Answer
$$\pi(\pi-2)$$
Work Step by Step
We need to integrate the integral to compute the volume.
We have:
$$Area =(R^2- r^2) \pi=\pi ((\sqrt 2)^2-\sec^2 x)$$
Now,
$$Volume = \int_{-\pi/4}^{\pi/4} (2-\sec^2 x) \space dx \times \pi \\ = [2x-\tan x]_{-\pi/4}^{\pi/4} \times \pi \\= \pi [\dfrac{\pi}{2}-1+\dfrac{\pi}{2}+(-1)] \\=\pi(\pi-2)$$
