Calculus 8th Edition

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

Chapter 15 - Multiple Integrals - 15.6 Triple Integrals - 15.6 Exercises - Page 1079: 46


$\dfrac{\pi kh^5}{10}$

Work Step by Step

Here, we have $I_z=\iiint_{E} (x^2+y^2) \rho(x,y,z) dV$ In the polar coordinates, we have $x=r \cos \theta ;y=r \sin \theta;z=z$ and $r^2=x^2+y^2 $ Thus, we have $k\int_{0}^{2 \pi} \int_{0}^{h}\int_{0}^{z}(r^2)r dr dz d\theta =k\int_{0}^{2 \pi} \int_{0}^{h}\int_{0}^{z}(r^3) dr dz d\theta $ This implies that $k \int_{0}^{2 \pi} d\theta \int_{0}^{h} \int_{0}^{z}(r^3) dr dz=2\pi k \int_0^h [\dfrac{z^4}{4}] dz $ Hence, $I_z=2\pi k [\dfrac{z^5}{20}]_0^h=\pi k [\dfrac{h^5-0}{10} ]=\dfrac{\pi kh^5}{10}$
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