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: 18


$$\dfrac{1}{2 \pi} $$

Work Step by Step

$I_{avg}=(\dfrac{1}{\frac{\pi}{4}}) \times \int_{0}^{1} \int_{0}^{\sqrt {1-x^2}} x \times y \ dy \ dx \\=\int_{0}^{1} \dfrac{4}{\pi} \times [\dfrac{xy^2}{2}]_{0}^{\sqrt {1-x^2}} \ dx \\= \int_{0}^{1} \dfrac{4}{\pi} \times [\dfrac{xy^2}{2}]_{0}^{\sqrt {1-x^2}} \ dx \\=\int_{0}^{1}\dfrac{2(x-x^3)}{\pi} \ dx \\= [\dfrac{2}{\pi} \dfrac{x^2}{2}]_{0}^{1} - [\dfrac{2}{\pi} \dfrac{x^3}{3}]_{0}^{1} \ dx \\=\dfrac{1}{2 \pi} $$
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