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: 18

Answer

$\pi $

Work Step by Step

$\int^1_{-1} \int^{\sqrt{1-x^2}}_\sqrt{1-x^2}\frac{2}{(1+x^2+y^2)}dydx $ =4$\int^{\pi/2}_0 \int^1_0 \frac{2r}{(1+r^2)^2}dr d\theta $ =4$\int^{\pi/2}_0 [\frac{-1}{1+r^2}]^1_0d\theta $ =$2\int^{\pi/2}_0 d\theta $ =$\pi $
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