University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 6 - Section 6.1 - Volumes Using Cross-Sections - Exercises - Page 356: 35


$$4 \pi \ln 4$$

Work Step by Step

We need to integrate the integral to compute the volume. We have: $$Area =\pi r^2=\pi (\dfrac{2}{\sqrt {y+1}})^2=\dfrac{4 \space \pi}{y+1}$$ Now, $$Volume= (4 \pi) \times \int_{0}^{3} \dfrac{1}{y+1} \space dy \\= 4 \pi [\ln |y+1|]_{0}^{3} \\=4 \pi (\ln 4 -0) \\=4 \pi \ln 4$$
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.