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: 52

Answer

$\frac{7\pi}{3}$

Work Step by Step

$ V=4\int_{0}^{\pi/2} \int^{\pi/4}_0 \int^{2\sec\phi}_{\sec\phi}p^2\sin\phi dp $ $ d\phi $ $ d\theta $ =$\frac{4}{3}\int_{0}^{\pi/2} \int^{\pi/4}_0 (8sec^3\phi-\sec^3\phi)\sin\phi d\phi d\theta $ =$\frac{28}{3}\int_{0}^{\pi/2} \int^{\pi/4}_0 \sec^3\phi \sin\phi d\phi d\theta $ =$\frac{28}{3}\int_{0}^{\pi/2} \int^{\pi/4}_0 \tan\phi \sec^2\phi d\phi d\theta $ =$\frac{28}{3}\int_{0}^{\pi/2} [\frac{1}{2}tan^2\phi]^{\pi/4}_0 d\theta $ =$\frac{14}{3}\int_{0}^{\pi/2} d\theta $ =$\frac{7\pi}{3}$
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