Thomas' Calculus 13th Edition

Published by Pearson

Chapter 15: Multiple Integrals - Section 15.4 - Double Integrals in Polar Form - Exercises 15.4 - Page 893: 39

Answer

$\frac{4}{3}+\frac{5\pi}{8}$

Work Step by Step

V=$2\int^{\pi/2}_0 \int^{1+cos\theta}_1 r^2cos\theta drd\theta$ =$\frac{2}{3} \int^{\pi/2}_0 (3cos^2\theta+3cos^3\theta +cos^4\theta) d\theta$ =$\frac{2}{3}[\frac{15\theta}{8}+sin2\theta+3sin\theta-sin^3\theta+\frac{sin4\theta}{32}]^{\pi/2}_0$ =$\frac{4}{3}+\frac{5\pi}{8}$

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