Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 322: 28

$4 \pi$

Area $=\pi r^2=\pi (y^{3/2} )^2= \pi y^3$ We integrate the integral to calculate the volume as follows: $V= \pi \int_{0}^{2} y^3 dy$ Now, $V=\pi [\dfrac{y^4}{4}]_{0}^{2}$ or, $= \pi [\dfrac{16}{4}]$ $V=4 \pi$