Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 322: 17

$(4-\pi) $

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