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: 43

Answer

$\frac{8\pi}{3}$

Work Step by Step

V=$4\int^{\pi/2}_0 \int^1_0 \int^{4-4r^2}_{r^4-1}dz $ r dr $ d\theta $ =$4\int^{\pi/2}_0 \int^1_0 (5r-4r^3-r^5)drd\theta $ =$4\int^{\pi/2}_0 (\frac{5}{2}-1-\frac{1}{6})d\theta $ =$4\int^{\pi/2}_0d\theta $ =$\frac{8\pi}{3}$
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