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: 20

Answer

$\pi(ln4-1)$

Work Step by Step

$\int^1_{-1} \int^{\sqrt{1-y^2}}_{\sqrt{1-y^2}}ln(x^2+y^2+1)dxdy $ =$4\int^{\pi/2}_0 \int^1_0 ln(r^2+1)rdrd\theta $ =2$\int^{\pi/2}_0(ln4-1)d\theta $ =$\pi(ln4-1)$
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