Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 6 - Applications of the Integral - 6.3 Volumes of Revolution - Exercises - Page 305: 36

Answer

$16 \pi $

Work Step by Step

The volume of a revolution can be calculated as: $V=\pi \int_{m}^{n} (R^2_{outside}-r^2_{inside}) \ dy$ where, $R_{outside}=2-0=2$ and $ R_{inside}=\sqrt{y-2}$ Now, $V=\pi \int_0^2 (2)^2 \ dy +\pi \int_2^6 [(2)^2-( \sqrt {y-2})^2 ] \ dy \\=\pi \int_0^2 4 \ dy + \pi \int_2^6 (6-y) \ dy \\=\pi [4y]_0^2 +\pi [6y -\dfrac{y^2}{2}]_2^6 \\=8 \pi +\pi (18-10) \\=16 \pi $
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.