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: 47

Answer

$\frac{-4+3\pi}{18}$

Work Step by Step

$ V=\int^{\pi/2}_{0} \int^{sin\theta}_0 \int^{\sqrt{1-r^2}}_0$ dz r dr $ d\theta $ =$\int^{\pi/2}_{0} \int^{sin\theta}_0 r\sqrt{1-r^2}dr d\theta $ =$\int^{\pi/2}_{0} [-\frac{1}{3}(1-r^2)^{3/2}]^{sin\theta}_0 d\theta $ =$-\frac{1}{3}\int^{\pi/2}_{0} [(1-sin^\theta)^{3/2}-1]d\theta $ =$-\frac{1}{3}\int^{\pi/2}_{0} (cos^3\theta-1)d\theta $ =$-\frac{1}{3}([\frac{cos^2\theta \sin \theta}{3}]^{\pi/2}_0+\frac{2}{3}\int^{\pi/2}_{0} \cos\theta d\theta)+[\frac{\theta}{3}]^{\pi/2}_0$ =$-\frac{2}{9}[\sin\theta]^{\pi/2}_0+\frac{\pi}{6}$ =$\frac{-4+3\pi}{18}$
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