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: 64

Answer

$\frac{3\pi}{16}$

Work Step by Step

We calculate the average as: average=$\frac{1}{(\frac{4\pi}{3})}\int^{2\pi}_0 \int^1_0 \int^{\sqrt{1-r^2}}_{-\sqrt{1-r^2}}r^2 $ dz dr d $\theta $ =$\frac{3}{4\pi} \int^{2\pi}_0 \int^1_0 2r^2 \sqrt{1-r^2}$ dr d $\theta $ =$\frac{3}{2\pi} \int^{2\pi}_0 [\frac{1}{8}sin^{-1}r-\frac{1}{8}r\sqrt{1-r^2}(1-2r^2)]^1_0d\theta $ =$\frac{3}{16\pi}\int^{2\pi}_0 (\frac{\pi}{2}+0)$ d $\theta $ =$\frac{3}{32}\int^{2\pi}_0 d\theta $ =$(\frac{3}{32})(2\pi)$ =$\frac{3\pi}{16}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.