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: 45

Answer

$\dfrac{\pi kha^4}{2}$

Work Step by Step

Consider $I_z=\iiint_{E} (x^2+y^2) \rho(x,y,z) dV$ 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}^{a}(r^2)r dr dz d\theta=k\int_{0}^{2 \pi} \int_{0}^{h}\int_{0}^{a}(r^3) dr dz d\theta $ This implies that $k \int_{0}^{2 \pi} d\theta \int_{0}^{h} dz \int_{0}^{a}(r^3) dr=2\pi k [z]_0^h [\dfrac{r^4}{4}]_0^a $ Hence, $I_z=2\pi k (h-0) [\dfrac{a^4}{4}-0] =\dfrac{\pi kha^4}{2}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.