Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 322: 25

$2.3014$

We integrate the integral to calculate the volume as follows: $V= \pi \int_{0}^{\pi/4} \pi (\sqrt 2-\sec x \tan x)^2 dx$ Now, $V=\pi \int_{0}^{\pi/4} \pi (2- 2\sqrt 2 \sec x \tan x+\tan^2 x \sec^2 x) dx$ or, $=(\pi) [2x-2\sqrt 2 \sec x+(1/3) \tan^3 x)]_0^{\pi/4}$ or, $=2.3014$