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: 35

Answer

$\frac{2a}{3}$

Work Step by Step

We calculate the average as follows: average=$\frac{1}{\pi a^2} \int^{a}_{-a} \int^{\sqrt{a^2-x^2}}_{-\sqrt{a^2-x^2}} \sqrt{x^2+y^2} dydx $ =$\frac{1}{\pi a^2} \int^{2\pi}_0 \int^a_0 r^2dr d\theta $ =$\frac{a}{3\pi }\int^{2\pi}_0 d\theta $ =$\frac{2a}{3}$
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