Multiple Integrals - Christian Brothers Universityfacstaff.cbu.edu/wschrein/media/M232...
Transcript of Multiple Integrals - Christian Brothers Universityfacstaff.cbu.edu/wschrein/media/M232...
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 thenorm of the partition kPk be the largest of the �xi’s. Then the definite integralof f on [a, b] is Z b
af(x) dx = lim
kPk!0
nXi=1
f(ci)�xi
providd the limit exists and is the same for all choices of the evaluation pointsci 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 smallerrectangles R1, R2, . . . , Rn (not necessarily the same size).
Let �Ai be the area of rectangle Ri and define the norm of the partition kPkto be the largest diagonal of any rectangle in the partition.
Definition. The double integral of f over R isZZR
f(x, y) dA = limkPk!0
nXi=1
f(ui, vi) �Ai
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 integrableover R.
114 13. MULTIPLE INTEGRALS
How to Compute
If f(x, y) � 0, the volume below the surface is
V =
ZZR
f(x, y) dA.
Using slices parallel to the y-axis (as in Calculus II),
V =
Z b
aA(x) dx
where A(x) is the area of the slice at x, which is
A(x) =
Z d
cf(x, y) dy
by a partial integration with respect to y. Then,
V =
Z b
aA(x) dx =
Z b
a
"Z d
cf(x, y) dy
#dx =
Z b
a
Z d
cf(x, y) dydx,
an iterated integral.
Symetrically, taking slices parallel to the x-axis,
V =
Z d
cA(y) dy =
Z d
c
"Z b
af(x, y) dx
#dy =
Z d
c
Z b
af(x, y) dxdy.
1. DOUBLE INTEGRALS 115
Theorem (1.1 – Fubini’s Theorem). Suppose f is integrable over therectangle R = {(x, y)|a x b, c y d}. ThenZZ
Rf(x, y) dA =
Z b
a
Z d
cf(x, y) dydx =
Z d
c
Z b
af(x, y) dxdy.
Example. If R = {(x, y)|2 x 4, 0 y 2} = [2, 4] ⇥ [0, 2], findZZR(x2y2 � 17) dA.
ZZR(x2y2 � 17) dA =
Z 4
2
Z 2
0(x2y2 � 17) dydx =
Z 4
2
hx2y3
3� 17y
���20
idx
=
Z 4
2
⇣8x2
3� 34
⌘dx =
8x3
9� 34x)
���42
=⇣512
9� 136
⌘�⇣64
9� 68
⌘=
448
9� 68 = �164
9.
Example.
ZZ[0,ln 2]⇥[0,ln 3]
ex+y dA =
Z ln 3
0
Z ln 2
0ex+y dxdy =
Z ln 3
0
hex+y
���ln 2
0
idy
=
Z ln 3
0
⇣eln 2+y � ey
⌘dy =
Z ln 3
0ey dy
= ey���ln 3
0= 3� 1 = 2
116 13. MULTIPLE INTEGRALS
Double Integrals over General Regions
For a general region R,we sum over inner partitions using the rectangles thatlie 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 isZZ
Rf(x, y) dA = lim
kPk!0
nXi=1
f(ui, vi)�Ai
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
Rf(x, y) dA =
Z b
a
Z g2(x)
g1(x)f(x, y) dydx.
(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
Rf(x, y) dA =
Z d
c
Z h2(y)
h1(y)f(x, y) dxdy.
118 13. MULTIPLE INTEGRALS
Theorem (1.4). Let f(x, y) and g(x, y) be integrable over R 2 R2 andlet c 2 R. Then
(1)
ZZR
cf(x, y) dA = c
ZZR
f(x, y) dA
(2)
ZZR
⇥f(x, y) + g(x, y)
⇤dA =
ZZR
f(x, y) dA +
ZZR
g(x, y) dA
(3) If R = R1 [R2 where R1 \R2 = �, thenZZR
f(x, y) dA =
ZZR1
f(x, y) dA +
ZZR2
f(x, y) dA
Note. This means any region can be decomposed into vertically and hori-zontally simple regions.
1. DOUBLE INTEGRALS 119
Example. Z 1
0
Z py
y(x + y) dxdy =
Z 1
0
⇣x2
2+ xy
���p
y
y
⌘dy =
Z 1
0
h⇣y
2+ y3/2
⌘�⇣y2
2+ y2
⌘idy =
Z 1
0
⇣y
2+ y3/2 � 3y2
2
⌘dy =
y2
4+
2y5/2
5� y3
2
���10
=1
4+
2
5� 1
2=
3
20This region is also vertically simple where x = y =) y = x and x =
py =)
y = x2. Thus Z 1
0
Z py
y(x + y) dxdy =
Z 1
0
Z x
x2(x + y) dydx.
120 13. MULTIPLE INTEGRALS
Example. Find
ZZR
xy dA where R is bounded by x = y2 and 3x+2y = 8.
Points of intersection:
(substitution) 3y2 + 2y = 8 =) 3y2 + 2y� 8 = 0 =) (3y� 4)(y + 2) = 0 =)y =
4
3or y = �2. Thus
⇣16
9,4
3
⌘and (4,�2) are points of intersection, giviving
us a region both vertically and horizontally simple.
horizontal: h1(y) = y2 and h2(y) = �2
3y +
8
3=)
ZZR
xy dA =
Z 4/3
�2
Z �23 y+8
3
y2xy dxdy =
Z 4/3
�2
⇣x2y
2
����23 y+8
3
y2
⌘dy =
Z 4/3
�2
hy2
⇣� 2
3y +
8
3
⌘2� y5
2
idy =
Z 4/3
�2
⇣2y3
9� 16y2
9+
32y
9� y5
2
⌘dy =
hy4
18� 16y3
27+
16y2
9� y6
12
i4/3
�2=
⇣128
729� 1024
729+
256
81� 1024
2187
⌘�⇣8
9+
128
27+
64
9� 16
3
⌘=
3200
2187� 200
27= �13000
2187.
vertical: ZZR
xy dA =
Z 16/9
0
Z px
�pxxy dydx +
Z 4
16/9
Z �32 x+4
�pxxy dydx.
1. DOUBLE INTEGRALS 121
Example. Find Z 3
0
Z 9
y2y sin(x2) dxdy
Zy sin(x2) dx has no elementary antiderivative.
The region is both horizotally and vertically simple. Also, x = y2 =) y =p
x.Z 3
0
Z 9
y2y sin(x2) dxdy =
Z 9
0
Z px
0y sin(x2) dydx =
Z 9
0
⇣y2
2sin(x2)
���p
x
0
⌘dx =
Z 9
0
x sin(x2)
2dx =
u = x2 du = 2xdxdu
4=
xdx
2
1
4
Z 81
0sin u du =
1
4
⇣� cos u
���81
0
⌘=
1
4
�� 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 thevolume of the region lying beneath the surface and above R.
For vertically simple, we can think:
V =
Z b
a
Z g2(x)
g1(x)f(x, y)| {z }
height
dy|{z}width
dx)|{z}length
.
For horizontally simple, we can think:
V =
Z d
c
Z h2(y)
h1(y)f(x, y)| {z }
height
dx|{z}width
dy)|{z}length
.
For any R 2 R2, volume under the surface z = 1 is
V =
ZZR
1 dA =
ZZR
dA = (1)(Area of R) = Area of 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:
A =
ZZR
dA =
Z 1
0
Z x2
x3dydx =
Z 1
0
hy���x2
x3
idx =
Z 1
0(x2 � x3) dx =
x3
3� x4
4
���10
=1
3� 1
4=
1
12.
124 13. MULTIPLE INTEGRALS
Problem (Page 931 #18). Find the volume of the solid bounded by z =2x + y + 1, z = �2x, x = y2, and x = 1.
The last two are cylinders whose cross sections in the x-y plane are shown inthe right half of the right graph. The first two are planes that intersect where2x + y + 1 = �2x or 4x + y = �1, whose projection on the x-y plane isshown by the decreasing line on right graph. On the side of the line where ourcylinders lie, since 2x + y + 1
��(0,0)
= 1 and �2x��(0,0)
= 0, z = 2x + y + 1 liesabove z = �2x. With the region bounded by the cylinders horizontally simple,
V =
Z 1
�1
Z 1
y2
⇥(2x + y + 1)� (�2x)
⇤dxdy =
Z 1
�1
Z 1
y2(4x + y + 1) dxdy =
Z 1
�1
⇣2x2 + yx + x
���1y2
⌘dy =
Z 1
�1
⇥(2 + y + 1)� (2y4 + y3 + y2)
⇤dy =
Z 1
�1(�2y4 � y3 � y2 + y + 3) dy = �2y5
5� y4
4� y3
3+
y2
2+ 3y
���1�1
=
⇣� 2
5� 1
4� 1
3+
1
2+ 3
⌘�⇣2
5� 1
4+
1
3+
1
2� 3
⌘=
151
60�⇣� 121
60
⌘=
272
60=
68
15.
Maple. See solidvolume(13.2).mw or solidvolume(13.2).pdf.
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 whosedensity 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) isalmost constant on Ri, whose mass is then
mi ⇡ ⇢(ui, vi)| {z }mass/unit area
�Ai|{z}area
where (ui, vi) is an arbitrary point in Ri. Summing
m ⇡nX
i=1
⇢(ui, vi)�Ai.
Then the exact mass is
m = limkPk!0
nXi=1
⇢(ui, vi)�Ai =
ZZR
⇢(x, y) dA.
126 13. MULTIPLE INTEGRALS
First Moments
With respect to the y-axis:
My = limkPk!0
nXi=1
ui⇢(ui, vi)�Ai =
ZZR
x⇢(x, y) dA.
With respect to the x-axis:
Mx = limkPk!0
nXi=1
vi⇢(ui, vi)�Ai =
ZZR
y⇢(x, y) dA.
Center of Mass
The center of mass is then (x, y) where
x =My
mand y =
Mx
m.
2. AREA, VOLUME, AND CENTER OF MASS 127
Problem (Page 932 #32). Find the mass and center of mass of the laminabounded by y = x2 � 4 and y = 5, ⇢(x, y) = the square of the distance fromthe y-axis.
x2 � 4 = 5 =) x2 = 9 =) x = ±3. Also, the region is vertically simple.
m =
Z 3
�3
Z 5
x2�4x2 dydx =
Z 3
�3
⇣x2y
���5x2�4
⌘dx =
Z 3
�3
h5x2 � (x4 � 4x2)
idx =
Z 3
�3
�9x2 � x4
�dx = 3x3 � x5
5
���3�3
=⇣81� 243
5
⌘�⇣� 81 +
243
5
⌘=
324
5
My =
Z 3
�3
Z 5
x2�4x3 dydx =
Z 3
�3
⇣x3y
���5x2�4
⌘dx =
Z 3
�3
h5x3 � (x5 � 4x3)
idx =
Z 3
�3
�9x3 � x5
�dx =
9
4x4 � x6
6
���3�3
=⇣729
4� 243
2
⌘�⇣729
4� 243
2
⌘= 0
128 13. MULTIPLE INTEGRALS
Thus x =My
m= 0.
Mx =
Z 3
�3
Z 5
x2�4x2y dydx =
Z 3
�3
⇣x2y2
2
���5x2�4
⌘dx =
Z 3
�3
h25x2
2� x2(x2 � 4)2
2
idx =
Z 3
�3
h25x2
2�⇣x6
2� 8x4
2+
16x2
2
⌘idx =
Z 3
�3
⇣� 1
2x6 + 4x4 +
9
2x2⌘i
dx = �x7
14+
4x5
5+
3x3
2
���3�3
=
⇣� 2187
14+
972
5+
81
2
⌘�⇣2187
14� 972
5� 81
2
⌘=
2754
35�⇣� 2754
35
⌘=
5508
35
Thus y =Mx
m=
5508353245
=17
7, and the center of mass is
⇣0, 17
7
⌘.
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 + y2 and abovethe first two quadrants of the circle x2 + y2 = 1.
This is a vertically simple region.
V =
Z 1
�1
Z p1�x2
0(x2 + y2) dydx =
Z 1
�1
⇣x2y +
y3
3
���p
1�x2
0
⌘dx =
Z 1
�1
⇣x2p
1� x2 +1
3(1� x2)3/2
⌘dx = · · · ugly
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 Rof 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) andagain use an inner partition.
Let r =1
2
�r1 + r2
�, the average radius of r1 and r2.
�A = area of outer sector� area of inner sector
=1
2r22�✓ � 1
2r21�✓ =
1
2(r2
2 � r21)�✓
=1
2(r2 + r1)(r2 � r1)�✓ = r�r�✓.
130 13. MULTIPLE INTEGRALS
Then the volume above the ith elementary polar region and below the surfaceis
Vi ⇡ f(ri, ✓i)| {z }height
�Ai|{z}area of base
= f(ri, ✓i)ri�ri�✓i
where (ri, ✓i) is a point in Ri and ri is the average radius in Ri.
Summing and taking the limit,
V = limkPk!0
nXi=1
f(ri, ✓i)ri�ri�✓i =
Z �
↵
Z g2(✓)
g1(✓)f(r, ✓) rdrd✓.
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 [↵,�]. ThenZZ
Rf(r, ✓) dA =
Z �
↵
Z g2(✓)
g1(✓)f(r, ✓) rdrd✓.
Example (continued). Since x2 + y2 = r2, 0 ✓ ⇡, and 0 r 1,
V =
Z 1
�1
Z p1�x2
0(x2 + y2) dydx =
Z ⇡
0
Z 1
0r2 rdrd✓ =
Z ⇡
0
⇣r4
4
���10
⌘d✓ =
Z ⇡
0
1
4d✓ =
1
4✓���⇡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 ✓.
2� 2 cos ✓ = 1 =) 2 cos ✓ = 1 =) cos ✓ =1
2=) ✓ = ±⇡
3.
A =
Z ⇡/3
�⇡/3
Z 1
2�2 cos ✓r drd✓ =
Z ⇡/3
�⇡/3
⇣r2
2
���12�2 cos ✓
⌘d✓ =
=
Z ⇡/3
�⇡/3
h1
2�⇣4� 8 cos ✓ + 4 cos2 ✓
2
⌘id✓
=1
2
Z ⇡/3
�⇡/3
⇣� 3 + 8 cos ✓ � 4 cos2 ✓
⌘d✓
=1
2
Z ⇡/3
�⇡/3
h� 3 + 8 cos ✓ � 2(1 + cos 2✓
⌘d✓ =
1
2
h� 5✓ + 8 sin ✓ � sin 2✓
i⇡/3
�⇡/3
=1
2
h⇣� 5⇡
3+ 4p
3�p
3
2
⌘�⇣5⇡
3� 4p
3 +
p3
2
⌘i= �5⇡
3+
7p
3
2
132 13. MULTIPLE INTEGRALS
Example. Find the volume of the solid that lies under the paraboloid z =x2 + y2, above the xy-plane, and inside the cylinder x2 + y2 = 2x.
Completing the square, (x� 1)2 + y2 = 1 is the shadow of the cylinder in thexy-plane.
Changing to polar coordinates, the shadow of the cylinder is r2 = 2r cos ✓ orr = 2 cos ✓, so
R =n
(r, ✓)���� ⇡
2 ✓ ⇡
2and 0 r 2 cos ✓
o.
V =
ZZR(x2 + y2) dA =
Z ⇡/2
�⇡/2
Z 2 cos ✓
0r2 rdrd✓ =
Z ⇡/2
�⇡/2
⇣r4
4
���2 cos ✓
0
⌘d✓ =
=
Z ⇡/2
�⇡/24 cos4 ✓ d✓ = 4
Z ⇡/2
�⇡/2cos4 ✓ d✓ =|{z}
#60
4h1
4cos3 ✓ sin ✓ +
3
4
Zcos2 ✓ d✓
i⇡/2
�⇡/2
=h
cos3 ✓ sin ✓ + 3⇣1
2cos ✓ sin ✓ +
1
2
Zd✓⌘i⇡/2
�⇡/2
=h
cos3 ✓ sin ✓ +3
2cos ✓ sin ✓ +
3
2✓i⇡/2
�⇡/2
=h3⇡
4�⇣� 3⇡
4
⌘i=
3⇡
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 tripleintegral of f over Q isZZZ
Qf(x, y, z) dV = lim
kPk!0
nXi=1
f(ui, vi, wi) �Vi,
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 integrableover 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 ZZZQ
f(x, y, z) dV =
Z r
s
Z d
c
Z b
af(x, y, z) dxdydz.
134 13. MULTIPLE INTEGRALS
Note. Any of the 6 possible orderings of dx, dy, and dz may be used, withthe 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]
ZZZQ(x2z � y2z) dV =
Z 4
2
Z 1
�1
Z 2
1(x2z � y2z) dxdydz =
Z 4
2
Z 1
�1
⇣1
3x3z � xy2z
���21
⌘dydz =
Z 4
2
Z 1
�1
h⇣8
3z � 2y2z
⌘�⇣1
3z � y2z
⌘idydz =
Z 4
2
Z 1
�1
⇣7
3z � y2z
⌘dydz =
Z 4
2
⇣7
3yz � 1
3y3z
���1�1
⌘dz =
Z 4
2
h⇣7
3z � 1
3z⌘�⇣� 7
3z +
1
3z⌘i
dz =
Z 4
2(4z) dz = 2z2
���42
= 32� 8 = 24
Note.
(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)
,
ZZZQ
f(x, y, z) dV =
ZZR
Z g2(x,y)
g1(x,y)f(x, y, z) dzdA.
5. TRIPLE INTEGRALS 135
Example. Find
ZZZQ(2x + 3y) dV where Q is the tetrahedron bounded
by 2x + 3y + z = 6 and the coordinate planes. The intercepts of the plane2x + 3y + z = 6 are x = 3, y = 2, and z = 6, as shown in the figure on the leftbelow.
Note that each point of the solid lies over the vertically simple triangular region
R in the xy-plane bounded by the x- and y-axes and the line y = �2
3x + 2.
ZZZQ(2x + 3y) dV =
ZZR
Z 6�2x�3y
0(2x + 3y) dzdA =
ZZR(2xz + 3yz)
���6�2x�3y
0dA =
ZZR
h2x(6� 2x� 3y) + 3y(6� 2x� 3y)
idA =ZZ
R(12x� 4x2 � 12xy + 18y � 9y2)dA =
Z 3
0
Z �23x+2
0(12x� 4x2 � 12xy + 18y � 9y2) dydx =
Z 3
0(12xy � 4x2y � 6xy2 + 9y2 � 3y3)
����23x+2
0dx =
Z 3
0(�8x2+24x+
8
3x3�8x2�8
3x2+16x2�24x+4x2+36+
8
9x3�8x2+24x�24) dx =
Z 3
0
⇣8
9x3 � 4x2 + 12
⌘dx =
2
9x4 � 4
3x3 + 12x
���30
= 18� 36 + 36 = 18.
136 13. MULTIPLE INTEGRALS
Example. Find the volume of the solid formed by the intersection of thecylinders x2 + z2 = 1 and y2 + z2 = 1. (Think of the intersection of tunnelsrunning 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
�p
1�z21 dy
2) Look along y-axis (figure on right above):
The region is horizontally simple =) we have the integral for “area”Z p1�z2
�p
1�z2
Z p1�z2
�p
1�z21 dydx
3) Integrate from z = �1 to z = 1.
V =
Z 1
�1
Z p1�z2
�p
1�z2
Z p1�z2
�p
1�z21 dydxdz =
Z 1
�1
Z p1�z2
�p
1�z2
⇣y���p
1�z2
�p
1�z2
⌘dxdz =
Z 1
�1
Z p1�z2
�p
1�z22p
1� z2 dxdz =
Z 1
�1
⇣2xp
1� z2���p
1�z2
�p
1�z2
⌘dz =
Z 1
�14(1� z2) dz = 4z � 4
3z3���1�1
=⇣4� 4
3
⌘�⇣� 4 +
4
3
⌘=
16
3.
5. TRIPLE INTEGRALS 137
Example. Find the solid whose volume is given byZ 1
0
Z p1�z2
0
Z p1�x2�z2
�p
1�x2�z2dydxdz
and rewrite it using a di↵erent innermost variable.
From the limits of integration for y, we get y = ±p
1� x2 � z2, equationsdefining the right and left hemispheres (in our standard view) of the spherex2 + y2 + z2 = 1 of radius 1 centered at the origin.
From the limits for x, x = 0 and x =p
1� z2, we are restricted to thehemisphere with x � 0.
Finally, the limits for z, z = 0 and z = 1, restrict us to the quarter sphere withx � 0 and above the xy-plane.
A di↵erent version of the integral isZ 1
0
Z p1�x2
�p
1�x2
Z p1�x2�y2
0dzdydx.
138 13. MULTIPLE INTEGRALS
Mass and Center of Mass
Suppose a solid Q has mass density function ⇢(x, y, z). As before,
m =
ZZZQ
⇢(x, y, z) dV
Myz =
ZZZQ
x⇢(x, y, z) dV
Mxz =
ZZZQ
y⇢(x, y, z) dV
Mxy =
ZZZQ
z⇢(x, y, z) dV
x =Myz
m, y =
Mxz
m, z =
Mxy
m
Maple. See triple integral(13.5).mw or triple integral(13.5).pdf.
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.
We still have x = r cos(✓), y = r sin(✓), and r =p
x2 + y2. These arecylindrical coordinates.
6. CYLINDRICAL COORDINATES 139
Surfaces obtained by setting one coordinate equal to a constant:
(1, ✓, z) (r,⇡
4, z) (r, ✓,�1)
(2, ✓, z) (r,3⇡
4, z) (r, ✓, 3)
The volume element in cylindrical coordinates.
�V = �A�z = rdrd✓dz
Other orders of integration are also possible.
140 13. MULTIPLE INTEGRALS
Example. Find the volume of the region R bounded by the paraboloidsz = x2 + y2 and z = 36� 3x2 � 3y2, or z = r2 and z = 36� 3r2.
Solving r2 = 36� 3r2 to get r = 3, the paraboloids intersect at z = 9 and theshadow of R on the r✓-plane is r = 3. Thus
V =
ZZZR
1 dV =
Z 2⇡
0
Z 3
0
Z 36�3r2
r2r dzdrd✓ =
Z 2⇡
0
Z 3
0
hrz���36�3r2
r2
idrd✓ =
Z 2⇡
0
Z 3
0(36r � 3r3 � r3) drd✓ =
Z 2⇡
0
Z 3
0(36r � 4r3) drd✓ =
Z 2⇡
0(18r2 � r4)
���30d✓ =
Z 2⇡
0(162� 81) d✓ = 81✓
���2⇡0
= 162⇡.
Problem (Page 962 #17). Set up
ZZZQ
f(x, y, z) dV in cylindrical coor-
dinates if Q is the region bounded by x = y2 + z2 and x = 2� y2 � z2.
Here, because of the y2 +z2, 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 theparaboloids intersect at x = 1 and the shadow of Q on the yz-plane is r = 1.
ThenZZZQ
f(x, y, z) dV =
Z 2⇡
0
Z 1
0
Z 2�r2
r2f(x, r cos ✓, r sin ✓)r dxdrd✓.
6. CYLINDRICAL COORDINATES 141
Problem (Page 962 #32). Find
ZZZQ(2x� y) dV , where Q is the tetra-
hedron bounded by 3x + y + 2z = 6 and the coordinate planes.
We get no help from cylindrical coordinates here.
ZZZQ(2x� y) dV =
Z 2
0
Z 6�3x
0
Z 3�32x�1
2y
0(2x� y) dzdydx =
Z 2
0
Z 6�3x
0(2xz � yz)
���3�32x�1
2y
0dydx =
Z 2
0
Z 6�3x
0
h⇣6x� 3x2 � xy
⌘�⇣3y � 3
2xy � 1
2y2⌘i
dydx =
Z 2
0
Z 6�3x
0
⇣6x� 3x2 +
1
2xy � 3y +
1
2y2⌘
dydx =
Z 2
0
⇣12xy � 3x2y +
1
4xy2 � 3y2 +
1
6y3���6�3x
0
⌘dx =
Z 2
0
⇣27x3
4� 63x2 + 135x� 72
⌘dx =
27x4
16� 21x3 +
135x2
2� 72x
���20
=
27� 168 + 270� 144 = �15
142 13. MULTIPLE INTEGRALS
Problem (Page 963 #44). Change to cylindrical coordiates and evaluate:Z 1
0
Z p1�x2
0
Z 4
1�x2�y2
px2 + y2 dzdydx =
Z ⇡2
0
Z 1
0
Z 4
1�r2rr dzdrd✓ =
Z ⇡2
0
Z 1
0
⇣r2z
���41�r2
⌘drd✓ =
Z ⇡2
0
Z 1
0
h4r2 � (r2 � r4)
idrd✓ =
Z ⇡2
0
Z 1
0(r4 + 3r2) drd✓ =
Z ⇡2
0
⇣r5
5+ r3
���10
⌘d✓ =
Z ⇡2
0
⇣1
5+ 1
⌘d✓ =
6
5✓���⇡
2
0=
3⇡
5.
Maple. See cylindrical(13.6).mw or cylindrical(13.6).pdf.