Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 15: Multiple Integrals - Practice Exercises - Page 933: 20


$[\ln (4) -1] \pi $

Work Step by Step

$I= \int_{-1}^{1} \int_{-\sqrt {1-y^2}}^{\sqrt {1-y^2}} \ln (x^2+y^2+1) \ dx \ dy$ or, $= \int_{0}^{2 \pi} \int_{0}^{1} r \ln (r^2+1) \ dr \ d \theta$ Consider $r^2+1 = u \implies r dr =du$ So, $I=\dfrac{1}{2} \int_{0}^{2 \pi} \int_{1}^2 (\ln u du) d \theta \\=\dfrac{1}{2} \int_{0}^{2 \pi} [u\ln (u) - u]_1^2 \ d\theta \\=\int_{0}^{2 \pi} \dfrac{(2 \ln 2 -1) )}{2} \ d\theta \\=[\ln (4) -1] \pi $
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