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: 56

Answer

$\frac{4\pi}{3}$

Work Step by Step

V=$8\int^{\pi/2}_0 \int^{\sqrt{2}}_1 \int^\sqrt{2-r^2}_0dz $ r dr $ d\theta $ =$8\int^{\pi/2}_0 \int^{\sqrt{2}}_1r\sqrt{2-r^2}dr $ $ d\theta $ =$8\int^{\pi/2}_0 [-\frac{1}{3}(2-r^2)^{3/2}]^{\sqrt{2}}_1 d\theta $ =$8\int^{\pi/2}_0 d\theta $ =$\frac{4\pi}{3}$
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