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: 28


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

Work Step by Step

The mass is given by: $m=\int_Bk\sqrt{x^2+y^2} dV=\int_0^{2\pi} \int_{0}^{a}\int_{-\sqrt{a^2-r^2}}^{\sqrt{a^2-r^2}} \times (kr) (r) dz dr d\theta$ $=\int_0^{2\pi} \int_{0}^{a}\int_{-\sqrt{a^2-r^2}}^{\sqrt{a^2-r^2}} kr^2 \times dz dr d\theta$ $=2\pi \int_{0}^{a} 2k \times r^2 \sqrt{a^2-r^2} dr$ Suppose $r=a\sin p \implies dr=a \cos p dp$ $=2\pi \times \int_{0}^{a} 2kr^2 \sqrt{a^2-r^2} dr$ $=2\pi \times \int_{0}^{\pi/2} 2ka^2 \times [\sin^2 p \sqrt{a^2-a^2 \sin^2 p} (a \cos p) ] dp$ $=\pi ka^4 \times \int_0^{\pi/2} \sin^2 (2p) dp$ $=(0.5) \pi \times ka^4 \times \int_0^{\pi/2} [1-\cos (4p)] dp$ $=(0.5) \times \pi ka^4 \times [p-\dfrac{\sin (4p) }{4}]_0^{\pi/2}$ $=\dfrac{\pi^2 ka^4}{4}$
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.