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: 34

Answer

$\frac{11\pi}{6}$

Work Step by Step

$ V=\int^{2\pi}_0 \int^{\pi/2}_0 \int^2_{cos\phi} p^2 \sin \phi d p $ $ d\phi $ $ d\theta $ =$\frac{1}{3} \int^{2\pi}_{0}\int^{\pi/2}_0 (3cos\phi+3cos^2\phi+cos^3\phi)sin \phi d\phi d\theta $ =$\frac{1}{3}\int^{2\pi}_0 [-\frac{3}{2}cos^2\phi-cos^3\phi-\frac{1}{4}cos^4\phi]^{\pi/2}_0 d\theta $ =$\frac{1}{3} \int^{2\pi}_0(\frac{3}{2}+1+\frac{1}{4}d\theta)$ =$\frac{11}{12}\int^{2\pi}_0d\theta $ =$(\frac{11}{12})(2\pi)$ =$\frac{11\pi}{6}$
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