Chapter 6 - Section 6.1 - Volumes Using Cross-Sections - Exercises - Page 356: 45

$$\dfrac{4\pi}{3}$$

We need to integrate the integral to compute the volume. We have: $$Area =(R^2- r^2) \pi=\pi (y^2+2y)$$ Now, $$Volume = \int_{0}^{1} (y^2+2y) \space dy \times \pi \\ = \pi [\dfrac{y^3}{3} +y^2]_{0}^{1} \\= \pi (\dfrac{1}{3} +1)_0^1 \\=\dfrac{4\pi}{3}$$