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: 6

Answer

$\frac{\pi}{3}$

Work Step by Step

$\int^{2\pi}_0 \int^1_0 \int^{1/2}_{-1/2}(r^2sin^2\theta+z^2)$ dz r dr $ d\theta $ =$\int^{2\pi}_0 \int^1_0 (r^3sin^2\theta+\frac{r}{12})drd\theta $ =$\int^{2\pi}_0 (\frac{sin^2\theta}{4}+\frac{1}{24})d\theta $ = $\frac{\pi}{3}$
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