Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 322: 33

$\pi(\pi-2)$

Area $=\pi r^2=\pi (1-\cos x)$ We integrate the integral to calculate the volume as follows: $V= \pi \times \int_{- \pi/2}^{\pi/2} (1-\cos x) dx$ Now, $V= \pi (x-\sin x)_{- \pi/2}^{\pi/2}$ or, $= \pi [\dfrac{\pi}{2}-\sin \dfrac{\pi}{2}-(-\dfrac{\pi}{2}) +\sin (-\dfrac{\pi}{2})])$ or, $=\pi(\pi-2)$