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 15.4 - Double Integrals in Polar Form - Exercises 15.4 - Page 893: 40

Answer

$\frac{6\pi\sqrt{2}+40\sqrt{2}-64}{9}$

Work Step by Step

V=$4\int^{\pi/4}_0 \int^{\sqrt{2cos2\theta}}r\sqrt{2-r^2}drd\theta $ =$-\frac{4}{3} \int^{\pi/4}_0 [(2-2cos2\theta)^{3/2}-2^{3/2}]d\theta $ =$\frac{2\pi\sqrt{2}}{3} -\frac{32}{3} \int^{\pi/4}_0 (1-cos^2\theta) sin\theta d\theta $ =$\frac{2\pi\sqrt{2}}{3}-\frac{32}{3} [\frac{cos^3\theta}{4}-cos\theta]^{\pi/4}_0$ =$\frac{6\pi\sqrt{2}+40\sqrt{2}-64}{9}$
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