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: 3

Answer

$\frac{17\pi}{5}$

Work Step by Step

$\int^{2\pi}_0 \int^{\theta/(2\pi)}_0 \int^{3+24r^3}_0 dz r d \theta $ =$\int^{2\pi}_0 \int^{\theta/(2\pi)}(3r+24r^3)dr d\theta$ =$\int^{2\pi}_0 [\frac{3}{2}r^2+6r^4] d\theta $ =$\frac{3}{2}\int^{2\pi}_0 (\frac{\theta^2}{4\pi^2}+\frac{4\theta^4}{16\pi^4})^{2\pi}_0$ =$\frac{17\pi}{5}$
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