Answer
$$\pi$$
Work Step by Step
We need to integrate the integral to compute the volume.
We have:
$$Area=(R^2- r^2) \pi=\pi (\sec^2 x-\tan^2 x)$$
Now,
$$Volume = \int_{0}^{1} (\sec^2 x-\tan^2 x) \space dx \times \pi \\ = \pi [\tan x-(\tan x-x)]_{0}^{1} \\= \pi (x)_0^1 \\=\pi$$