Chapter 6 - Section 6.1 - Volumes Using Cross-Sections - Exercises - Page 356: 46

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

We need to integrate the integral to compute the volume. We have: $$ Area =(R^2- r^2) \pi=\pi (1^2-(1-y)^2) =\pi (2y-y^2)$$ Now, $$Volume = \pi \times \int_{0}^{1} (2y-y^2) \space dy \\ = \pi [y^2-\dfrac{y^3}{3}]_{0}^{1} \\= \dfrac{ 2\pi}{3}$$