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: 5

Answer

$\pi(6\sqrt{2}-8)$

Work Step by Step

$\int^{2\pi}_0 \int^1_0 \int^{(2-r^2)^{-1/2}}_r $3 dz r dr $ d\theta $ =$3\int^{2\pi}_0 \int^1_0 [r(2-r^2)^{-1/2}-r^2]$ dr $ d\theta $ =$3\int^{2\pi}_0[-(2-r^2)^{1/2}-\frac{r^3}{3}]^1_0 $ d $ d \theta $ =$3\int^{2\pi}_0 (\sqrt{2}-\frac{4}{3})d\theta $ =$\pi(6\sqrt{2}-8)$
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