Chapter 6: Applications of Definite Integrals - Section 6.2 - Volumes Using Cylindrical Shels - Exercises 6.2 - Page 330: 20

$\dfrac{ 8\pi}{3}$

We need to use the shell model as follows: $V=\int_p^{q} (2 \pi) \cdot (\space radius \space of \space shell) ( height \space of \space Shell) \space dx$ $ \implies V= \int_{0}^{2} (2 \pi) \cdot y (y-\dfrac{y}{2}) dy$ Now, $V= 2\pi \times [\dfrac{y^3}{6}]_{0}^{2}$ or, $=2 \pi (\dfrac{8}{6})$ or, $=\dfrac{ 8\pi}{3}$