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 15.7 - Triple Integrals in Cylindrical and Spherical Coordinates - Exercises 15.7 - Page 921: 63

Answer

$\frac{2}{3}$

Work Step by Step

We calculate the average as: average=$\frac{1}{2\pi} \int^{2\pi}_0 \int^1_0 \int^1_{-1} 2r^2drd\theta $ =$\frac{1}{2\pi}\int^{2\pi}_0 \int^1_0 2r^2dr $ d $\theta $ =$\frac{1}{3\pi} \int^{2\pi}_0 d\theta $ =$\frac{2}{3}$
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