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 894: 45

Answer

$\frac{1}{2}(a^2+2h^2)$

Work Step by Step

We calculate the average as follows: average=$\frac{1}{\pi a^2 }\int^{2\pi}_0 \int^a_0 [(rcos\theta -h)^2+r^2sin^\theta]rdrd\theta $ =$\frac{1}{\pi r^2}\int^{2\pi}_0 \int^a_0 (r^3-2r^2hcos\theta+rh^2)drd\theta $ =$\frac{1}{\pi a^2} \int^{2\pi}_0 (\frac{a^4}{4}-\frac{2a^3hcos\theta}{3}+\frac{a^2 h^2}{2})d\theta $ =$\frac{1}{\pi}\int^{2\pi}_0 (\frac{a^2}{4}-\frac{2ahcos\theta}{3}+\frac{h^2}{2})d\theta $ =$\frac{1}{\pi}[\frac{a^2\theta}{4}-\frac{2ahsin\theta}{4}+\frac{h^2\theta}{2}]^{2\pi}_0$ =$\frac{1}{2}(a^2+2h^2)$
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