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 920: 44

Answer

$\pi $

Work Step by Step

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