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: 58

Answer

$16\pi $

Work Step by Step

$ V=\int^{2\pi}_0 \int^2_0 \int^{4-r\cos\theta-r\sin\theta} $ dz r dr $ d\theta $ =$\int^{2\pi}_0 \int^{2}_0 [4r-r^2(cos\theta+\sin\theta)]dr d\theta $ =$\frac{8}{3} \int^{2\pi}_0(3-\cos\theta)-\sin\theta d\theta $ =$16\pi $
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