Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 322: 39

$\pi(\pi-2)$

Area $=R^2\pi- r^2 \pi=\pi ((\sqrt 2)^2-\sec^2 x)$ We integrate the integral to calculate the volume as follows: $V= \pi \times \int_{-\pi/4}^{\pi/4} (2-\sec^2 x) dx$ Now, $V= \pi [2x-\tan x]_{-\pi/4}^{\pi/4}$ or, $= \pi (\dfrac{\pi}{2}-1+\dfrac{\pi}{2}+(-1))$ or, $=\pi(\pi-2)$