Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 322: 30

$4$

Area $=\pi r^2=\pi (\sqrt {\cos (\pi y/4)})^2=\pi \cos (\pi y /4)$ We integrate the integral to calculate the volume as follows: $V= \pi \int_{-2}^{0} \cos (\pi y /4) dy$ Now, $V= \pi [(4/pi) \sin (\pi y/4)]_{-2}^{0}$ or, $=4 [\sin (0) -\sin (\dfrac{-\pi}{2})]$ $V=4$