James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 15.6 Triple Integrals 1037
Because of the symmetry of E and about the xz-plane, we can immediately say that
M xz − 0 and therefore y − 0. The other moments are
M yz − y yy x dV − y 1 y x
21 y1 x dz dx dy
y 2 0
E
− y 1 21 y1 x 1
2 dx dy −
y y
21F x 3 x−1
2 3Gx−y 2
dy
− 2 3 y1 s1 2 y 6 d dy − 2 Fy 2 y 7 1
− 4
0
3 7 7G0
M xy − y yy z dV − y 1 y x
21 y1 z dz dx dy
y 2 0
E
− y
1
z−x
z2
21 y1
y 2F
2Gz−0
dx dy − 2 y1 21 y1 y 2 x 2 dx dy
Therefore the center of mass is
− 3 y1 0 s1 2 y 6 d dy − 2 7
sx, y, zd −S M yz
m , M xz
m , M xy
m
D − ( 5 7 , 0, 5
14 ) ■
1. Evaluate the integral in Example 1, integrating first with
respect to y, then z, and then x.
2. Evaluate the integral yyy E
sxy 1 z 2 d dV, where
E − 5sx, y, zd | 0 < x < 2, 0 < y < 1, 0 < z < 36
using three different orders of integration.
3–8 Evaluate the iterated integral.
3. y 2
4. y 1
5. y 2
6. y 1
7. y
8. y 1
0 yz2 0 yy2z 0
0 y2y y
yx1y 0
x
1 y2z 0 yln 0
0 y1 0 ys12z 2
0
0 y1 0 ys12z 2
0
2y 2
0 y1 0 y22x2 0
s2x 2 yd dx dy dz
6xy dz dx dy
xe 2y dy dx dz
z
dx dz dy
y 1 1
z sin x dy dz dx
xye z dz dy dx
9–18 Evaluate the triple integral.
9. yyy E
y dV, where
E − hsx, y, zd | 0 < x < 3, 0 < y < x, x 2 y < z < x 1 yj
10. yyy E
e zyy dV, where
E − 5sx, y, zd | 0 < y < 1, y < x < 1, 0 < z < xy6
z
11. yyy E
dV, where
x 2 2
1 z
E − hsx, y, zd | 1 < y < 4, y < z < 4, 0 < x < zj
12. yyy E
sin y dV, where E lies below the plane z − x and above
the triangular region with vertices s0, 0, 0d, s, 0, 0d, and
s0, , 0d
13. yyy E
6xy dV, where E lies under the plane z − 1 1 x 1 y
and above the region in the xy-plane bounded by the curves
y − sx , y − 0, and x − 1
14. yyy E
sx 2 yd dV, where E is enclosed by the surfaces
z − x 2 2 1, z − 1 2 x 2 , y − 0, and y − 2
15. yyy T
y 2 dV, where T is the solid tetrahedron with vertices
s0, 0, 0d, s2, 0, 0d, s0, 2, 0d, and s0, 0, 2d
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