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: 17

Answer

$(1-ln2)\pi $

Work Step by Step

$\int ^0_{-1} \int ^0_{-\sqrt{1-x^2}}\frac{2}{1+\sqrt{x^2+y^2}} dydx $ =$\int^{3\pi/2}_{\pi} \int^1_0\frac{2r}{1+r}drd\theta $ =$2\int^{3\pi/2}_{\pi} \int^{1}_0(1-\frac{1}{1+r})drd\theta $ =$2\int ^{3\pi/2}_\pi (1-ln2)d\theta $ =$(1-ln2)$\pi $
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