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 921: 51

Answer

$\frac{5\pi}{3}$

Work Step by Step

$ V=\int^{2\pi}_0 \int^{\pi/3}_0 \int^2_{\sec\phi}p^2\sin\phi $ dp $ d\phi $ $ d\theta $ =$\frac{1}{3}\int^{2\pi}_0 \int^{\pi/3}_0(8\sin\phi-\tan\phi \sec^2\phi)d\phi d\theta $ =$\frac{1}{3}\int^{2\pi}_0 [-8\cos\phi-\frac{1}{2}tan^2\phi]^{\pi/3}_0d\theta $ =$\frac{1}{3}\int^{2\pi}_0 [-4-\frac{1}{2}(3)+8]d\theta $ =$\frac{1}{3}\int^{2\pi}_0 \frac{5}{2}d\theta $ =$\frac{5}{6}(2\pi)$ =$\frac{5\pi}{3}$
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