SECTION 12.5
12.5
TRIPLE INTEGRALS
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1
TRIPLE INTEGRALS
A Click here for answers.
S Click here for solutions.
1. Evaluate the integral xxxE x 2 yz dV , where
11.
E x, y, z 0 x 2, 3 y 0, 1 z 1
where E is bounded by the planes x 0, y 0,
z 0, y z 1, and x z 1
xxxE z dV,
12–15
using three different orders of integration.
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Use a triple integral to find the volume of the given solid.
12. The tetrahedron bounded by the coordinate planes and the
2–6
z
y
0
0
yyy
4.
y yy
6.
yyy
xyz dx dy dz
2
s4z 2
0
0
0
2
x
1y
1
0
0
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7–11
7.
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z sin y dx dz dy
1
2x
3.
yy y
5.
yy
0
x
3
s9x 2
0
0
xy
y
x
0
14. The solid bounded by the cylinder x y 2 and the planes z 0
yz dy dz dx
and x z 1
15. The solid enclosed by the paraboloids z x 2 y 2 and
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z 18 x 2 y 2
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Evaluate the triple integral.
where
E x, y, z 0 z 1, 0 y 2z, 0 x z 2
xxxE e x dV,
9.
where E lies under the plane z x 2y and
above the region in the xy-plane bounded by the curves
y x 2, y 0, and x 1
where
E x, y, z 0 y 1, 0 x y, 0 z x y
xxxE y dV,
where E is bounded by the planes x 0, y 0,
z 0, and 3x 2y z 6
xxxE x dV,
the planes y 0 and y z 2
x 3y 2z dz dy dx
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xxxE yz dV,
13. The solid bounded by the elliptic cylinder 4x 2 z 2 4 and
2xy dz dy dx
0
8.
10.
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1
0
2.
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plane 2x 3y 6z 12
Evaluate the iterated integral.
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CAS
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16. Evaluate the triple integral exactly:
2
sin x
yy y
0
1
zx
zx
e 3x5y 2z dy dz dx
17. Set up, but do not evaluate, integral expressions for
(a) the mass, (b) the center of mass, and (c) the moment of
inertia about the z -axis of the solid bounded by the paraboloid
x 4y 2 4z 2 and the plane x 4 with density function
x, y, z x 2 y 2 z 2.
18. Find the average value of the function f x, y, z x y z
over the tetrahedron with vertices 0, 0, 0, 1, 0, 0, 0, 1, 0,
and 0, 0, 1.
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2
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SECTION 12.5
12.5
TRIPLE INTEGRALS
ANSWERS
E Click here for exercises.
S Click here for solutions.
1. 16
8
14. 15
1
2. 48
15. 81π
3. 23
15
16. e
4. 16
3
749
− 511
cos 4 − 140
sin 4 − 1521
− 35
18
338
169
√
1 1−y2 4
2
√
17. (a) −1
x + y 2 + z 2 dx dz dy
2 4y 2 +4z 2
6
−
5. 162
5
7. 75
8. 12 (7 − 2e)
5
9. 28
10. 3
1
11. 12
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13. 4π
1−y
(b) (x, y, z) where
1 √1−y2 4
1
√
x= m
x x2 + y 2 + z 2 dx dz dy
−1 − 1−y2 4y 2 +4z 2
1 √1−y2 4
1
√
y= m
y x2 + y 2 + z 2 dx dz dy
−1 − 1−y2 4y 2 +4z 2
1 √1−y2 4
1
√
z= m
z x2 + y 2 + z 2 dx dz dy
−1 − 1−y 2 4y2 +4z 2
1 √1−y2 4
(c) −1 √ 2 4y2 +4z2 x2 + y 2 x2 + y 2 + z 2 dx dz dy
7387
6. 10,080
12. 8
−
18.
3
4
1−y
SECTION 12.5
12.5
TRIPLE INTEGRALS
■
3
SOLUTIONS
E Click here for exercises.
1.
2.
3.
4.
5.
Copyright © 2013, Cengage Learning. All rights reserved.
6.
x2 + yz dz dy dx
2 0
2 0 z=1
= 0 −3 x2 z + 12 yz 2 z=−1 dy dx = 0 −3 2x2 dy dx
y=0
2
2
2
= 0 2x2 y y=−3 dx = 0 6x2 dx = 2x3 0 = 16
1 0 2 2
x + yz dx dy dz
−1 −3 0
x=2
1 0 = −1 −3 13 x3 + xyz x=0 dy dz
1 0 1 y=0
= −1 −3 83 + 2yz dy dz = −1 83 y + y 2 z y=−3 dz
1
1
= −1 (8 − 9z) dz = 8z − 92 z 2 −1 = 16
1 2 0 2
x + yz dy dx dz
−1 0 −3
y=0
1 2
= −1 0 x2 y + 12 y 2 z y=−3 dx dz
x=2
1 2
1 = −1 0 3x2 − 92 z dx dz = −1 x3 − 92 xz x=0 dz
1
1
= −1 (8 − 9z) dz = 8z − 92 z 2 −1 = 16
2 0 1 0
−3 −1
1 z y
0
0
0
xyz dx dy dz =
y 3 z dy dz
1
= 0 81 z 5 dz =
2
1 6 1
z 0
48
=
2xy dz dy dx
2x
1
= 0 x 2x2 y + 2xy 2 dy dx = 0 x2 y 2 + 23 xy 3 x dx
1
1
x4 dx = 23
x4 0 = 23
= 0 23
3
15
15
0
x
0
01
0
ex dz dx dy =
1 y
1) ex ]y0
0
3x + 2y + z = 6 and above the region
in the xy-plane bounded by the lines x = 0, y = 0 and
3x + 2y = 6, so
2 3−3x/2 6−3x−2y
x dV = 0 0
x dz dy dx
0
E
2 3−3x/2 2
6x − 3x − 2xy dy dx
= 0 0
2 2 = 0 6x − 3x2 3 − 32 x − x 3 − 32 x
dx
2
2
1 3
= 9 0 x − x + 4 x dx
1 4 2
x 0=3
= 9 12 x2 − 13 x3 + 16
1
48
1 2x x+y
11.
1 2x π 2 √4−z2
0
0
0
z sin y dx dz dy
π 2 √
= 0 0 z 4 − z 2 sin ydz dy
π
π
3/2 2
sin y dy = 0
= 0 − 13 4 − z 2
0
π
= − 83 cos y 0 = 16
3
8
3
sin y dy
3 √9−x2 x
0
3 √9−x2 1 2 yz dy dz dx = 0 0
x z dz dx
2
3 1 2
3 1 2 2 √9−x2
dx = 0 4 9x − x4 dx
= 0 4x z 0
3
= 3x3 − 15 x5 0 = 162
5
0
0
2 x 1−y
1
0
0
0
0
z=1−y
x3 y 2 z 2 z=0
2
2
x y z dz dy dx = 1 0
dy dx
2 x 1 3 2
= 1 0 2 x y (1 − y) dy dx
2 x = 1 0 12 x3 y 2 − x3 y 3 + 12 x3 y 4 dy dx
2
1 3 5 y=x
x y y=0 dx
= 1 16 x3 y 3 − 14 x3 y 4 + 10
2
1 8
x dx
= 1 16 x6 − 14 x7 + 10
1 7
1 8
1 9 2
= 42
x − 32
x + 90
x 1
1 2z z+2
0
2 x 1
3 2
=
7.
0
0
10. Here E is the region that lies under the plane
1 z 1
0
1 y x+y
(x + y) ex dx dy
0 1
dy = 0 [(2y − 1) ey − (y − 1)] dy
= 0 [(x + y −
1
= 2yey − 3ey − 12 y 2 + y 0 = 12 (7 − 2e)
1 x2 1 x2 x+2y
9. 0 0 0
y dz dy dx = 0 0 yx + 2y 2 dy dx
1
1
x2
= 0 12 xy 2 + 23 y 3 0 dx = 0 12 x5 + 23 x6 dx
1 6
2 7 1
5
x + 21
x 0 = 28
= 12
8.
128
42
−
256
32
yz dx dy dz =
+
512
90
−
1
42
+
1
32
−
1
90
=
7387
10,080
1 2z
0
0
yz (z + 2) dy dz
1
= 0 2z 4 + 4z 3 dz =
7
5
By symmetry
z dV = 2
z dV where E is the
E
E
part of E to the left [as viewed from (10, 10, 0)] of the plane
x = y. So
1 1 1−x
1 1
z dV = 0 y 0
2z dz dx dy = 0 y (1 − x)2 dx dy
E
x=1
1
1
= 0 − 13 (1 − x)3 x=y dy = 0 31 (1 − y)3 dy
1
1
1
= 12
(1 − y)4 0 = 12
12. The plane 2x + 3y + 6z = 12 intersects the xy-plane when
2x + 3y + 6 (0) = 12 ⇒ y = 4 − 23 x. So
E = (x, y, z) | 0 ≤ x ≤ 6, 0 ≤ y ≤ 4 − 23 x,
0≤z≤
1
6
(12 − 2x − 3y)
and
6 4−2x/3 (12−2x−3y)/6
V = 0 0
dz dy dx
0
1 6 4−2x/3
(12 − 2x − 3y) dy dx
= 6 0 0
y=4−2x/3
6
= 16 0 12y − 2xy − 32 y 2 y=0
dx
6
(12 − 2x)2
1
3 12 − 2x
=
−
dx
6 0
3
2
9
1 1
6
6
1
(12 − 2x)2 dx = 36
− 6 (12 − 2x)3 0 = 8
= 36
0
4
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SECTION 12.5
TRIPLE INTEGRALS
13.
16. A CAS gives
2 sin x z+x
e3x (5y + 2z) dy dz dx
749
− 511
cos 4 − 140
sin 4 − 1521
.
= e6 − 35
18
338
169
In Maple, we use
int(int(int(f,y=z-x..z+x),z=-1..sin(x)),
x=0..2);.
0
V =
√
4−4x2
=2
=2
dy dz dx
0
√
−
0
−
4−4x2
1 √4−4x2
√
1
4−4x2
dy dz dx
(z + 2) dz dx
z=2√1−x2
−
1 2
dx
z + 2z
√
2
z=−2 1−x2
1 √
= 2 0 8 1 − x2 dx
1
√
= 16 12 x 1 − x2 + 12 sin−1 x 0 = 4π
=2
−
1 √x
dz dy dx = 0 −√x (1 − x) dy dx
1 √
1 √
x − x3/2 dx
= 0 2 x (1 − x) dx = 0 2
1
8
= 2 23 x3/2 − 25 x5/2 = 2 23 − 25 = 15
=
0
2
2
15. The paraboloids z = x + y and z = 18 − x − y
2
2
2
2
2
intersect when x + y = 18 − x − y ⇒
2x2 + 2y 2 = 18 ⇒ x2 + y 2 = 9. Thus,
E = (x, y, z) | x2 + y 2 ≤ 9,
x2 + y 2 ≤ z ≤ 18 − x2 − y 2
2
2
Let D = (x, y) | x + y ≤ 9 . Then
18−x2 −y2 V =
dV
=
dz dA
E
D
x2 +y 2
2
2
= D 18 − 2x − 2y dA
2π 3 = 0 0 18 − 2r2 r dr dθ
2π
r=3
2π dθ = 81π
= 0 9r2 − 12 r4 r=0 dθ = 0 81
2
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dx dz dy
1−y
(1)(1)(1)
6
= 16 . The equation of the plane through
the last three vertices is x + y + z = 1, so
1 1−x 1 −x−y
1
(x + y + z) dz dy dx
fave = 1/6
0 0
0
1 1−x (x + y) (1 − x − y) + 12 (1 − x − y)2 dy dx
=6 0 0
1 1−x 1 − 2xy − x2 − y 2 dy dx
=3 0 0
1 1−x 1 − (x + y)2 dy dx
=3 0 0
y=1−x
1
= 3 0 y − 13 (x + y)3 y=0 dx
1
1
= 3 0 1 − x − 13 + 13 x3 dx = 0 x3 − 3x + 2 dx
√
− x 0
2
2
1−y
18. V (E) =
1 √x 1−x
0
2
(c)
1 √1−y2 4
Iz = −1 √ 2 4y2 +4z2 x2 + y 2 x2 + y 2 + z 2 dx dz dy
0
14.
V =
2
(b) (x, y, z) where
1 √1−y2 4
1
√
x= m
x x2 + y 2 + z 2 dx dz dy
−1 − 1−y2 4y2 +4z 2
1 √1−y2 4
1
√
y= m
y x2 + y 2 + z 2 dx dz dy
−1 − 1−y2 4y2 +4z 2
1 √1−y2 4
1
√
z= m
z x2 + y 2 + z 2 dx dz dy
−1
2 4y 2 +4z 2
z+2
0
√
17. (a) m = −1
2
2 x + y + z
− 1−y2 4y +4z
z+2
1 √4−4x2 0
z−x
1 √1−y2 4
1 √4−4x2 −1 −
−1
1
4
−
3
2
+2=
3
4