Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 322: 22

$\dfrac{\pi}{30}$

Area $=\pi r^2=\pi (x-x^2)^2=\pi (x^2-2x^3+x^4)$ We integrate the integral to calculate the volume as follows: $V= \int_{0}^{1} \pi (x^2-2x^3+x^4) dx$ Now, $V=[\dfrac{x^3}{3}-\dfrac{2x^4}{4}+\dfrac{x^5}{5}]_0^1 $ or, $=(\pi)(\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{5})$ or, $=\dfrac{\pi}{30}$