CHAPTER 13
Multiple Integrals
1. Double Integrals
Definition (1.1 – Riemann Integral). Suppose f is continuous on [a, b].
Partition [a, b] into n subintervals
a = x0 < x1 < x2 < · · · < xn
1
< xn = b.
Let xi = xi xi 1 be the width of the i’th subinterval [xi 1, xi] and let the
norm of the partition kP k be the largest of the xi’s. Then the definite integral
of f on [a, b] is
Z b
n
X
f (x) dx = lim
f (ci) xi
a
kP k!0
i=1
providd the limit exists and is the same for all choices of the evaluation points
ci 2 [xi 1, xi] for i = 1, 2, . . . , n. In this case, we say f is integrable on [a, b].
112
1. DOUBLE INTEGRALS
113
Double Integrals over a Rectangle
Let f (x, y) be continuous on the rectangle R = {(x, y)|a  x  b, c  y  d}.
Partition R by laying a rectangular grid on top of R consisting of n smaller
rectangles R1, R2, . . . , Rn (not necessarily the same size).
Let Ai be the area of rectangle Ri and define the norm of the partition kP k
to be the largest diagonal of any rectangle in the partition.
Definition. The double integral of f over R is
ZZ
n
X
f (x, y) dA = lim
f (ui, vi) Ai
R
kP k!0
i=1
provided the limit exists and is the same for every choice of the evaluation points
(ui, vi) in Ri, for i = 1, 2, . . . , n. When this happens, we say f is integrable
over R.
114
13. MULTIPLE INTEGRALS
How to Compute
If f (x, y)
0, the volume below the surface is
ZZ
V =
f (x, y) dA.
R
Using slices parallel to the y-axis (as in Calculus II),
Z b
V =
A(x) dx
a
where A(x) is the area of the slice at x, which is
Z d
A(x) =
f (x, y) dy
c
by a partial integration with respect to y. Then,
#
Z b
Z b"Z d
Z bZ d
V =
A(x) dx =
f (x, y) dy dx =
f (x, y) dydx,
a
a
c
a
c
an iterated integral.
Symetrically, taking slices parallel to the x-axis,
#
Z d
Z d"Z b
Z dZ b
V =
A(y) dy =
f (x, y) dx dy =
f (x, y) dxdy.
c
c
a
c
a
1. DOUBLE INTEGRALS
115
Theorem (1.1 – Fubini’s Theorem). Suppose f is integrable over the
rectangle R = {(x, y)|a  x  b, c  y  d}. Then
ZZ
Z bZ d
Z dZ b
f (x, y) dA =
f (x, y) dydx =
f (x, y) dxdy.
R
a
c
c
a
Z ZExample. If R = {(x, y)|2  x  4, 0  y  2} = [2, 4] ⇥ [0, 2], find
(x2y 2 17) dA.
R
ZZ
(x2y 2
R
Example.
ZZ
Z 4Z
Z 4h 2 3
2i
xy
2 2
17) dA =
(x y
17) dydx =
17y dx
0
3
2
0
2
Z 4⇣ 2
⌘
4
8x
8x3
=
34 dx =
34x)
2
3
9
2
⇣ 512
⌘ ⇣ 64
⌘ 448
164
=
136
68 =
68 =
.
9
9
9
9
x+y
e
dA =
[0,ln 2]⇥[0,ln 3]
2
Z
0
=
Z
0
= ey
ln 3 Z ln 2
x+y
e
dxdy =
0
ln 3 ⇣
ln 2+y
e
ln 3
0
=3
Z
y
e
1=2
dy =
x+y
e
0
⌘
ln 3 h
Z
0
ln 3
ln 2 i
0
ey dy
dy
116
13. MULTIPLE INTEGRALS
Double Integrals over General Regions
For a general region R,we sum over inner partitions using the rectangles that
lie entirely within the region R.
Definition. For any function f (x, y) defined on a bounded region R 2 R2,
the double integral of f over R is
ZZ
n
X
f (x, y) dA = lim
f (ui, vi) Ai
R
kP k!0
i=1
provided the limit exists and is the same for every choice of evaluation points
(ui, vi) in Ri, for i = 1, 2, . . . , n. In this case, we say f is integrable over R.
1. DOUBLE INTEGRALS
117
How to Compute
(1) Vertically simple region:
Theorem (1.2). Suppose f (x, y) is continuous on R defined by
R = {(x, y)|a  x  b, g1(x)  y  g2(x)}
for continuous functions g1 and g2 where g1(x)  g2(x) for all x 2 [a, b].
Then
ZZ
Z b Z g2(x)
f (x, y) dA =
f (x, y) dydx.
R
a
g1 (x)
(1) Horizontally simple region:
Theorem (1.3). Suppose f (x, y) is continuous on R defined by
R = {(x, y)|c  y  d, h1(y)  x  h2(y)}
for continuous functions h1 and h2 where h1(y)  h2(y) for all y 2 [c, d].
Then
ZZ
Z d Z h2(y)
f (x, y) dA =
f (x, y) dxdy.
R
c
h1 (y)
118
13. MULTIPLE INTEGRALS
Theorem (1.4). Let f (x, y) and g(x, y) be integrable over R 2 R2 and
let c 2 R. Then
ZZ
ZZ
(1)
cf (x, y) dA = c
f (x, y) dA
R
R
ZZ
ZZ
ZZ
⇥
⇤
(2)
f (x, y) + g(x, y) dA =
f (x, y) dA +
g(x, y) dA
R
R
R
(3) If R = R1 [ R2 where R1 \ R2 = , then
ZZ
ZZ
ZZ
f (x, y) dA =
f (x, y) dA +
f (x, y) dA
R
R1
R2
Note. This means any region can be decomposed into vertically and horizontally simple regions.
1. DOUBLE INTEGRALS
Example.
Z 1Z
0
p
y
y
Z 1⇣ 2
x
(x + y) dxdy =
+ xy
2
0
119
p ⌘
y
y
dy =
Z 1 h⇣
Z 1⇣
⌘ ⇣ y2
⌘i
y
y
3y 2 ⌘
3/2
2
3/2
+y
+y
dy =
+y
dy =
2
2
2
2
0
0
y 2 2y 5/2 y 3 1 1 2 1
3
+
= +
=
4
5
2 0 4 5 2 20
p
This region is also vertically simple where x = y =) y = x and x = y =)
y = x2. Thus
Z 1 Z py
Z 1Z x
(x + y) dxdy =
(x + y) dydx.
0
y
0
x2
120
13. MULTIPLE INTEGRALS
ZZ
Example. Find
xy dA where R is bounded by x = y 2 and 3x+2y = 8.
R
Points of intersection:
(substitution) 3y 2 + 2y = 8 =) 3y 2 + 2y 8 = 0 =) (3y 4)(y + 2) = 0 =)
⇣ 16 4 ⌘
4
y = or y = 2. Thus
,
and (4, 2) are points of intersection, giviving
3
9 3
us a region both vertically and horizontally simple.
2
8
horizontal: h1(y) = y 2 and h2(y) =
y + =)
3
3
ZZ
Z 4/3 Z 2 y+ 8
Z 4/3 ⇣ 2
3
3
x y 23 y+ 83 ⌘
xy dA =
xy dxdy =
dy =
2
y
2
2
R
2
y
2
Z 4/3 h ⇣
Z
4/3 ⇣ 3
y
2
8 ⌘2 y 5 i
2y
16y 2 32y y 5 ⌘
y+
dy =
+
dy =
2
3
3
2
9
9
9
2
2
2
h y 4 16y 3 16y 2 y 6 i4/3
+
=
18
27
9
12 2
⇣ 128 1024 256 1024 ⌘ ⇣ 8 128 64 16 ⌘ 3200 200
13000
+
+
+
=
=
.
729
729
81
2187
9
27
9
3
2187
27
2187
vertical:
ZZ
Z 16/9 Z px
Z 4 Z 3 x+4
2
xy dA =
xy
dydx
+
xy dydx.
p
p
R
0
x
16/9
x
1. DOUBLE INTEGRALS
Example. Find
Z
Z 3Z
0
9
121
y sin(x2) dxdy
y2
y sin(x2) dx has no elementary antiderivative.
The region is both horizotally and vertically simple. Also, x = y 2 =) y =
Z 3Z 9
Z 9 Z px
y sin(x2) dxdy =
y sin(x2) dydx =
0
Z
y2
9⇣ 2
p ⌘
x
0
y
sin(x2)
dx =
0
2
0
du xdx
u = x2 du = 2xdx
=
4
2
Z 81
81 ⌘
1
1⇣
1
sin u du =
cos u
=
0
4 0
4
4
p
x.
0
Z
9
0
x sin(x2)
dx =
2
cos 81
( 1) =
1
cos 81
.
4
122
13. MULTIPLE INTEGRALS
2. Area, Volume, and Center of Mass
Consider a function f (x, y)
0 over a region R in the x-y plane and the
volume of the region lying beneath the surface and above R.
For vertically simple, we can think:
V =
Z bZ
a
g2 (x)
g1 (x)
For horizontally simple, we can think:
V =
Z dZ
c
f (x, y) |{z}
dy dx) .
| {z }
|{z}
height
width length
h2 (y)
h1 (y)
f (x, y) |{z}
dx dy) .
| {z }
|{z}
height
width length
For any R 2 R2, volume under the surface z = 1 is
ZZ
ZZ
V =
1 dA =
dA = (1)(Area of R) = Area of R.
R
R
2. AREA, VOLUME, AND CENTER OF MASS
123
Problem (Page 931 #6). Find the area bounded by y = x3 and y = x2.
The curves intersect where x3 = x2 or x3 x2 = x2(x 1) = 0, i.e., at (0, 0)
and (1, 1). We have a vertically (and horizontally) simple region:
ZZ
Z 1 Z x2
Z 1 h x2 i
A=
dA =
dydx =
y 3 dx =
x
3
R
0
x
0
Z 1
x3 x4 1 1 1
1
2
3
(x
x ) dx =
=
= .
3
4 0 3 4 12
0
124
13. MULTIPLE INTEGRALS
Problem (Page 931 #18). Find the volume of the solid bounded by z =
2x + y + 1, z = 2x, x = y 2, and x = 1.
The last two are cylinders whose cross sections in the x-y plane are shown in
the right half of the right graph. The first two are planes that intersect where
2x + y + 1 = 2x or 4x + y = 1, whose projection on the x-y plane is
shown by the decreasing line on right graph. On the side of the line where our
cylinders lie, since 2x + y + 1 (0,0) = 1 and 2x (0,0) = 0, z = 2x + y + 1 lies
above z = 2x. With the region bounded by the cylinders horizontally simple,
Z 1Z 1
Z 1Z 1
⇥
⇤
V =
(2x + y + 1) ( 2x) dxdy =
(4x + y + 1) dxdy =
Z
Z
1
1
1
1
y2
⇣
2x2 + yx + x
1
y2
⌘
dy =
Z
1
1
1
⇥
(2 + y + 1)
y2
⇤
(2y 4 + y 3 + y 2) dy =
2y 5 y 4 y 3 y 2
( 2y
y
y + y + 3) dy =
+ + 3y
5
4
3
2
1
⇣ 2 1 1 1
⌘ ⇣2 1 1 1
⌘
+ +3
+ +
3 =
5 4 3 2
5 4 3 2
151 ⇣ 121 ⌘ 272 68
=
= .
60
60
60
15
Maple. See solidvolume(13.2).mw or solidvolume(13.2).pdf.
4
3
2
1
1
=
2. AREA, VOLUME, AND CENTER OF MASS
125
Moments and Center of Mass
Consider a lamina (a thin, flat plate) in the shape of a region R 2 R2 whose
density varies throughout the plate.
Our goal is to find the center of mass. But first we need to find the total mass.
Assume we have a mass density function ⇢(x, y).
We partition the region, and if a subregion Ri is small enough, then ⇢(x, y) is
almost constant on Ri, whose mass is then
mi ⇡ ⇢(ui, vi) |{z}
Ai
| {z }
mass/unit area area
where (ui, vi) is an arbitrary point in Ri. Summing
n
X
m⇡
⇢(ui, vi) Ai.
i=1
Then the exact mass is
m = lim
kP k!0
n
X
i=1
⇢(ui, vi) Ai =
ZZ
R
⇢(x, y) dA.
126
13. MULTIPLE INTEGRALS
First Moments
With respect to the y-axis:
ZZ
n
X
My = lim
ui⇢(ui, vi) Ai =
x⇢(x, y) dA.
kP k!0
i=1
R
With respect to the x-axis:
ZZ
n
X
Mx = lim
vi⇢(ui, vi) Ai =
y⇢(x, y) dA.
kP k!0
i=1
Center of Mass
The center of mass is then (x, y) where
My
Mx
x=
and y =
.
m
m
R
2. AREA, VOLUME, AND CENTER OF MASS
127
Problem (Page 932 #32). Find the mass and center of mass of the lamina
bounded by y = x2 4 and y = 5, ⇢(x, y) = the square of the distance from
the y-axis.
x2
4 = 5 =) x2 = 9 =) x = ±3. Also, the region is vertically simple.
m=
Z 3
Z
Z
3
5
2
x dydx =
x2
3
4
2
4
Z
3
x dx = 3x
9x
3
Z
My =
Z 3
3
3
3
Z
5
x2
3
9x
3
x dydx =
4
Z
9
x dx = x4
4
5
3
3
⇣
x2y
x5
5
3
3
3
3
5
x2 4
dx =
⇣
= 81
⇣
x3y
x6 3
6
⌘
3
5
x2 4
=
⌘
Z
243 ⌘
5
dx =
⇣ 729
4
Z
3
3
h
5x2
3
3
⇣
h
5x3
243 ⌘
2
4
(x
i
4x ) dx =
2
243 ⌘ 324
81 +
=
5
5
5
(x
⇣ 729
4
i
4x ) dx =
3
243 ⌘
=0
2
128
13. MULTIPLE INTEGRALS
Thus x =
My
= 0.
m
Mx =
Z
3
Z
5
2
Z
3
⇣ x2y 2
5
⌘
x y dydx =
dx =
2 4
x
2
3
4
3
Z 3h
Z
3 h
25x2 x2(x2 4)2 i
25x2 ⇣ x6 8x4 16x2 ⌘i
dx =
+
dx =
2
2
2
2
2
2
3
3
Z 3⇣
1 6
9 2⌘i
x7 4x5 3x3 3
4
x + 4x + x
dx =
+
+
=
3
2
2
14
5
2
3
⇣ 2187 972 81 ⌘ ⇣ 2187 972 81 ⌘ 2754 ⇣ 2754 ⌘ 5508
+
+
=
=
14
5
2
14
5
2
35
35
35
5508
⇣
⌘
Mx
17
17
35
Thus y =
= 324 = , and the center of mass is 0, 7 .
m
7
5
x2
Maple. See center of mass(13.2).mw or center of mass(13.2).pdf.
3. Double Integrals in Polar Coordinates
Example. Find the volume below the surface f (x, y) = x2 + y 2 and above
the first two quadrants of the circle x2 + y 2 = 1.
This is a vertically simple region.
p
Z 1 Z p1 x2
Z 1⇣
3 1 x2 ⌘
y
V =
(x2 + y 2) dydx =
x2y +
dx =
0
3
1 0
1
Z 1⇣ p
⌘
1
2
2
3/2
x 1 x2 + (1 x )
dx = · · · ugly
3
1
3. DOUBLE INTEGRALS IN POLAR COORDINATES
129
We try polar coordinates instead.
Suppose we wish to integrate f (r, ✓), converted from f (x, y), over a region R
of the type
R = (r, ✓)|↵  ✓ 
and g1(✓)  r  g2(✓) ,
where 0  g1(✓)  g2(✓) for all ✓ 2 [↵, ], as seen below.
We make a grid of elementary polar regions (shown above on the right) and
again use an inner partition.
1
Let r = r1 + r2 , the average radius of r1 and r2.
2
A = area of outer sector area of inner sector
1
1 2
1
= r22 ✓
r1 ✓ = (r22 r12) ✓
2
2
2
1
= (r2 + r1)(r2 r1) ✓ = r r ✓.
2
130
13. MULTIPLE INTEGRALS
Then the volume above the ith elementary polar region and below the surface
is
Vi ⇡ f (ri, ✓i) |{z}
Ai = f (ri, ✓i)ri ri ✓i
| {z }
height
area of base
where (ri, ✓i) is a point in Ri and ri is the average radius in Ri.
Summing and taking the limit,
Z
n
X
V = lim
f (ri, ✓i)ri ri ✓i =
kP k!0
↵
i=1
Z
g2 (✓)
f (r, ✓) rdrd✓.
g1 (✓)
Theorem (3.1 – Fubini). Suppose f (r, ✓) is continuous on the region
R = (r, ✓)|↵  ✓ 
and g1(✓)  r  g2(✓)
where 0  g1(✓)  g2(✓) for all ✓ 2 [↵, ]. Then
ZZ
Z Z g2(✓)
f (r, ✓) dA =
f (r, ✓) rdrd✓.
R
↵
g1 (✓)
Example (continued). Since x2 + y 2 = r2, 0  ✓  ⇡, and 0  r  1,
Z 1 Z p1 x2
Z ⇡Z 1
V =
(x2 + y 2) dydx =
r2 rdrd✓ =
1 0
0
0
Z ⇡ ⇣ 4 1⌘
Z ⇡
r
1
1 ⇡ ⇡
d✓ =
d✓ = ✓ = .
4 0
4 0
4
0
0 4
3. DOUBLE INTEGRALS IN POLAR COORDINATES
131
Problem (Page 939 #6). Find the area inside r = 1 and outside r =
2 2 cos ✓.
1
⇡
2 2 cos ✓ = 1 =) 2 cos ✓ = 1 =) cos ✓ = =) ✓ = ± .
2
3
A=
Z
Z
⇡/3
Z
1
⇡/3 2 2 cos ✓
⇣
⇡/3 h
r drd✓ =
Z
⇡/3
⇡/3
⇣ r2
2
1
2 2 cos ✓
⌘
d✓ =
1
4 8 cos ✓ + 4 cos2 ✓ ⌘i
=
d✓
2
2
⇡/3
Z ⇡/3 ⇣
⌘
1
2
=
3 + 8 cos ✓ 4 cos ✓ d✓
2 ⇡/3
Z
⌘
i⇡/3
1 ⇡/3 h
1h
=
3 + 8 cos ✓ 2(1 + cos 2✓ d✓ =
5✓ + 8 sin ✓ sin 2✓
⇡/3
2 ⇡/3
2
p ⌘ ⇣
p ⌘i
p
p
p
1 h⇣ 5⇡
3
5⇡
3
5⇡ 7 3
=
+4 3
4 3+
=
+
2
3
2
3
2
3
2
132
13. MULTIPLE INTEGRALS
Example. Find the volume of the solid that lies under the paraboloid z =
x + y 2, above the xy-plane, and inside the cylinder x2 + y 2 = 2x.
2
Completing the square, (x
xy-plane.
1)2 + y 2 = 1 is the shadow of the cylinder in the
Changing to polar coordinates, the shadow of the cylinder is r2 = 2r cos ✓ or
r = 2 cos ✓, so
n
o
⇡
⇡
R = (r, ✓)
 ✓  and 0  r  2 cos ✓ .
2
2
ZZ
Z ⇡/2 Z 2 cos ✓
Z ⇡/2 ⇣ 4 2 cos ✓ ⌘
r
V =
(x2 + y 2) dA =
r2 rdrd✓ =
d✓ =
0
4
R
⇡/2 0
⇡/2
Z ⇡/2
Z ⇡/2
Z
h1
i⇡/2
3
4
4
3
2
=
4 cos ✓ d✓ = 4
cos ✓ d✓ |{z}
= 4 cos ✓ sin ✓ +
cos ✓ d✓
⇡/2
4
4
⇡/2
⇡/2
#60
Z
h
⇣1
⌘i⇡/2
1
3
= cos ✓ sin ✓ + 3 cos ✓ sin ✓ +
d✓
⇡/2
2
2
h
3
3 i⇡/2
3
= cos ✓ sin ✓ + cos ✓ sin ✓ + ✓
⇡/2
2
2
h 3⇡ ⇣ 3⇡ ⌘i 3⇡
=
= .
4
4
2
Maple. See polarint(13.3).mw or polarint(13.3).pdf.
5. TRIPLE INTEGRALS
133
5. Triple Integrals
As a first step, we wish to integrate f (x, y, z) over a box
Q = (x, y, z)|a  x  b, c  y  d, r  z  s .
We partition Q into sub-boxes by slicing it with planes parallel to the xy-, xz-,
and yz-planes. The volume of a sub-box Qi is then
Vi =
xi yi zi.
Definition. For any function f (x, y, z) defined on the box Q, the triple
integral of f over Q is
ZZZ
n
X
f (x, y, z) dV = lim
f (ui, vi, wi) Vi,
Q
kP k!0
i=1
provided the limit exists and is the same for every choice of evaluation points
(ui, vi, wi) in Qi for i = 1, 2 . . . , n. When this happens, we say f is integrable
over Q.
Theorem (5.1–Fubini). Suppose f (x, y, z) is continuous on the box
Q = (x, y, z)|a  x  b, c  y  d, r  z  s .
Then
ZZZ
Q
f (x, y, z) dV =
Z rZ dZ
s
c
a
b
f (x, y, z) dxdydz.
134
13. MULTIPLE INTEGRALS
Note. Any of the 6 possible orderings of dx, dy, and dz may be used, with
the limits of integration changing accordingly.
Example. For
Q = (x, y, z)|1  x  2, 1  y  1, 2  z  4 = [1, 2] ⇥ [ 1, 1] ⇥ [2, 4]
Z 4Z
1
ZZZ
⇣1
2
(x z
Q
3
2
2
y z) dV =
2⌘
x z xy z
1
2
1
Z 4Z 1 ⇣
7
z
2
1 3
Z 4 h⇣
7
1 ⌘ ⇣
z
z
3
3
2
Note.
3
Z 4Z
2
Z 4Z
1
Z
2
(x2z
y 2z) dxdydz =
1 1
1 h⇣
⌘ ⇣1
⌘i
8
2
2
dydz =
z 2y z
z y z dydz =
3
3
2
1
Z
⌘
4⇣
7
1 3 1 ⌘
2
y z dydz =
yz
y z
dz =
1
3
3
2
Z 4
4
7
1 ⌘i
2
z + z dz =
(4z) dz = 2z = 32 8 = 24
2
3
3
2
(1) If the region Q is not a box, the integral is the limit over all inner boxes.
(2) If Q = (x, y, z)|(x, y 2 R and g1(x, y)  z  g2(x, y) ,
ZZZ
Z Z Z g2(x,y)
f (x, y, z) dV =
f (x, y, z) dzdA.
Q
R
g1 (x,y)
Example. Find
ZZZ
5. TRIPLE INTEGRALS
135
(2x + 3y) dV where Q is the tetrahedron bounded
Q
by 2x + 3y + z = 6 and the coordinate planes. The intercepts of the plane
2x + 3y + z = 6 are x = 3, y = 2, and z = 6, as shown in the figure on the left
below.
Note that each point of the solid lies over the vertically simple triangular region
2
R in the xy-plane bounded by the x- and y-axes and the line y =
x + 2.
3
ZZZ
Z Z Z 6 2x 3y
(2x + 3y) dV =
(2x + 3y) dzdA =
R 0
ZZ Q
ZZ h
i
6 2x 3y
(2xz + 3yz)
dA =
2x(6 2x 3y) + 3y(6 2x 3y) dA =
0
R
R
ZZ
(12x 4x2 12xy + 18y 9y 2)dA =
Z 3Z
0
Z
0
Z
0
3
R
2 x+2
3
(12x
4x2
9y 2) dydx =
12xy + 18y
0
3
(12xy
2
4x y
2
6xy + 9y
2
2 x+2
3
3
3y )
0
dx =
8
8
8
( 8x2+24x+ x3 8x2 x2+16x2 24x+4x2+36+ x3 8x2+24x 24) dx =
3
3
9
Z 3⇣
⌘
3
8 3
2 4 4 3
2
x
4x + 12 dx = x
x + 12x = 18 36 + 36 = 18.
0
9
9
3
0
136
13. MULTIPLE INTEGRALS
Example. Find the volume of the solid formed by the intersection of the
cylinders x2 + z 2 = 1 and y 2 + z 2 = 1. (Think of the intersection of tunnels
running along the y- and x-axes.)
1) Look along x-axis (figure on left below):
The region is horizontally simple =) we have the integral for “length”
Z p1 z2
1 dy
p
1 z2
2) Look along y-axis (figure on right above):
The region is horizontally simple =) we have the integral for “area”
Z p1 z2 Z p1 z2
1 dydx
p
p
1 z2
1 z2
3) Integrate from z = 1 to z = 1.
Z 1 Z p1 z2 Z p1 z2
Z
V =
1 dydxdz =
p
p
1
Z
1
1
Z
Z
1 z2
p
1 z2
p
1 z2
1
4(1
1
1 z2
p
2 1
2
z 2 dxdz =
z ) dz = 4z
Z
4 3
z
3
1
1
1
1
1
Z
p
1 z2
p
1 z2
⇣ p
2x 1
1
⇣
= 4
p
1 z2 ⌘
dxdz =
p
1 z2
p
1 z2 ⌘
z 2 p 2 dz =
1 z
⇣
y
4⌘
3
⇣
4 ⌘ 16
4+
= .
3
3
5. TRIPLE INTEGRALS
137
Example. Find the solid whose volume is given by
Z 1 Z p1 z2 Z p1 x2 z2
dydxdz
p
0
1 x2 z 2
0
and rewrite it using a di↵erent innermost variable.
p
From the limits of integration for y, we get y = ± 1 x2 z 2, equations
defining the right and left hemispheres (in our standard view) of the sphere
x2 + y 2 + z 2 = 1 of radius 1 centered at the origin.
p
From the limits for x, x = 0 and x = 1 z 2, we are restricted to the
hemisphere with x 0.
Finally, the limits for z, z = 0 and z = 1, restrict us to the quarter sphere with
x 0 and above the xy-plane.
A di↵erent version of the integral is
Z 1 Z p1 x2 Z p1
0
p
1 x2
0
x2 y 2
dzdydx.
138
13. MULTIPLE INTEGRALS
Mass and Center of Mass
Suppose a solid Q has mass density function ⇢(x, y, z). As before,
ZZZ
m=
⇢(x, y, z) dV
Z Z ZQ
Myz =
x⇢(x, y, z) dV
Z Z ZQ
Mxz =
y⇢(x, y, z) dV
Z Z ZQ
Mxy =
z⇢(x, y, z) dV
Q
Myz
Mxz
Mxy
, y=
, z=
m
m
m
Maple. See triple integral(13.5).mw or triple integral(13.5).pdf.
x=
6. Cylindrical Coordinates
Just as (x, y, z) can represent each point in space, so can the coordinates
(r, ✓, z), with 0  r < 1, 0  ✓  2⇡, 1 < z < 1.
p
We still have x = r cos(✓), y = r sin(✓), and r =
x2 + y 2. These are
cylindrical coordinates.
6. CYLINDRICAL COORDINATES
139
Surfaces obtained by setting one coordinate equal to a constant:
⇡
(r, , z)
4
3⇡
(2, ✓, z)
(r, , z)
4
The volume element in cylindrical coordinates.
(1, ✓, z)
V = A z = rdrd✓dz
Other orders of integration are also possible.
(r, ✓, 1)
(r, ✓, 3)
140
13. MULTIPLE INTEGRALS
Example. Find the volume of the region R bounded by the paraboloids
z = x2 + y 2 and z = 36 3x2 3y 2, or z = r2 and z = 36 3r2.
Solving r2 = 36 3r2 to get r = 3, the paraboloids intersect at z = 9 and the
shadow of R on the r✓-plane is r = 3. Thus
ZZZ
Z 2⇡ Z 3 Z 36 3r2
Z 2⇡ Z 3 h 36 3r2 i
V =
1 dV =
r dzdrd✓ =
rz 2
drd✓ =
r
2
R
0
0
r
0
0
Z 2⇡ Z 3
Z 2⇡ Z 3
(36r 3r3 r3) drd✓ =
(36r 4r3) drd✓ =
0
0
0
0
Z 2⇡
Z
2⇡
3
2⇡
2
4
(18r
r ) d✓ =
(162 81) d✓ = 81✓ = 162⇡.
0
0
0
ZZZ
Problem (Page 962 #17). Set up
Q
2
0
f (x, y, z) dV in cylindrical coor-
dinates if Q is the region bounded by x = y + z 2 and x = 2
y2
z 2.
Here, because of the y 2 + z 2, we will use cylindrical coordinates in the yz-plane,
i.e., y = r cos ✓, z = r sin ✓, x = x.
Then we have x = r2 and x = 2 r2. Solving r2 = 2 r2, we get r = 1. So the
paraboloids intersect at x = 1 and the shadow of Q on the yz-plane is r = 1.
Then
ZZZ
Q
f (x, y, z) dV =
Z
0
2⇡
Z 1Z
0
2 r2
r2
f (x, r cos ✓, r sin ✓)r dxdrd✓.
6. CYLINDRICAL COORDINATES
Problem (Page 962 #32). Find
ZZZ
(2x
141
y) dV , where Q is the tetra-
Q
hedron bounded by 3x + y + 2z = 6 and the coordinate planes.
We get no help from cylindrical coordinates here.
ZZZ
y) dV =
(2x
Q
Z 2Z
0
Z 2Z
0
Z 2Z
0
6 3x h⇣
6 3x Z 3 32 x 12 y
0
6 3x
(2x
y) dzdydx =
0
yz)
(2xz
0
0
⇣
3y
dydx =
3
1 2⌘i
6x 3x
xy
xy
y
dydx =
2
2
0
Z 2 Z 6 3x ⇣
1
1 2⌘
2
6x 3x + xy 3y + y dydx =
2
2
0
0
Z 2⇣
1 2
1 3 6 3x⌘
2
2
12xy 3x y + xy
3y + y
dx =
0
4
6
0
Z 2⇣
⌘
27x3
2
63x + 135x 72 dx =
4
0
2
27x4
135x2
3
21x +
72x =
0
16
2
27 168 + 270 144 = 15
2
⌘
3 32 x 12 y
142
13. MULTIPLE INTEGRALS
Problem (Page 963 #44). Change to cylindrical coordiates and evaluate:
Z 1 Z p1 x2 Z 4
p
x2 + y 2 dzdydx =
0
Z
0
Z
0
⇡
2
0
Z 1Z
0
1 x2 y 2
4
1 r2
rr dzdrd✓ =
Z
0
⇡
2
Z 1⇣
r2z
0
4
1 r2
⌘
drd✓ =
Z 1h
Z ⇡Z 1
i
2
2
2
4
4r
(r
r ) drd✓ =
(r4 + 3r2) drd✓ =
0
0
0
⇡
Z ⇡⇣ 5
Z
⇡
⌘
1⌘
2
2 ⇣1
r
6
3⇡
2
+ r3 d✓ =
+ 1 d✓ = ✓ = .
0
5
5
5 0
5
0
0
⇡
2
Maple. See cylindrical(13.6).mw or cylindrical(13.6).pdf.