Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 322: 15

$\dfrac{2 \pi}{3}$

We integrate the integral to calculate the volume as follows: $V= (\pi) \times \int_{0}^{2} (1-x+\dfrac{x^2}{4}) dx$ Now, $V=(\pi) (x-\dfrac{x^2}{2}+\dfrac{x^3}{12})]{0}^{2}$ or, $=\pi (2-2+\dfrac{2}{3})$ or, $=\dfrac{2 \pi}{3}$