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: 74

Answer

$\frac{8\pi a^5}{15}$

Work Step by Step

$ I_2=\int^{2a}_0 \int^{a}_0 \int^{\sqrt{a^2-r^2}}_{-\sqrt{a^2-r^2}}r^3 dz $ dr d $\theta $ =$\int^{2\pi}_0 \int^a_0 2r^3\sqrt{a^2-r^2}$ dr d $\theta $ =$2 \int^{2\pi}_0 [(-\frac{r^2}{5}-\frac{2a^2}{15})(a^2-r^2)^{3/2}]^a_0d\theta $ =$2\int^{2\pi}_0 \frac{2}{15}a^5d\theta $ =$\frac{8\pi a^5}{15}$
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