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: 48

Answer

$\frac{3\pi-4}{6}$

Work Step by Step

$ V=\int^{\pi/2}_0 \int^{\cos\theta}_0 \int^{3\sqrt{1-r^2}}_0dz $ r dr $ d\theta $ =$\int^{\pi/2}_0 \int^{\cos\theta}_0 3r\sqrt{1-r^2}drd\theta $ =$\int^{\pi/2}_0 [-(1-r^2)^{3/2}]^{cos\theta}_0 d\theta $ =$\int^{\pi/2}_0[-(1-cos^2\theta)^{3/2}+1]d\theta $ =$\int^{\pi/2}_0 (1-sin^3\theta) d\theta $ =$[\theta+\frac{sin^2\theta\cos\theta}{3}]^{\pi/2}_0-\frac{2}{3}\int^{\pi/2}_0 \sin\theta d\theta $ =$\frac{\pi}{2}+\frac{2}{3}[cos\theta]^{\pi/2}_0 $ =$\frac{\pi}{2}-\frac{2}{3}$ =$\frac{3\pi-4}{6}$
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