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: 19



Work Step by Step

$I_{avg}=\int_{-1}^{1} \int_{-\sqrt {1-x^2}}^{\sqrt {1-x^2}} \dfrac{2}{(1+x^2+y^2)} \ dy \ dx ...(1)$ We will write in the above equation (1) polar coordinates. $$I_{avg}= \int_{0}^{2 \pi} \int_{0}^{1} \dfrac{2r}{(1+r^2)} \ dr \ d \theta\\= \int_{0}^{2 \pi} [- \dfrac{1}{(1+r^2)}]_{0}^{1} \ d \theta \\= \int_{0}^{2 \pi} \dfrac{1}{2}\ d\theta \\=\dfrac{1}{2} [2\pi -0] \\= \pi $$
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