## University Calculus: Early Transcendentals (3rd Edition)

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