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: 19

Answer

$\dfrac{8\pi}{3}+\dfrac{128}{15}$

Work Step by Step

$I=\int_0^{(\pi/2)} \int_{0}^{2}\int_{0}^{4-r^2} (r \cos \theta+r \sin \theta+z) \times (r dr dz d\theta)$ $=\int_0^{(\pi/2)} \int_{0}^{2}\int_{0}^{4-r^2}[r^2( \cos \theta+ \sin \theta+rz) dz dr d\theta$ $=\int_0^{(\pi/2)} \int_{0}^{2}[r^2( \cos \theta+ \sin \theta)\times (z)+\dfrac{z^2 r}{2}]_{0}^{(4-r^2)} dr d\theta$ $=\int_0^{(\pi/2)} (\dfrac{64}{15}) \times (\cos \theta+ \sin \theta)+(\dfrac{16}{3}) d\theta$ $=\dfrac{8\pi}{3}+\dfrac{128}{15}$
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