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: 4

Answer

$\frac{37\pi}{15}$

Work Step by Step

$\int^{\pi}_0 \int^{\theta/\pi}_0 \int^{3\sqrt{4-r^2}}_{-\sqrt{4-r^2}} $ z dz dr $ d\theta $ =$\int^\pi_0 \int^{\theta/\pi} _0 \frac{1}{2}(9-r^2)-(4-r^2)]r $ dr d $\theta $ =$\int^{\pi}_0 \int^{\theta/\pi} _0[4r-r^3]^drd\theta $ =$4\int^\pi_0 [2r^2-\frac{r^4}{4}]^{\theta/\pi}_0$ =$4\int^{\pi}_0 (\frac{2\theta^2}{\pi}-\frac{\theta^4}{4\pi^4})d\theta $ =$\frac{37\pi}{15}$
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