Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 322: 42

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

Area $=R^2\pi- r^2 \pi=\pi (1^2-(1-y)^2) =\pi (2y-y^2)$ We integrate the integral to calculate the volume as follows: $V= \pi \times \int_{0}^{1} (2y-y^2) dy$ Now, $V= \pi [y^2-\dfrac{y^3}{3}]_{0}^{1}$ or, $= \pi (1-\dfrac{1}{3})$ or, $=\dfrac{ 2\pi}{3}$