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: 36

Answer

$\frac{\pi}{6}$

Work Step by Step

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