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: 1

Answer

$\frac{4\pi(\sqrt{2}-1)}{3}$

Work Step by Step

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