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: 10

Answer

$8\pi $

Work Step by Step

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