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

Answer

$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 $
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