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: 26

Answer

$\frac{\pi}{4}$

Work Step by Step

=$\int^{2\pi}_0 \int^{4\pi}_0 \int^{sec\phi}_0 p^3 \sin \phi \cos\phi $ $ dp $ $ d\phi $ $ d\theta $ =$\frac{1}{4}\int^{2\pi}_0 \int^{\pi/4}_0 tan\phi \sec^2\phi d \phi $ $ d\theta $ =$\frac{1}{4}\int^{2\pi}_0 [\frac{1}{2}tan^2\phi]^{\pi/4}_0 d\theta $ =$\frac{1}{8}\int^{2\pi}_0 d\theta $ =$\frac{\pi}{4}$
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