Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 15: Multiple Integrals - Section 15.7 - Triple Integrals in Cylindrical and Spherical Coordinates - Exercises 15.7 - Page 921: 75

Answer

$\frac{\pi h a^4}{10}$

Work Step by Step

$ I_2= \int^{2\pi}_0 \int^a_0 \int^h _{\frac{h}{a}r}(x^2+y^2) $ dz r dr d $\theta $ =$\int ^{2\pi}_0 \int^a_0 \int^h _{\frac{hr}{a}}r^3$ dz dr d $\theta $ =$\int^{2\pi}_0 \int^{a}_0 (hr^3-\frac{hr^4}{a})$ dr d $\theta $ =$\int^{2\pi}_0 h[\frac{r^4}{4}-\frac{r^5}{5a}]^a_0 d\theta $ =$\int^{2\pi}_0 h(\frac{a^4}{4}-\frac{a^5}{5a})d\theta $ =$\frac{ha^4}{20} \int^{2\pi}_0 d\theta $ =$\frac{\pi h a^4}{10}$
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.