Thomas' Calculus 13th Edition

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 920: 33

Answer

$\frac{31\pi}{6}$

Work Step by Step

$ V=\int^{2\pi}_0 \int^{\pi/2}_0 \int^2_ {cos\phi} p^2 sin \phi dp d\phi d\theta $ =$\frac{1}{3} \int^{2\pi}_0 \int^{\pi/2}_0 (8-cos^3\phi) sin\phi $ $ d\phi $ $ d\theta $ =$\frac{1}{3}\int^{2\pi}_0 [-8cos\phi+\frac{cos^4\phi}{4}]^{\pi/2}_0 d\theta $ =$\frac{1}{3}\int^{2\pi}_0 (8-\frac{1}{4})d\theta $ =$(\frac{31}{12}(2\pi))$ =$\frac{31\pi}{6}$
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