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

1048 Chapter 15 Multiple Integrals
This formula can be extended to include more general spherical regions such as
E − hs, , d | < < , c < < d, t 1s, d < < t 2 s, dj
In this case the formula is the same as in (3) except that the limits of integration for are
t 1 s, d and t 2 s, d.
Usually, spherical coordinates are used in triple integrals when surfaces such as cones
and spheres form the boundary of the region of integration.
ExamplE 3 Evaluate yyy B
e sx2 1y 2 1z 2 d 3y2 dV, where B is the unit ball:
B − hsx, y, zd | x 2 1 y 2 1 z 2 < 1j
SOLUtion Since the boundary of B is a sphere, we use spherical coordinates:
B − hs, , d | 0 < < 1, 0 < < 2, 0 < < j
In addition, spherical coordinates are appropriate because
Thus (3) gives
x 2 1 y 2 1 z 2 − 2
y y e sx2 1y 2 1z 2 d 3y2 dV − y
y 1 d 3y2
0 y2 2 sin d d d
0 0 es2
B
− y
sin d y 2
0 0
d y 1
2 e 3 d
0
− f2cos g 0 s2d f 1 3 e 1
g 3 0− 4
3 se 2 1d ■
Note It would have been extremely awkward to evaluate the integral in Example 3
without spherical coordinates. In rectangular coordinates the iterated integral would have
been
y 1 21 ys12x 2
2s12x 2 y s12x 2 2y 2
2s12x 2 2y 2
e sx2 1y 2 1z 2 d 3y2 dz dy dx
ExamplE 4 Use spherical coordinates to find the volume of the solid that lies above the
cone z − sx 2 1 y 2 and below the sphere x 2 1 y 2 1 z 2 − z. (See Figure 9.)
z
(0, 0, 1)
≈+¥+z@=z
π
4
z=œ„„„„„ ≈+¥
FIGURE 9
x
y
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.