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 919: 24

Answer

$\frac{5\pi}{2}$

Work Step by Step

$\int^{3\pi/2}_0 \int^{\pi}_0 \int^1_0 5p^3sin \phi $ $ dp $ $ d\phi $ $ d\theta $ =$\frac{5}{4} \int^{3\pi/2}_0 \int^{\pi}_0 sin^3\phi $ $ d\phi $ $ d\theta $ =$\frac{5}{4}\int^{3\pi/2}_0 ([-\frac{-sin^2\phi \cos\phi}{3}]^\pi_0+\frac{2}{3}\int^\pi_0 \sin \phi$ $ d\phi) d\theta$ =$\frac{5}{6}\int^{3\pi/2}_0 [-cis\phi]^\pi_0 d\theta $ =$\frac{5}{3}\int^{3\pi/2}_0 d\theta $ =$\frac{5\pi}{2}$
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