Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 322: 29

$2 \pi$

Area $=\pi r^2=\pi (\sqrt {2 \sin 2y})^2=2 \pi \sin 2y$ We integrate the integral to calculate the volume as follows: $V= 2 \pi \int_{0}^{\pi/2} \sin 2y dy$ Now, $V= 2 \pi [(-1/2) \cos (2y)]_{0}^{\pi/2}$ or, $=- \pi [\cos \pi -\cos (0)]$ $V=2 \pi$