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: 8


$12\pi $

Work Step by Step

$\int^1_{-1} \int^\sqrt{z}_0 \int^{1+cos\theta}_0 4r $ dr d $\theta $ dz =$\int^1_{-1} \int^{2\pi}_0 2(1+cos\theta)^2$ $ d\theta $ dz =$\int^1_{-1}6\pi $ d $\theta $ =$12\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.