Answer
$16\pi $
Work Step by Step
$ V=\int^{2\pi}_0 \int^2_0 \int^{4-r\sin\theta}_0 $ dz r dr $ d\theta $
=$\int^{2\pi}_0 \int^2_0 (4r-r^2\sin\theta) dr $ $ d\theta $
=$8\int^{2\pi}_0 (1-\frac{\sin\theta}{3})d\theta $
=$16\pi $
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 15: Multiple Integrals - Section 15.7 - Triple Integrals in Cylindrical and Spherical Coordinates - Exercises 15.7 - Page 921: 57
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