15.6.7 We integrate the triple integral directly.
Z π/2 0
Z y 0
Z x 0
cos (x + y + z)dzdxdy = Z π/2
0
Z y 0
(sin (2x + y) − sin (x + y))dxdy
= Z π/2
0
(cos 2y − 1
2cos 3y − cos y +1
2cos y)dy = −1 3
15.6.17 Viewed the integral as the mass of the solid with density x at height x. Then Z Z Z
E
x dV = Z 4
0
πx
4 · xdx = 16π 3 15.6.33 (a)
Z 1 0
Z y2 0
Z 1−y 0
f (x, y, z)dzdxdy (b)
Z 1 0
Z 1−y 0
Z y2 0
f (x, y, z)dxdzdy (c)
Z 1 0
Z 1−z 0
Z y2 0
f (x, y, z)dxdydz (d)
Z 1 0
Z (1−z)2 0
Z 1−z
√x
f (x, y, z)dydxdz (e)
Z 1 0
Z 1−√ x
0
Z 1−z
√x
f (x, y, z)dydzdx 15.6.34 (a)
Z 1 0
Z
√1−z
0
Z 1−x 0
f (x, y, z)dydxdz (b)
Z 1 0
Z 1 0
Z 1−y 0
f (x, y, z)dxdydz (c)
Z 1 0
Z 1 0
Z 1−y 0
f (x, y, z)dxdzdy (d)
Z 1 0
Z 1−x 0
Z 1−x2 0
f (x, y, z)dzdydz (e)
Z 1 0
Z 1−y 0
Z 1−x2 0
f (x, y, z)dzdzdy
1
15.6.44 Let R be the solid conepx2+ y2 ≤ z ≤ h. Then I =
Z Z Z
R
(x2+ y2)dV Using the cylinder coordinate. We get
I = Z 2π
0
Z h 0
Z z 0
r3drdzdθ = h5π 10 15.6.49 (a) Computing
Z 2 0
Z 2 0
Z 2 0
Cxyzdxdydz = 1 We get C = 1/8.
(b)
Z 1 0
Z 1 0
Z 1 0
1
8xyzdxdydz = 1 16
(c) The region T is bounded by x + y + z = 1, x = 0, y = 0, z = 0. Thus Z
T
xyz
8 dxdydz = Z 1
0
Z 1−z 0
Z 1−y−z 0
xyz
8 dxdydz = 1 60
15.6.53 The value attends to its maximum when E is the region bounded by 1 = x2+ 2y2+ 3z2. Since inside E, the integrand is always positive, and clearly, the integrand is negative outside E.
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