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: 18


$$\dfrac{\pi^2}{16} $$

Work Step by Step

We need to integrate the integral to compute the volume. $$Volume = \int_{0}^{\pi/2} (\dfrac{\pi}{4}) \sin^2 (2x) dx \\=(\dfrac{\pi}{4}) \int_{0}^{\pi/2} (1- \cos 4x) \space dx \\=(\dfrac{\pi}{8}) \dfrac{\pi}{2}- (\dfrac{\pi}{8}) \cdot \dfrac{\sin (2 \pi) }{4}-(\dfrac{\pi}{8}) (0)+(\dfrac{\pi}{8}) (0)] \space dy \\= (\dfrac{\pi}{8})[\dfrac{\pi}{2}-0] \\=\dfrac{\pi^2}{16} $$
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.