Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 322: 34

$\pi(\dfrac{\pi}{2}-1)$

Area $=\pi r^2=\pi (1-\tan^2 y)$ We integrate the integral to calculate the volume as follows: $V= \pi \times \int_{0}^{\pi/4} (1-tan^2 y) dy$ Now, $V= \pi (y-(\tan y -y))]_{0}^{\pi/4}$ or, $= \pi [2y-\tan y]_{0}^{\pi/4}$ or, $=\pi(\dfrac{\pi}{2}-1)$