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: 37

Answer

$$\pi(\pi-2)$$

Work Step by Step

We need to integrate the integral to compute the volume. We have: Area $=\pi r^2=\pi (1-\cos x)$ $$Volume = \pi \times \int_{- \pi/2}^{\pi/2} (1-\cos x) \space dx \\= \pi (x-\sin x)_{- \pi/2}^{\pi/2} \\ = \pi [\dfrac{\pi}{2}-\sin (\dfrac{\pi}{2}) -(-\dfrac{\pi}{2}) +\sin (-\dfrac{\pi}{2})]) \\=\pi \times \dfrac{\pi}{2}- \pi \times \sin (\dfrac{\pi}{2}) + \pi \times (\dfrac{\pi}{2}) +\pi \times \sin (-\dfrac{\pi}{2})]) \\=\pi(\pi-2)$$
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