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: 61



Work Step by Step

We calculate the volume as follows: V=$8\int^{2\pi}_0 \int^1_0 \int^{\sqrt{4-r^2}}_0$ dz r dr $ d\theta $ =$8\int^{2\pi}_0 \int^1_0 r(4-r^2)^{1/2}dr $ d $\theta $ =$8\int^{2\pi}_0 [-\frac{1}{3}(4-r^2)^{3/2}]^1_0 d\theta $ =$\frac{8}{3} \int^{2\pi}_0 (3^{3/2}-8)d\theta $ =$\frac{4\pi(8-3\sqrt{3})}{3}$
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