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: 29

Answer

$0$

Work Step by Step

$x=\sqrt{4-y^2}$ $x^2=4-y^2$ $x^2+y^2=4 $ That is, $ r=2$ In cylindrical coordinate system, we have $I=\int_0^{2\pi} \int_{0}^{2}\int_{r}^{2} (z) \times (r^2) \cos \theta \times dz dr d\theta$ $=\int_0^{2\pi} \cos \theta d\theta \times [ \int_{0}^{2} (1) \times \int_{r}^{2} (z)(r^2) dz dr ] $ $=[\sin (\theta) ]_0^{2 \pi} \times [ \int_{0}^{2}(1) \times \int_{r}^{2} \times (z) \times (r^2) dz dr ] $ $=[\sin (2 \pi)-\sin (0) ] \times [ \int_{0}^{2}\int_{r}^{2} (z) \times (r^2) dz dr ] $ $=0 \times [ \int_{0}^{2}\int_{r}^{2} (z) \times (r^2) dz dr ] $ $=0$
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