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: 24


$(\dfrac{8\sqrt 2-7}{6}) \pi$

Work Step by Step

Consider $V=\int_0^{2\pi} \int_{0}^{1}\int_{r^2}^{\sqrt{2-r^2}} r dz dr d\theta$ $=\int_0^{2\pi} \int_{0}^{1}[rz]_{r^2}^{\sqrt{2-r^2}} r \times dz dr d\theta$ $=\int_0^{2\pi} \int_{0}^{1}(r\sqrt{2-r^2}-r^3) \times dr d\theta$ Plug $a=2-r^2 \implies -2rdr=da$ $=-\int_0^{2\pi} [\int_{2}^{1} (1/2) \times a^{1/2} da -\int_0^1 r^3 dr] d\theta$ $=\int_0^{2\pi} [\dfrac{1}{3}(2 \sqrt 2 -1-\dfrac{1}{4}) d\theta$ $=\int_0^{2\pi} (\dfrac{8\sqrt 2-7}{12}) d\theta$ $=(\dfrac{8\sqrt 2-7}{12}) \times [\theta]_0^{2\pi}$ $=(\dfrac{8\sqrt 2-7}{6}) \pi$
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