Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 15 - Multiple Integrals - 15.7 Triple-Integrals in Cylindrical Coordinates - 15.7 Exercises - Page 1084: 23


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

Work Step by Step

Consider $V=\int_0^{2\pi} \int_{0}^{1}\int_{r}^{\sqrt{2-r^2}} r \times dz dr d\theta$ $=\int_0^{2\pi} \int_{0}^{1}[rz]_{r}^{\sqrt{2-r^2}} \times r dz dr d\theta$ $=\int_0^{2\pi} \int_{0}^{1}(r \times \sqrt{2-r^2}-r^2) dr d\theta$ Plug $a=2-r^2 \implies -2rdr=da $ $=\int_0^{2\pi} [\int_{1}^{2} (1/2) \times a^{1/2} dk -\int_0^1 r^2 dr] d\theta$ $=\int_0^{2\pi} [\dfrac{2}{3}(\sqrt 2 -1) d\theta$ $=\dfrac{2}{3}(\sqrt 2 -1) \times [\theta]_0^{2\pi}$ $=\dfrac{4\pi}{3}(\sqrt 2 -1)$
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