 5). Consider the function f(x, y, z) = 2x
p
y + z2
and the path
x(t) =
(
(t, t2
, 0) if 0  t  1
(1, 1, t 1) if 1  t  3
Compute the line integral Z
x
f ds .
Solution: The path x is the concatenation of the paths
x1(t) = (t, t2
, 0) for 0  t  1
and
x2(t) = (1, 1, t 1) for 1  t  3 .
The integral
R
x
f ds can be broken up as the sum
Z
x
f ds =
Z
x1
f ds +
Z
x2
f ds .
34

35. We see that
Z
x1
f ds =
Z 1
0
f(x1(t)) ||x0
1(t)|| dt
=
Z 1
0
(2t t) ||(1, 2t, 0)|| dt
=
Z 1
0
t
p
1 + 4t2 dt
=
Z 5
1
p
u
8
du
⇥
substitute u = 1 + 4t2
=) du = 8t dt
⇤
=
p
125 1
12
.
Also
Z
x2
f ds =
Z 3
1
f(x2(t)) ||x0
2(t)|| dt
=
Z 3
1
(1 + (t 1)2
) ||(0, 0, 1)|| dt
=
Z 3
1
(1 + (t 1)2
) dt
=
14
3
.
Therefore,
Z
x
f ds =
p
125 1
12
+
14
3
.
Vector line integrals
This time, the integrand will be a vector field.
Let x : [a, b] ! Rn
be a path. Let F : Rn
! Rn
be a vector field. Then the vector line
integral of F along x is
Z
x
F · ds =
Z b
a
F(x(t)) · x0
(t) dt .
Example. Let F(x, y, z) = 3xi + y2
j + 6zk = (3, y2
, 6z). Consider the path x(t) = (t, 3, 5t2
)
defined for 0  t  1. Compute
R
x
F · ds .
Solution:
F(x(t)) = (3t, 9, 30t2
)
35

36. and
x0
(t) = (1, 0, 10t) .
Therefore,
Z
x
F · ds =
Z b
a
F(x(t)) · x0
(t) dt
=
Z 1
0
(3t, 9, 30t2
) · (1, 0, 10t) dt
=
Z 1
0
3t + 300t3
dt
=
153
2
.
Di↵erential form
Suppose that we are given a vector field F : R2
! R2
defined by
F(x, y) = M(x, y)i + N(x, y)j
and a path x : [a, b] ! R2
. We can express the formula of the line integral
Z
x
F · ds
as Z
x
(M(x, y), N(x, y)) · (x0
(t), y0
(t)) dt =
Z
x
[M(x, y)x0
(t) + N(x, y)y0
(t)] dt .
If we multiply dt into the rest of integrand and use the di↵erentials
dx = x0
(t) dt , and dy = y0
(t) dt
we get the di↵erential form
Z
x
F · ds =
Z
x
M dx + N dy .
The same is true for a vector field on R3
F(x, y, z) = M(x, y, z)i + N(x, y, z)j + P(x, y, z)k
and a path
x : [a, b] ! R3
.
The vector line integral of F along x has the di↵erential form
Z
x
F · ds =
Z
x
M dx + N dy + P dz .
36

37. Example. Consider the path
x : [0, 1] ! R3
defined by the formula x(t) = (t, t2
, t3
). Compute
Z
x
2x dx + z dy + y dz .
Solution: Compute the di↵erentials
dx = x0
(t) dt = 1 dt , dy = y0
(t) dt = 2t dt , dz = z0
(t) dt = 3t2
dt .
Then substitute
Z
x
2x dx + z dy + y dz =
Z 1
0
2t dt + (t3
)(2t dt) + t2
(3t2
dt) =
Z 1
0
(2t + 2t4
+ 3t4
) dt = 2 .
Example (Compare with problem 22). Calculate
Z
C
z dx x dy + 2y dz
where C is the curve obtained by intersecting the surfaces z = x2
and x2
+ y2
= 4 and
oriented counterclockwise around the z–axis (as seen from the positive z–axis).
Solution:
Since C is the intersection of two surfaces, we want a parametrization
x : [a, b] ! R3
, x(t) = (x(t), y(t), z(t))
which satisfies both equations
x2
+ y2
= 4 and z = x2
.
Start with the equation x2
+ y2
= 4 because the shadow of C in the xy–plane will be the
circle x2
+ y2
= 4. The standard (counterclockwise) parametrization of that circle in R2
is
(2 cos t, 2 sin t) , 0  t  2⇡ .
Since C also lies on the surface z = x2
, we just set z(t) = (x(t))2
= (2 cos t)2
= 4 cos2
t .
It follows that the parametrization for C should be
x : [0, 2⇡] ! R3
, x(t) = (2 cos t, 2 sin t, 4 cos2
t) .
The di↵erentials are
dx = 2 sin t dt , dy = 2 cos t dt , dz = 8 cos t sin t dt .
37

38. Therefore,
Z
C
z dx x dy + 2y dz =
Z 2⇡
0
(4 cos2
t)( 2 sin t) dt (2 cos t)(2 cos t) dt + 2(2 sin t)( 8 cos t sin t) dt
=
Z 2⇡
0
( 8 cos2
sin t 4 cos2
t 32 sin2
t cos t) dt
= 4⇡ .
The e↵ect of reparametrization
Example. The image of each of the parametrizations
x : [0, 2⇡] ! R2
, x(t) = (cos t, sin t)
and
y : [0, ⇡] ! R2
, y(t) = (cos 2t, sin 2t)
is the unit circle {(x, y) | x2
+ y2
= 1}.
Definition. A reparametrization of a path
x : [a, b] ! Rn
is another path
y : [c, d] ! Rn
for which there exists a one-to-one and onto function u : [c, d] ! [a, b] with the
property y(t) = x(u(t)).
Furthermore, if u(c) = a, then y is an orientation-preserving reparametrization
of x; if u(c) = b, then y is an orientation-reversing reparametrization of x.
Two quick and easy way to see if the reparametrization preserves or reverses orientation is:
1. If you know u(t), then u0
(t) is negative if and only if y reverses orientation.
2. y reverses orientation if and only if the tangent (velocity) vectors for y and x point in
di↵erent directions at the same point on the image curve.
Example. In the previous example, we see that y is a reparametrization of x with
u : [0, ⇡] ! [0, 2⇡] , u(t) = 2t .
Notice that y preserves orientation.
38

39. Definition (Opposite path). Suppose that
x : [a, b] ! Rn
is a path. The opposite path
xopp : [a, b] ! Rn
is defined by
xopp(t) = x(a + b t) .
Example. The opposite path of
x : [0, 2⇡] ! R2
, x(t) = (cos t, sin t)
is
xopp : [0, 2⇡] ! R2
, x(t) = (cos(2⇡ t), sin(2⇡ t)) = (cos t, sin t).
The path x traces the unit circle in the counter-clockwise direction while the opposite path
xopp traces the unit circle in the clockwise direction.
Theorem (Reparametrization and scaler line integrals). Let
x : [a, b] ! Rn
be a path. Let f : Rn
! R. If
y : [c, d] ! Rn
is a reparametrization of x, then
Z
x
f ds =
Z
y
f ds .
Since scaler line integrals are not a↵ected by reparametrizations, we can define scaler line
integrals over curves without having to refer to parametrizations until we need them.
Example. Let be the upper half of the unit circle
= {(x, y) | x2
+ y2
= 1, y 0} .
We would like to integrate the function f(x, y) = y along . Consider the parametrization
x : [0, ⇡] ! R2
, x(t) = (cos t, sin t) .
So x0
(t) = ( sin t, cos t), and
Z
x
y ds =
Z ⇡
0
sin t
p
sin2
t + cos2 t dt =
Z ⇡
0
sin t dt = 2 .
39

40. Now consider another parametrization
y : [ 1, 1] ! R2
, y(t) = (t,
p
1 t2) .
So y0
(t) = (1, t/
p
1 t2), and
Z
y
y ds =
Z 1
1
p
1 t2
r
1 +
t2
1 t2
!
dt =
Z 1
1
dt = 2 .
Theorem (Reparametrization and vector line integrals). Let
x : [a, b] ! Rn
be a path. Let F : Rn
! Rn
be a vector field. Suppose that
y : [c, d] ! Rn
is a reparametrization of x.
1. If y preserves orientation, then
R
y
F · ds =
R
x
F · ds.
2. If y reverses orientation, then
R
y
F · ds =
R
x
F · ds.
Example. Let F : R2
! R2
be the vector field F(x, y) = xyi + y2
j. Consider the path
x : [0, 4] ! R2
, x(t) = (t, t2
) .
Then Z
x
F · ds =
Z 4
0
(t3
, t4
) · (1, 2t) dt =
Z 4
0
t3
+ 2t5
dt =
4288
3
.
Consider the reparametrization
y : [ 4, 0] ! R2
, y(t) = ( t, t2
) .
Notice that y reverses orientation. Then
Z
y
F · ds =
Z 0
4
( t3
, t4
) · ( 1, 2t) dt =
Z 0
4
t3
+ 2t5
dt =
4288
3
.
40

41. Section 6.2. Green’s Theorem
All paths (curves) under consideration will be piecewise C1
, unless otherwise specified. Also,
all scaler functions f : Rn
! R and vector fields F : Rn
! Rn
will be assumed to be C1
unless otherwise specified.
Green’s Theorem
Definitions (simple, closed). A curve in Rn
is simple in if it has no self-
intersections. A curve in Rn
is closed if it is a loop.
Definition (positive orientation). Let C be collection of disjoint simple closed
curves in R2
. Suppose that C is the boundary of a region D. The orientation
for C in which D lies to the left of C as one traverses C is called the positive
orientation; when C is a single curve, then positive means counter-clockwise. The
opposite orientation is called the negative orientation.
Special Notation: When expressing a line integral over a closed curve C, we use the special
integral symbol I
C
in place of the ordinary symbol
R
C
.
Theorem (Green’s Theorem). Let D ⇢ R2
be a closed, bounded region whose
boundary @D consists of simple closed curves. Give @D the positive orientation.
Then I
@D
M dx + N dy =
ZZ
D
✓
@N
@x
@M
@y
◆
dx dy .
Example (Problem 2). Consider the vector field F : R2
! R2
given by
F(x, y) = M i + N j
where M = x2
y and N = x+y2
. Let D be the rectangle with verticies (0, 0), (2, 0), (2, 1),
and (0, 1). Verify the formula in Green’s Theorem by calculating each of the integrals
I
@D
M dx + N dy and
ZZ
D
✓
@N
@x
@M
@y
◆
dx dy .
Solution:
We start with the double integral
ZZ
D
✓
@N
@x
@M
@y
◆
dx dy .
41

42. D is just the rectangle [0, 2] ⇥ [0, 1], so
ZZ
D
✓
@N
@x
@M
@y
◆
dx dy =
ZZ
D
(1 ( 1)) dx dy
=
Z 1
0
Z 2
0
2 dx dy
= 4 .
Now take care of the line integral
I
@D
M dx + N dy .
This is more labor-intensive:
The boundary @D consists of four line-segments 1, 2, 3, and 4 where
• 1 connects (0, 0) to (2, 0)
• 2 connects (2, 0) to (2, 1)
• 3 connects (2, 1) to (0, 1)
• 4 connects (0, 1) to (0, 0).
The simplest parametrizations for these segments (with the appropriate orientations) are the
following.
• 1 is given by x1 : [0, 2] ! R2
, x1(t) = (t, 0)
• 2 is given by x2 : [0, 1] ! R2
, x2(t) = (2, t)
• 3 is given by x3 : [0, 1] ! R2
, x3(t) = (1 t)2, 1
• 4 is given by x4 : [0, 1] ! R2
, x4(t) = (0, 1 t)
Then
I
@D
M dx + N dy =
4X
i=1
✓Z
xi
M dx + N dy
◆
= 4 .
Example (Problem 8). Find the work done by the force field
F = (4y 3x)i + (x 4y)j
along the positively oriented ellipse
x2
+ 4y2
= 4 .
42

43. Solution: For this problem, let D be the region bounded by the ellipse:
D = {(x, y) | x2
+ 4y2
 4} .
The work done by F along @D is the line integral
I
@D
F · ds .
By Green’s Theorem, we have
I
@D
F · ds =
ZZ
D
✓
@N
@x
@M
@y
◆
dx dy
=
ZZ
D
(1 4) dx dy
=
ZZ
D
3 dx dy
= 3 Area(D) .
The standard form for the equation of @D is
⇣x
2
⌘2
+ y2
= 1 .
Recall that the area of the region D enclosed by the ellipse
⇣x
a
⌘2
+
⇣y
b
⌘2
= 1
is ab⇡. This is most easily seen by using the polar-like substitution:
x
a
= r cos ✓ ,
y
b
= r sin ✓
with Jacobian
@(x, y)
@(r, ✓)
= abr .
So
Area(D) =
ZZ
D
dx dy =
Z 2⇡
0
Z 1
0
abr dr d✓ = ab⇡ .
It follows that the area of D is 2⇡.
Therefore, the work done by F along @D is 6⇡.
43

44. Area formula
The following is an easy consequence of Green’s Theorem.
If D is a region to which Green’s Theorem applies, then
Area(D) =
I
@D
x dy =
I
@D
y dx =
1
2
I
@D
x dy y dx .
The Divergence Theorem
The following theorem is useful for studying the flux of a vector field across a closed curve
in R2
.
Theorem (Divergence Theorem in the plane). If D is a region to which Green’s
Theorem applies, n is the outward unit normal vector to D, and
F = M(x, y)i + N(x, y)j
is a vector field on D, then
I
@D
F · n ds =
ZZ
D
r · F dA .
Recall that the outward unit normal vector n at a point (x(t), y(t)) can be given by the
formula
n = n(t) =
y0
(t)i x0
(t)j
p
(x0(t))2 + (y0(t))2
.
The quantity
r · F
is also known as the divergence of the vector field F, and is usually denoted by div F.
In practice, the equivalent formula
I
@D
F · n ds =
ZZ
D
✓
@M
@x
+
@N
@y
◆
dA
in the Divergence Theorem will be more useful.
Example (Problem 17 (a)). Use the Divergence Theorem to show that
I
C
F · n ds = 0
where F = 2yi 3xj and C is the (positively oriented) unit circle.
44

45. Solution: By the Divergence Theorem, it is enough to show that
ZZ
D
✓
@M
@x
+
@N
@y
◆
dA = 0 .
Since M = 2y and N = 3x, we see that
@M
@x
+
@N
@y
= 0 .
Therefore I
@D
F · n ds =
ZZ
D
✓
@M
@x
+
@N
@y
◆
dA = 0 .
Without the Divergence Theorem, we would have compute the integral directly. Parameter-
ize the unit circle in the standard way:
x(t) = (cos t, sin t) , 0  t  2⇡ .
So
x0
(t) = ( sin t, cos t) =) ||x0
(t)|| = 1 .
To compute the integral, put F and n in terms of x(t):
F(x(t)) = (2 sin t, 3 cos t) , n(t) = (cos t, sin t) .
Therefore
I
@D
F · n ds =
Z 2⇡
0
F(t) · n(t) ||x0
(t)|| dt
=
Z 2⇡
0
(2 sin t, 3 cos t) · (cos t, sin t) dt
=
Z 2⇡
0
cos t sin t dt
= 0 .
45

46. Section 6.3. Conservative vector fields
All paths (curves) under consideration will be piecewise C1
, unless otherwise specified. Also,
all scaler functions f : Rn
! R and vector fields F : Rn
! Rn
will be assumed to be C1
unless otherwise specified. The same assumptions apply if we restrict the domains to regions
R ⇢ Rn
.
Definition. Let R ⇢ Rn
be a region. We say that R is connected if every pair of
points in R can be joined by a continuous path ⇢ R. The idea is that R consists
of only one “piece”.
Definition (Path independent line integrals). Let F : R ! Rn
be a vector field on
R. We say that F has path independent line integrals on R if
Z
C1
F · ds =
Z
C2
F · ds
for any paths C1 and C2 in R with the same initial and final points.
Example. We consider two paths C1 and C2 in R2
starting at (1, 0) and ending at (0, 1):
Let C1 be quarter-circle given by formula x1(t) = (cos t, sin t) for 0  t  ⇡/2.
Let C2 be the line-segment given by the formula x2(t) = (1 t, t) for 0  t  1.
Consider the vector field F(x, y) = yi + xj. Then
Z
C1
F · ds =
Z ⇡/2
0
( sin t, cos t) · ( sin t, cos t) dt =
Z ⇡/2
0
dt = ⇡/2 ,
while Z
C2
F · ds =
Z 1
0
( t, 1 t) · ( 1, 1) dt =
Z 1
0
dt = 1 .
We see that F is does not have path-independent line integrals.
Theorem. Let F : R ! Rn
be a vector field. Then F has path independent line
integrals on R if and only if I
C
F · ds = 0
for any simple closed curve C in R.
46

47. Gradient fields and line integrals
Definition (gradient (or conservative) field, scaler potential). Let R ⇢ Rn
be a
region. Suppose that F : R ! Rn
is a vector field. We say that F is a gradient
field on R (or conservative field) on R if
F = rf
for some scaler function f : R ! R (of class C2
). In this case, we call the function
f a scaler potential function for F. In particular, we realize the component
functions of F as the partial derivatives of f.
Example. A simple example of a conservative vector field in R ⇢ R2
is anything of the form
F(x, y) = M(x)i + N(y)j .
where M and N are C1
functions. Set f(x, y) =
R
M(x) dx +
R
N(y) dy. It is easy to see
that
F = rf .
This construction can be applied in R3
in the obvious way. In particular, F(x, y, z) =
2x i + 3y2
j 3 sin 3z k and f(x, y, z) = x2
+ y3
+ cos(3z) satisfy F = rf.
Theorem (Fundamental Theorem of Line Integrals). Let R ⇢ Rn
be an open
connected region. Suppose that F : R ! Rn
is a continuous vector field. Then F is
conservative on R if and only if F has path independent line integrals on R.
Furthermore, if F = rf and the path C ⇢ R has initial point A and final point B,
then Z
C
F · ds = f(B) f(A) .
Example. Let C be a continuous path in R2
beginning at A = (1, 2) and ending at B =
( 4, 0). Consider the vector field F(x, y) = 2xi + 3y2
j. Compute the vector line integral
Z
C
F · ds .
Solution:
Notice that F = rf where f(x, y) = x2
+ y3
; compute f(A) = 9 and f(B) = 16. By the
Fundamental Theorem of Line Integrals, we have
Z
C
F · ds = f(B) f(A) = 16 9 = 7 .
47

48. Criterion for conservative vector fields
Definition (Simply-connected region). A region R ⇢ Rn
is simply-connected if
it is connected and every simple closed curve in R can be continuously shrunk in
R to a point.
Example. The following regions are simply-connected.
• Rn
; to see this, shrink along straight lines to any desired point P 2 Rn
.
• The closed (or open) unit disk in R2
; again, shrink along straight lines.
• The closed (or open) unit ball in R3
; again, shrink along straight lines.
• The unit sphere in R3
; this is not obvious.
• Any “continuously deformed” version of the above examples. For more information
and to see a co↵ee cup deformed into a doughnut, see the website
http://en.wikipedia.org/wiki/Topology
Recall the following definition (see Chapter 3).
Definition (Curl of a vector field). Suppose that F(x, y, z) = M(x, y, z) i +
N(x, y, z) j + P(x, y, z) k is a vector field on R3
. The curl of F is the vector
field
r⇥F = det
2
6
4
i j k
@
@x
@
@y
@
@z
M N P
3
7
5 =
✓
@P
@y
@N
@z
◆
i
✓
@P
@x
@M
@z
◆
j+
✓
@N
@x
@M
@y
◆
k .
If F(x, y) = M(x, y)i + N(x, y)j is a vector field on R2
, then extend it to a vector
field ¯F(x, y, z) on R3
by the formula
¯F(x, y, z) = M(x, y)i + N(x, y)j + 0k
and define the curl of F as
r ⇥ F = r ⇥ ¯F =
✓
@N
@x
@M
@y
◆
k .
This is the formula that shows up in the vector reformulation of Green’s Theorem.
Theorem (Conservative = zero curl). Let R ⇢ Rn
be a simply-connected region.
Let F : R ! Rn
be a vector field. Then F is conservative on R if and only if
r ⇥ F = 0 on R.
48

49. Remark. When solving problems, if the region R ⇢ Rn
is not specified, then assume that R
is the largest simply-connected region (usually Rn
itself) such that F is defined and of class
C1
.
Example. Consider the vector field
F(x, y) = 2x sin(7y) i + 7x2
cos(7y) + 2y j .
Is F conservative?
Solution: We set M = 2x sin(7y) and N = 7x2
cos(7y) + 2y. Then Compute
@N
@x
= 14x cos(7y) and
@M
@y
= 14x cos(7y) .
Therefore F is conservative.
Example. Consider the vector field
F(x, y, z) = 3x2
y i + (x3
+ 2y) j + 6z k .
Is F conservative?
Solution: We set M = 3x2
, N = x3
+ 2y and P = 6z and compute
r⇥F = det
2
6
4
i j k
@
@x
@
@y
@
@z
M N P
3
7
5 = det
2
6
4
i j k
@
@x
@
@y
@
@z
3x2
x3
+ 2y 6z
3
7
5 = (0 0)i (0 0)j+(3x2
3x2
)k = 0 .
Therefore F is conservative.
Example. Consider the vector field
F(x, y, z) = y i x j + x2
yz k .
Is F conservative?
Solution: We set M = y, N = x and P = x2
yz and compute
r⇥F = det
2
6
4
i j k
@
@x
@
@y
@
@z
M N P
3
7
5 = det
2
6
4
i j k
@
@x
@
@y
@
@z
y x x2
yz
3
7
5 = (x2
z 0)i (2xyz 0)j+( 1 1)k 6= 0 .
Therefore F is not conservative.
49

50. Finding scaler potentials on R2
Suppose that F = M i + N j is a conservative vector field on a simply-connected
region R ⇢ R2
. We want to find a scaler potential function f for F. This amounts
to solving the system of partial di↵erential equations
@f
@x
= M and
@f
@y
= N .
The general procedure for solving the system is to
• Integrate the equation @f
@x
= M to get f(x, y) = (
R
M dx) + g(y)
• Di↵erentiate this f(x, y) with respect to y to get another formula for @f
@y
;
compare to N and deduce a formula for g0
(y).
• Integrate one more time: g(y) =
R
g0
(y) dy
• Combine the previous formulas to obtain an explicit formula for f(x, y).
There is a variation of this procedure.
Example. Earlier, we showed that
F(x, y) = 2x sin(7y) i + 7x2
cos(7y) + 2y j
is a conservative vector field on R2
. Find a scaler potential for F.
Solution: We follow the standard procedure with M = 2x sin(7y) and N = 7x2
cos(7y) + 2y.
• Integrate @f
@x
= 2x sin(7y) to get
f(x, y) = x2
sin(7y) + g(y) .
• Di↵erentiate f with respect to y:
@f
@y
=
@
@y
x2
sin(7y) + g(y) = 7x2
cos(7y) + g0
(y) ,
and set this equal to
N = 7x2
cos(7y) + 2y .
Deduce that g0
(y) = 2y.
• Integrate g0
(y) to get g(y):
g(y) =
Z
g0
(y) dy = y2
.
• Conclude that
f(x, y) = x2
sin(7y) + y2
.
50

51. Finding scaler potentials on R3
The method here is a generalization of the method in R2
.
Suppose that F = M i + N j + P k is a conservative vector field on a simply-
connected region R ⇢ R3
. We want to find a scaler potential function f for F. This
amounts to solving the system of partial di↵erential equations
@f
@x
= M ,
@f
@y
= N , and
@f
@z
= P .
The general procedure for solving the system is to
• Integrate the equation @f
@x
= M to get f(x, y, z) = (
R
M dx) + g(y, z)
• Di↵erentiate this f(x, y, z) with respect to y to get another formula for @f
@y
;
compare to N. Deduce a formula for @g
@y
.
• Integrate @g
@y
with respect to y to get g(y, z) = (
R @g
@y
dy) + h(z); substitute
this g(y, z) into the formula for f(x, y, z) and simplify.
• Di↵erentiate this f(x, y, z) with respect to z to get another formula for @f
@z
;
compare to P and deduce a formula for h0
(z).
• Integrate one more time: h(z) =
R
h0
(z) dz
• Combine the previous formulas to obtain an explicit formula for f(x, y, z).
There are variations of this procedure.
Example. Earlier, we showed that F(x, y, z) = 3x2
y i+(x3
+2y) j+6z k is a conservative
vector field on R3
. Find a scaler potential for F.
Solution: We follow the standard procedure with M = 3x2
y, N = x3
+ 2y and P = 6z.
• Integrate the equation @f
@x
= 3x2
y to get f(x, y, z) = x3
y + g(y, z)
• Di↵erentiate this f(x, y, z) with respect to y:
@
@y
x3
y + g(y, z) = x3
+
@g
@y
,
and set this equal to N = x3
+ 2y. Deduce that @g
@y
= 2y.
• Integrate @g
@y
= 2y with respect to y to get
g(y, z) = y2
+ h(z) .
51

52. It follows that
f(x, y, z) = x3
y + y2
+ h(z) .
• Di↵erentiate this f(x, y, z) with respect to z to get
@f
@z
= 0 + 0 + h0
(z) ;
compare to P = 6z and deduce that h0
(z) = 6z.
• Integrate h0
(z) = 6z:
h(z) =
Z
6z dz = 3z2
.
• Combine the previous formulas to obtain an explicit formula for f(x, y, z):
f(x, y, z) = x3
y + y2
+ 3z2
.
52

53. Section 7.1. Parametrized surfaces
Basic definitions and examples
Definition (Parametrized surface in R3
). Let D be a connected region in R2
. A
parametrized surface in R3
is a continuous function
X : D ! R3
that is one-to-one on the interior of D. The image
S = X(D) ⇢ R3
is called the underlying surface of X. We can also refer to the function X as a
parametrization of the surface S.
See Figure 7.1 in the textbook.
We can write X(s, t) as
X(s, t) = x(s, t), y(s, t), z(s, t)
where x(s, t), y(s, t) and z(s, t) are the coordinate functions or component functions
of X.
It be helpful to express the parametrization X : D ! R3
as
X : Dst ! R3
xyz
to emphasize the names of all of the variables involved.
Remark. For the purposes of integration, we will generally want D to be an elementary
region (as in chapter 5).
Examples (Graphs of functions z = f(x, y)). A major example of surfaces arise from
functions of two variables. See Figure 7.5 in the textbook.
1. The graph of z = x2
+ y2
is the image of the parametrization
X(s, t) = (s, t, s2
+ t2
)
where D is any connected region in R2
.
2. The graph of z = x2
is the image of the parametrization
X(s, t) = (s, t, s2
)
where D is any connected region in R2
.
53

54. 3. In general, the graph of a continuous function z = f(x, y) over a connected region
D ⇢ R2
is the image of the parametrization
X : D ! R3
, X(s, t) = (s, t, f(s, t)) .
Example (Cylinder). The cylinder x2
+ y2
= 4 has the parametrization
X : D✓z ! R3
xyz
defined by
X(✓, z) = (2 cos ✓, 2 sin ✓, z)
where
D✓z = {(✓, z) | 0  ✓  2⇡} .
Here, we are using the more appropriate variables ✓ instead of s and z instead of t. See
Figure 7.4 in the textbook.
Example (Sphere). The sphere x2
+ y2
+ z2
= 25 has the parametrization
X : D✓' ! R3
xyz
defined by
X(✓, ') = (5 sin ' cos ✓, 5 sin ' sin ✓, 5 cos ')
where
D✓' = {(✓, ') | 0  ✓  2⇡ , 0  '  ⇡} .
Here, we are using the more appropriate variables ✓ instead of s and ' instead of t. See
Figure 7.3 in the textbook.
Coordinate curves
The main idea here is that we can fix one of variables, s or t, of X(s, t) to produce special
curves on the surface S = X(D).
54

55. Definition (Coordinate curves). Let X : D ! R3
be a parametrized surface. Let
(s0, t0) 2 D.
• The s–coordinate curve through X(s0, t0) is the curve
s 7! X(s, t0)
defined for s–values close to s0.
• The t–coordinate curve through X(s0, t0) is the curve
t 7! X(s0, t)
defined for t–values close to t0.
Of course, the largest domain for these curves depends on the region D.
See Figure 7.6 in the textbook.
Tangent vector, tangent plane and normal vector
Let X : Dst ! R3
be a parametrization for the surface S = X(D). Let (s0, t0) 2 D. Consider
the coordinate curves
X(s, t0) and X(s0, t)
near the point X(s0, t0) 2 S. The velocity vectors of these coordinate curves at the point
X(s0, t0) are
Ts(s0, t0) =
@X
@s
(s0, t0) and Tt(s0, t0) =
@X
@t
(s0, t0) .
These vectors are tangent vectors to the surface S. The plane determined by Ts(s0, t0)
and Tt(s0, t0) is the tangent plane to S at X(s0, t0). See Figure 7.10 in the textbook.
The vectors Ts and Tt determine a standard normal vector to the tangent plane:
N(s0, t0) = Ts(s0, t0) ⇥ Tt(s0, t0) .
Then the unit normal vector is defined as
n(s0, t0) =
N(s0, t0)
||N(s0, t0)||
.
If N(s0, t0) 6= 0, then the tangent plane to S at the point X(s0, t0) = (x0, y0, z0) has the
equation
N(s0, t0) · (x x0, y y0, z z0) = 0 .
55

56. Definition (Smooth point). Let X : D ! R3
is a parametrized surface that is
C1
near the point X(s0, t0) and N(s0, t0) 6= 0, then X(s0, t0) is called a smooth
point. If S = X(D) consists of all smooth points, then S is called a smooth
parametrized surface.
Example. Consider the surface S ⇢ R3
given by the parametrization
X(s, t) = (4es
, t2
e2s
, 3e s
+ t)
for (s, t) 2 R2
. Find an equation for the tangent plane to S at the point (4, 9, 0).
Solution: First find the point (s, t) 2 R2
that corresponds to the point (4, 9, 0). To do this,
solve the equation
X(s, t) = (4, 9, 0) .
This amounts to solving the system of equations
4es
= 4
t2
e2s
= 9
3e s
+ t = 0 .
We see that the first equation 4es
= 4 implies s = 0. Then plug this into the third equation
3e s
+ t = 0 to get t = 3. The tangent vectors to S have the formulas
Ts =
@X
@s
= (4es
, 2t2
e2s
, 3e s
) and Tt =
@X
@t
= (0, 2te2s
, 1) .
At the point X(0, 3), the tangent vectors are
Ts(0, 3) = (4, 18, 3) and Tt(0, 3) = (0, 6, 1) .
Now compute the normal vector at X(0, 3):
N(0, 3) = Ts(0, 3) ⇥ Tt(0, 3) = (4, 18, 3) ⇥ (0, 6, 1) = (0, 4, 24) .
Therefore, the tangent plane to S at the point X(0, 3) = (4, 9, 0) is
(0, 4, 24) · (x 4, y 9, z 0) = 0 =) 4(y 9) + 24z = 0 .
Example (Graphs of z = f(x, y)). In the special case of a graph of a C1
function f(x, y)
with domain D ⇢ R2
, we can make the special parametrization
X : D ! R3
, X(s, t) = (s, t, f(s, t)) .
With this parametrization, the tangent vector formulas are
Ts =
✓
1, 0,
@f
@s
◆
and Tt =
✓
0, 1,
@f
@t
◆
.
56

57. So the normal vector is
N = Ts ⇥ Tt =
✓
@f
@s
,
@f
@t
, 1
◆
.
For example, consider the graph of the function f(x, y) = x2
+ xey
on the domain D = R2
.
Give the graph of f(x, y) the standard parametrization
X(s, t) = (s, t, s2
+ set
) .
Then
N =
✓
@f
@s
,
@f
@t
, 1
◆
= (2s + et
), set
, 1 .
At the point (1, 0, 2) on the graph of f, the equation of the tangent plane is
N(1, 0)·(x 1, y, z 2) = 0 =) ( 3, 1, 1)·(x 1, y, z 2) = 0 =) 3(x 1) y+(z+2) = 0 .
This simplifies to
3x + y z = 1 .
Piecewise smooth parametrized surfaces
Definition (Piecewise parametrized surface). A piecewise parametrized sur-
face is the union of parametrized surfaces X : Di ! R3
, i = 1, . . . , m where
• Each Xi is of class C1
, except possibly along @Di.
• Each Si = X(Di) is smooth, except possibly at finitely many points.
Examples. The following are examples of piecewise smooth parametrized surfaces.
• The boundary of the solid cube [0, 1] ⇥ [0, 1] ⇥ [0, 1] ⇢ R3
. See Figure 7.12 in the
textbook.
• The boundary of nearly every solid that appears in section 5.4. In particular, see
Figures 5.47, 5.56, 5.63, 5.65 and 5.67 in the textbook.
Example. Consider the cylindrical solid
{(x, y, z) | x2
+ y2
 1, 0  z  1} ⇢ R3
.
Let S denote the boundary of the solid. Then S consists of three smooth pieces S1, S2, S3:
• S1 = {(x, y, 0) | x2
+ y2
 1} is the bottom of the cylinder.
57

58. • S2 = {(x, y, 1) | x2
+ y2
 1} is the top of the cylinder.
• S3 = {(x, y, z) | x2
+ y2
= 1, 0  z  1} is the lateral portion of S.
See Figure 7.29 in the textbook. Notice that each Si is a smooth parametrized surface. So
when you need to calculate something over S, you can use parametrizations such as the
following.
• For S1, use X1(r, ✓) = (r cos ✓, r sin ✓, 0) , (r, ✓) 2 [0, 1] ⇥ [0, 2⇡].
• For S2, use X2(r, ✓) = (r cos ✓, r sin ✓, 1) , (r, ✓) 2 [0, 1] ⇥ [0, 2⇡].
• For S3, use X3(✓, z) = (cos ✓, sin ✓, z) , (✓, z) 2 [0, 2⇡] ⇥ [0, 1].
Area of a parametrized surface
Let X : D ! R3
be a parametrized surface. The surface area of S = X(D) is
defined to be ZZ
D
||Ts ⇥ Tt|| ds dt =
ZZ
D
||N(s, t)|| ds dt .
Example. Let S be the portion of the paraboloid z = 25 x2
y2
that lies over the xy–plane.
Find the area of S.
Solution 1: We can realize S as the graph of f(x, y) = 25 x2
y2
on the domain D ⇢ R2
where D is the shadow of S in the xy–plane. It is easy to see that D is the disk
D = {(x, y) | x2
+ y2
 25} .
So consider the standard parametrization
X(s, t) = (s, t, 25 s2
t2
) .
Then
Ts = (1, 0, 2s) and Tt = (0, 1, 2t) .
So
||Ts ⇥ Tt|| = ||(2s, 2t, 1)|| =
p
(2s)2 + (2t)2 + 1 .
The area of S is
ZZ
D
||Ts ⇥ Tt|| ds dt =
ZZ
D
p
(2s)2 + (2t)2 + 1 ds dt .
To evaluate this integral, we should use polar coordinates for D. Make the substitutions
s = r cos ✓ , t = r sin ✓ ( 0  r  5 , 0  ✓  2⇡ )
58

59. to get
ZZ
D
p
(2s)2 + (2t)2 + 1 ds dt =
Z 2⇡
0
Z 5
0
p
(2r cos ✓)2 + (2r sin ✓)2 + 1 r dr d✓
=
Z 2⇡
0
Z 5
0
p
4r2 + 1 r dr d✓
=
⇡
6
⇣
101
p
101 1
⌘
.
Solution 2: Consider the following (cylindrical) parametrization of S:
X : [0, 5] ⇥ [0, 2⇡] ! R3
, X(r, ✓) = (r cos ✓, r sin ✓, 25 r2
) .
This time, D is just the rectangle [0, 5] ⇥ [0, 2⇡]. Then
Tr = (cos ✓, sin ✓, 2r) and T✓ = ( r sin ✓, r cos ✓, 0) .
So
Tr ⇥ T✓ = (2r2
cos ✓, 2r2
sin ✓, r cos2
✓ + r sin2
✓) = (2r2
cos ✓, 2r2
sin ✓, r) .
The area of S is
ZZ
D
||Tr ⇥ T✓|| dr d✓ =
Z 2⇡
0
Z 5
0
p
4r4 cos2 ✓ + 4r4 sin2
✓ + r2 dr d✓
=
Z 2⇡
0
Z 5
0
p
4r4 + r2 dr d✓
=
Z 2⇡
0
Z 5
0
r
p
4r2 + 1 dr d✓
=
⇡
6
⇣
101
p
101 1
⌘
.
Remark. If S is a piecewise smooth parametrized surface where the smooth pieces S1, . . . , Sm
of S meet along curves, then calculate the area of S in a piecewise way:
Area(S) = Area(S1) + · · · + Area(Sm) .
59

60. Section 7.2. Surface integrals
Scaler surface integrals
Definition. Let X : D ! R3
be a smooth parametrized surface, where D ⇢ R2
is
a bounded region. Let f be a continuous scaler function on S = X(D). The scaler
surface integral of f along X is
ZZ
X
f dS =
ZZ
D
f X(s, t) ||Ts ⇥ Tt|| ds dt =
ZZ
D
f X(s, t) ||N(s, t)|| ds dt .
The definition extends to piecewise smooth parametrized surfaces.
Example. Let D = [0, 1] ⇥ [0, 2] and consider the parametrized surface
X(s, t) = (s, s + t, t) .
Compute
RR
X
x2
+ y2
+ z2
dS.
Solution: The coordinate tangent vectors are
Ts = (1, 1, 0) and Tt = (0, 1, 1) ,
so the normal vector Ts ⇥ Tt is
Ts ⇥ Tt = (1, 1, 1) ; so ||Ts ⇥ Tt|| =
p
3 .
Therefore
ZZ
X
x2
+ y2
+ z2
dS =
Z 2
0
Z 1
0
(s2
+ (s + t)2
+ t2
)
p
3 ds dt =
26
p
3
.
Vector surface integrals
Definition. Let X : D ! R3
be a smooth parametrized surface, where D ⇢ R2
is
a bounded region. Let F(x, y, z) be a continuous vector field on S = X(D). Then
the vector surface integral of F along X is
ZZ
X
F · dS =
ZZ
D
F X(s, t) · N(s, t) ds dt .
This integral is also known as a flux integral because it computes the flux of a
vector field F in R3
though the surface S.
60

61. Example. Consider the vector field F(x, y, z) = 2x i+y j z k on the parametrized surface
X : [0, 1] ⇥ [0, 2] ! R3
, X(s, t) = (s + t, t, st) .
Compute ZZ
X
F · dS .
Solution: The coordinate tangent vectors are
Ts = (1, 0, t) and Tt = (1, 1, s) ,
so the normal vector is
N(s, t) = Ts ⇥ Tt = ( t, t s, 1) .
Therefore
ZZ
X
F · dS =
Z 2
0
Z 1
0
2(s + t), t, st · t, t s, 1 ds dt
=
Z 2
0
Z 1
0
2t(s + t) + t(t s) st ds dt
=
Z 2
0
Z 1
0
t2
4st ds dt
=
20
3
.
The e↵ect of reparametrization
The situation is similar to the one for line integrals, the scaler integrals do not depend on
reparametrization, and the same is true for vector integrals (up to ± sign).
Definition. A reparametrization of a surface
X : D1 ! R3
is another parametrization
Y : D2 ! R3
for which there exists a di↵erentiable coordinate transformation H : D2 ! D1 that
is one-to-one and onto, and Y(s, t) = X(H(s, t)).
61

62. The e↵ect of reparametrization for scaler surface integrals
Theorem (Reparametrization and scaler surface integrals). Let
X : D1 ! R3
be a parametrized surface. Let f be a continuous scaler function on S = X(D1). If
Y : D2 ! R3
is a reparametrization of X, then
ZZ
X
f dS =
ZZ
Y
f dS .
Since scaler surface integrals are not a↵ected by reparametrizations, we can define scaler
surface integrals ZZ
S
f dS
without having to refer to parametrizations until we need to calculate them.
Orientation
Definition (Orientable surface). Let S ⇢ R3
be a smooth parametrized surface.
We say that S is orientable if it is possible to define a continously varying unit
normal vector n to every point on S. Otherwise, S is called nonorientable.
Remark. A surface S ⇢ R3
is orientable if and only if it has two sides. See Figure 7.21 in
the textbook. The only surfaces that we will care about in this class will be orientable.
Definition (Orientation-preserving reparametrizations). Let X : D1 ! R3
be a
parametrization for an orientable surface S ⇢ R3
. Suppose that Y : D2 ! R3
is a
reparametrization of X. Then Y is an orientation-preserving reparametriza-
tion if the normal vectors from X and Y point in the same direction at correspond-
ing points on S. Otherwise Y is an orientation-reversing reparametrization.
More precisely, if Y(s, t) = X(H(s, t)) where H(s, t) is a coordinate transformation, then Y
is orientation-preserving precisely when the Jacobian determinant of H is positive.
Example (Variable swap). Suppose that X : D1 ! R3
is a parametrization of the surface
S = X(D1). Let D2 be the region obtained by swapping the variables in D1:
D2 = {(s, t) | (t, s) 2 D1} .
62

63. Then define
Y : D2 ! R3
, Y(s, t) = X(t, s) .
The Jacobian matrix of
H(s, t) = (t, s)
is 
0 1
1 0
;
the determinant of this matrix is 1. Therefore Y is a orientation-reversing reparametriza-
tion of X. You should pictorially compare the tangent vectors and normal vectors of X and
Y. You can also see Figure 7.20 in the textbook.
The e↵ect of reparametrization for vector surface integrals
Theorem (Reparametrization and vector surface integrals). Let
X : D1 ! R3
be a parametrized surface. Let F be a continuous vector field on S = X(D1). If
Y : D2 ! R3
is a reparametrization of X, then
•
RR
X
F · dS =
RR
Y
F · dS if Y is orientation-preserving.
•
RR
X
F · dS =
RR
Y
F · dS if Y is orientation-reversing.
So if we have an oriented surface S, then then the integral
ZZ
S
F · dS
is defined to be ZZ
X
F · dS
where X(s, t) is a parametrization whose orientation agrees with the prescribed orientation.
Outward (pointing) normal
The orientation that we usually prescribe for a surface will one that agrees with the standard
normal vectors of the graph of a function z = f(x, y). The main idea is that we want to define
the “top-side” of the surface to be what we would observe from high above the xy–plane.
63

64. Definition (Outward (pointing) normal vector). Let S ⇢ R3
be an orientable
surface that bounds a solid W with finite volume. Then the outward (pointing)
normal vector can be defined in the obvious way, that is, the normal vector should
point out of the solid.
Remark. In the case that S is the graph of a function z = f(x, y), the upward (pointing)
normal will be ✓
@f
@x
,
@f
@y
, 1
◆
.
In addition, if f(x, y) is nonnegative, then this upward (pointing) normal is the same as the
outward (pointing) normal for the solid between S and the xy–plane.
Example (Cylinder). Define S ⇢ R3
to be the boundary of the solid cylindrical region
{(x, y, z) | x2
+ y2
 1 , 0  z  1} .
Give S the outward normal orientation. So S consists of three smooth pieces S1, S2, S3:
• S1 = {(x, y, 1) | x2
+ y2
 1} is the bottom of the cylinder.
• S2 = {(x, y, 0) | x2
+ y2
 1} is the top of the cylinder.
• S3 = {(x, y, z) | x2
+ y2
= 1, 0  z  1} is the lateral portion of S.
We want to parametrize each of the Si in a way that is consistent with the orientation of S.
Let D ⇢ R2
be the closed unit disk. See Figure 7.29 in the textbook.
• A suitable parametrization for S1 is
X1 : D ! R3
, X1(s, t) = (t, s, 0) .
Notice that
@X1
@s
= (0, 1, 0) and
@X1
@t
= (1, 0, 0) =) N1 =
@X1
@s
⇥
@X1
@t
= (0, 0, 1) .
The outward normal vector is N1 = (0, 0, 1).
• A suitable parametrization for S2 is
X2 : D ! R3
, X2(s, t) = (s, t, 1) .
Notice that
@X2
@s
= (1, 0, 0) and
@X2
@t
= (0, 1, 0) =) N2 =
@X2
@s
⇥
@X2
@t
= (0, 0, 1) .
The outward normal vector is N2.
64

65. • A suitable parametrization for S3 is
X3 : [0, 2⇡] ⇥ [0, 1] ! R3
, X3(✓, z) = (cos ✓, sin ✓, z) .
Notice that
@X3
@✓
= ( sin ✓, cos ✓, 0) and
@X3
@z
= (0, 0, 1) =) N3 =
@X3
@✓
⇥
@X3
@z
= (cos ✓, sin ✓, 0) .
The outward normal vector is N3.
Now consider the vector field
F(x, y, z) = x i 2y j + (x2
+ z) k
on R3
. Then the flux of F through S is
ZZ
S
F · dS =
ZZ
S1
F · dS +
ZZ
S2
F · dS +
ZZ
S3
F · dS
=
ZZ
D
(s, 2t, s2
) · (0, 0, 1) ds dt +
ZZ
D
(s, 2t, s2
) · (0, 0, 1) ds dt
+
Z 2⇡
0
Z 1
0
(cos ✓, 2 sin ✓, cos2
✓ + z) · (cos ✓, sin ✓, 0) dz d✓
=
ZZ
D
s2
ds dt +
ZZ
D
s2
ds dt +
Z 2⇡
0
Z 1
0
(cos2
✓ 2 sin2
✓) dz d✓
= 0 ⇡/2
= ⇡/2 .
65

66. Section 7.3. Stokes’s and Gauss’s Theorems
Stokes’s Theorem
Definition (Boundary orientation). Let S be a piecewise smooth oriented surface.
Suppose that @S consists of simple closed curves. Let C be one of the one of these
boundary circles. The boundary orientation (a.k.a. induced orientation) on
C is the orientation such that the “top of S” lies to the left of C as one traverses
C. See Figures 7.30 and 7.31 in the textbook.
Theorem (Stokes’s Theorem). Let S be a bounded piecewise smooth oriented sur-
face. Suppose that the boundary @S consists of simple closed curves with the bound-
ary orientation. Let F be a C1
vector field on S. Then
ZZ
S
(r ⇥ F) · dS =
I
@S
F · ds .
Remark: Stokes’s Theorem is a generalization of (the vector reformulation of) Green’s
Theorem since
r ⇥ (Mi + Nj + 0k) =
✓
@N
@x
@M
@y
◆
k .
Example (Compare with Example 1 in the textbook). Verify Stokes’s Theorem for the
vector field
F(x, y, z) = (z2
, 2x, y)
and S the portion of the paraboloid z = 1 (x2
+ y2
) with z 0; orient S with upward
pointing normal vectors.
Solution: First we calculate ZZ
S
(r ⇥ F) · dS .
The curl of F is
r ⇥ F = (1, 2z, 2) .
Notice that S is the graph of the function
f(x, y) = 1 (x2
+ y2
)
with domain the unit disk in D ⇢ R2
. Give S the standard parametriztion
X(s, t) = (s, t, 1 (s2
+ t2
)) , (s, t) 2 D .
Then calculate the standard normal vector of X:
N = (2s, 2t, 1) .
66

67. Then evaluate
ZZ
S
(r ⇥ F) · dS =
ZZ
D
(r ⇥ F) · N ds dt
=
ZZ
D
(1, 2(1 (s2
+ t2
)), 2) · (2s, 2t, 1) ds dt
=
ZZ
D
2s + 4t(1 (s2
+ t2
)) + 2 ds dt
=
Z 2⇡
0
Z 1
0
2r cos ✓ + 4r sin ✓(1 r2
) + 2 r dr d✓
=
Z 2⇡
0

2
3
r3
cos ✓ +
4
3
r3
sin ✓
4
5
r5
sin ✓ + r2
r=1
r=0
d✓
=
Z 2⇡
0
✓
2
3
cos ✓ +
4
3
sin ✓
4
5
sin ✓ + 1
◆
d✓
= 2⇡ .
Next we calculate
H
@S
F·ds. Notice that @S is just the unit circle in the xy–plane, and the ap-
propriate orientation for @S is the counterclockwise orientation. So give @S the parametriza-
tion
x(✓) = (cos ✓, sin ✓, 0) , 0  ✓  2⇡ .
The velocity (tangent) vector is
x0
(✓) = ( sin ✓, cos ✓, 0) .
Therefore
I
@S
F · ds =
Z 2⇡
0
F(x(✓)) · x0
(✓) d✓
=
Z 2⇡
0
(0, 2 cos ✓, sin ✓) · ( sin ✓, cos ✓, 0) d✓
=
Z ⇡
0
2 cos2
✓ d✓
= 2⇡ .
Example (Compare with Example 2 in the textbook). Consider the vector field
F(x, y, z) = (2y, ez2
, ey2
) .
67

68. Let S be the portion of the paraboloid z = 1 (x2
+ y2
) with z 0; orient S with upward
pointing normal vectors. Compute
ZZ
S
(r ⇥ F) · dS .
Solution: First compute the curl of F:
r ⇥ F = (2yey2
2zez2
, 0, 2) .
To compute
RR
S
(r ⇥ F) · dS directly, we would parametrize S by
X(s, t) = (s, t, 1 (s2
+ t2
)) , (s, t) 2 D = {(s, t) | s2
+ t2
 1} .
The normal vector would be
N = (2s, 2t, 1) .
Then the integral would be
ZZ
S
(r ⇥ F) · dS =
ZZ
D
(2tet2
2(1 (s2
+ t2
))e(1 (s2+t2))2
, 0, 2) · (2s, 2t, 1) ds dt
=
ZZ
D
(2tet2
2(1 (s2
+ t2
))e(1 (s2+t2))2
)(2s) 2 ds dt
Then to evaluate the integral, use polar coordinates on D. This computation would be
extremely unpleasant to do by-hand or even with a computer.
The remedy for this complication is to pass to a simpler surface S0
that has the same oriented
boundary as S, that is, @S0
= @S as oriented curves. By applying Stokes’s Theorem to S
and S0
we have
ZZ
S
(r ⇥ F) · dS =
I
@S
F · dS =
I
@S0
F · dS =
ZZ
S0
(r ⇥ F) · dS .
It is clear that the simpler surface S0
should be the disk {(x, y, 0) | x2
+ y2
 1} with the
natural parametrization (s, t, 0). The normal vector for S0
is N = (0, 0, 1), so
ZZ
S0
(r⇥F)·dS =
ZZ
S0
(2tet2
, 0, 2)·(0, 0, 1) ds dt =
ZZ
S0
2 ds dt = 2 Area(S0
) = 2⇡ .
Gauss’s Theorem (Divergence Theorem)
Definition (Closed surface). Let S be a union of finitely many bounded
parametrized surfaces. We call S closed if it has no boundary points.
68

69. Examples. The following surfaces are closed.
• The sphere; more generally, an ellipsoid.
• The torus.
• The boundary of the solid region bounded by the sphere x2
+ y2
+ z2
= 25 and the
cone z =
p
x2 + y2; this generalizes to many regions that we see in this clas.
Theorem (Gauss’s Theorem or the Divergence Theorem). Let W ⇢ R3
be a
bounded solid whose boundary @W consists of finitely many piecewise smooth, closed
surfaces with outward normal orientations. Let F be a C1
vector field on W. Then
ZZZ
W
r · F dV =
ZZ
@W
F · n dS .
Notice that the integral on the right calculates the flux of F through @W.
Keep in mind that the integral ZZ
@W
F · n dS
that appears in the Divergence Theorem is the same as the integral
ZZ
@W
F · dS .
Example. Let W be the solid between the paraboloid z = 9 x2
y2
and the xy–plane.
Verify the Divergence Theorem for the vector field
F(x, y, z) = (2x, 5y, 3z) .
Solution: First, we calculate
RRR
W
r · F dV :
ZZZ
W
r · F dV =
ZZZ
W
(2 + 5 + 3) dV =
ZZZ
W
10 dV .
Then evaluate the integral in cylindrical coordinates
ZZZ
W
10 dV =
Z 2⇡
0
Z 3
0
Z 9 r2
0
10r dz dr d✓ = 405⇡ .
Now we calculate
RR
@W
F · n dS. Keep in mind that this is just the usual flux integral
ZZ
@W
F · dS .
69

70. The boundary of W consists of two pieces
S1 = {(x, y, 0) | x2
+ y2
 9} and S2 = {(x, y, z) | z = 9 x2
y2
, z 0}
With S1 and S2 properly oriented, we can compute
RR
@W
F · dS as the sum
ZZ
@W
F · dS =
ZZ
S1
F · dS +
ZZ
S2
F · dS .
Let D dentote the closed disk {(s, t) | s2
+ t2
 9} ⇢ R2
.
For S1, we can use the parametrization
X1 : D ! R3
, X1(s, t) = (s, t, 0)
as long as we use the downward pointing normal vector
N1 = (0, 0, 1) .
Thus ZZ
S1
F · dS =
ZZ
D
(2s, 5t, 0) · (0, 0, 1) ds dt =
ZZ
D
0 ds dt = 0 .
For S2, we can use the parametrization
X2 : D ! R3
, X2(s, t) = (s, t, 9 s2
t2
)
with the upward pointing normal vector
N2 = (2s, 2t, 1) .
This yields
ZZ
S2
F · dS =
ZZ
S2
2s, 5t, 3(9 s2
t2
) · (2s, 2t, 1) ds dt
=
ZZ
D
(s2
+ 7t2
+ 27) ds dt
Then change to polar coordinates
s = r cos ✓ , t = r sin ✓
for (r, ✓) 2 [0, 3] ⇥ [0, 2⇡].
70

71. The integral is now
ZZ
S2
F · dS =
ZZ
D
(s2
+ 7t2
+ 27) ds dt
=
Z 2⇡
0
Z 3
0
(r2
cos2
✓ + 7r2
sin2
✓ + 27) r dr d✓
=
Z 2⇡
0
Z 3
0
(r2
+ 6r2
sin2
✓ + 27) r dr d✓
= 405⇡ .
Therefore ZZ
@W
F · dS = 0 + 405⇡ = 405⇡ .
Example. Let S be the boundary of the cube C = [ 1, 1]3
; orient S with normal vectors
that point into C. Consider the vector field
F(x, y, z) =
✓
x
⇢3
,
y
⇢3
,
z
⇢3
◆
where ⇢ = ⇢(x, y, z) =
p
x2 + y2 + z2. Use the Divergence Theorem to compute the surface
integral ZZ
S
F · dS .
Solution:
The denominator ⇢3
in the components of F is problematic. It makes F is not defined at the
origin, so we can’t just set
RR
S
F · dS equal to
RRR
C
r · F dV .
The good news is that the Divergence Theorem allows us to integrate F over large spheres
instead of over S:
First, we compute the divergence of F: Use the quotient rule to compute the necessary
partial derivatives:
@
@x
✓
x
⇢3
◆
=
⇢2
3x2
⇢5
@
@y
✓
y
⇢3
◆
=
⇢2
3y2
⇢5
@
@z
✓
z
⇢3
◆
=
⇢2
3z2
⇢5
.
71

72. It follows that
r · F =
@
@x
✓
x
⇢3
◆
+
@
@y
✓
y
⇢3
◆
+
@
@z
✓
z
⇢3
◆
=
3⇢2
3⇢2
⇢5
= 0 .
Let S0
⇢ R3
be the sphere
S0
= {(x, y, z) | ⇢ = 2} .
Let W ⇢ R3
be the solid region between S and S0
along with the surfaces S and S0
. Give
S0
the outward normal orientation.
Since the origin is not contained in W, the Divergence Theorem gives
ZZ
@W
F · dS =
ZZZ
W
r · F dV =
ZZZ
W
0 dV = 0 .
This implies the equality
ZZ
@W
F · dS =
ZZ
S
F · dS +
ZZ
S0
F · dS = 0 .
Hence ZZ
S
F · dS =
ZZ
S0
F · dS .
It is actually possible to avoid parametrizing S0
to calculate the integral because
the unit normal to S0
at a point (x, y, z) 2 S0
is just
n =
✓
x
⇢
,
y
⇢
,
z
⇢
◆
.
Then integral is
72

73. ZZ
S0
F · dS =
ZZ
S0
(F · n) dS
=
ZZ
S0
✓
x
⇢3
,
y
⇢3
,
z
⇢3
◆
·
✓
x
⇢
,
y
⇢
,
z
⇢
◆
dS
=
ZZ
S0
✓
x2
+ y2
+ z2
⇢4
◆
dS
=
ZZ
S0
✓
⇢2
⇢4
◆
dS
=
✓
1
⇢2
◆ ZZ
S0
dS
=
✓
1
⇢2
◆
Area(S0
)
=
✓
1
⇢2
◆
4⇡⇢2
= 4⇡ .
Therefore ZZ
S
F · dS = 4⇡ .
Remark: We know that ⇢ = 2 on S0
, but if we want to increase ⇢ to do another problem,
the calculations are completely similar. The point here is that ⇢ is a constant function on a
sphere centered at the origin.
73