Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 322: 18

$\dfrac{\pi^2}{16} $

We integrate the integral to calculate the volume as follows: $V= (\pi/4) \times \int_{0}^{\pi/2} \sin^2 (2x) dx $ Now, $V=(\pi/4) \int_{0}^{\pi/2} (1- \cos 4x) dx$ or, $=(\pi/8) (\dfrac{\pi}{2}-\dfrac{\sin (2 \pi) }{4}-0+0) dy$ or, $= (\pi/8)[\dfrac{\pi}{2}-0]$ or, $=\dfrac{\pi^2}{16} $