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: 22

Answer

$\frac{\pi}{16}$

Work Step by Step

$\int^2_1 \int^{\sqrt{2x-x^2}}_0 \frac{1}{(x^2+y^2)^2}dydx $ =$\int^{\pi/4}_0 \int^{2cos\theta}_{sec\theta}\frac{1}{r^4}drd\theta $ =$\int^{\pi/4}_0[\frac{-1}{2r^2}]^{2cos\theta}_{sec\theta}d\theta $ =$\int^{\pi/4}_0(\frac{1}{2}cos^2\theta-\frac{1}{8}sec^2\theta) d\theta $ =$[\frac{1}{4}\theta+\frac{1}{8}sin2\theta-\frac{1}{8}tan\theta]^{\pi/4}_0$ =$\frac{\pi}{16}$
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