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: 55

Answer

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

Work Step by Step

$ V=8\int^{\pi/2}_0 \int^\sqrt{2}_1 \int^r_0 $ dz r dr $ d\theta $ =$8\int^{\pi/2}_0 \int^{\sqrt{2}}_1 r^2 dr d\theta $ =$8(\frac{2\sqrt{2}-1}{3})\int^{\pi/2}_0d\theta $ =$\frac{4\pi(2\sqrt{2}-1)}{3}$
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