Chapter 6 - Section 6.1 - Volumes Using Cross-Sections - Exercises - Page 356: 32

$$4 \pi$$

We need to integrate the integral to compute the volume. We have: Area $=\pi r^2=\pi (y^{3/2} )^2= \pi y^3$ Now, $$Volume = \pi \int_{0}^{2} y^3 \space dy \\ =\pi [\dfrac{y^4}{4}]_{0}^{2} \\= \pi [\dfrac{16}{4}] \\ =4 \pi$$