Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 322: 24

$2 \pi$

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