Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 322: 23

$\pi$

Area $=\pi r^2=\pi (\sqrt {\cos x})^2=\pi \cos (x)$ We integrate the integral to calculate the volume as follows: $V= \int_{0}^{\pi/2} \pi \cos (x)$ Now, $V=(\pi) [(\sin (x)]_0^{\pi/2}$ or, $=(\pi) [\sin \dfrac{\pi}{2}-\sin (0)]$ or, $=\pi$