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: 57

Answer

$16\pi $

Work Step by Step

$ V=\int^{2\pi}_0 \int^2_0 \int^{4-r\sin\theta}_0 $ dz r dr $ d\theta $ =$\int^{2\pi}_0 \int^2_0 (4r-r^2\sin\theta) dr $ $ d\theta $ =$8\int^{2\pi}_0 (1-\frac{\sin\theta}{3})d\theta $ =$16\pi $
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