Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 322: 40

$\pi$

Area $=R^2\pi- r^2 \pi=\pi (\sec^2 x-\tan^2 x)$ We integrate the integral to calculate the volume as follows: $V= \pi \times \int_{0}^{1} (\sec^2 x-\tan^2 x) dx$ Now, $V= \pi [\tan x-(\tan x-x)]_{0}^{1}$ or, $= \pi (x)_0^1$ or, $=\pi$