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 918: 2

Answer

$\frac{9\pi(8\sqrt{2}-7)}{2}$

Work Step by Step

$\int^{2\pi}_0 \int^3_0 \int^{\sqrt{18-r^2}}_{r^2/3} dz $ r dr d $\theta $ =$\int^{2\pi}_0 \int^{3}_0[r(18-r^2)^{1/2}-\frac{r^3}{3}]dr $ d $\theta $ =$\int^{2\pi}_0 [\frac{-1}{3}(18-r^2)+\frac{r^4}{12}]^{3}_0 d\theta $ =$\frac{9\pi(8\sqrt{2}-7)}{2}$
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.