Chapter 6 - Section 6.1 - Volumes Using Cross-Sections - Exercises - Page 356: 33

$$2 \pi$$

We need to integrate the integral to compute the volume. We have: Area $=\pi r^2=(\pi) (\sqrt {2 \sin 2y})^2=2 \pi \sin 2y$ $$Volume = 2 \pi \int_{0}^{\pi/2} \sin (2y) \space dy \\= 2 \pi [(-\dfrac{1}{2} ) \cos (2y)]_{0}^{\pi} \\=- \pi [\cos \pi -\cos (0)] \\=2 \pi$$