Answer
$$V = \frac{{76}}{3}\pi $$
Work Step by Step
$$\eqalign{
& {\text{We have }}x = 2y,\,\,\,y = 2,{\text{ }}\,y = 3\,\,x = 0 \cr
& {\text{The volume of the solid can be calculated using cylindrical shells}} \cr
& V = \int_c^d {2\pi y\left[ {f\left( y \right) - g\left( y \right)} \right]dy} \cr
& {\text{Let }}f\left( y \right) = 2y{\text{ and }}g\left( y \right) = 0{\text{ and the interval }}2 \leqslant y \leqslant 3 \cr
& V = \int_2^3 {2\pi y\left( {2y - 0} \right)dy} \cr
& V = 4\pi \int_2^3 {{y^2}dy} \cr
& V = \frac{{4\pi }}{3}\left[ {{y^3}} \right]_2^3 \cr
& {\text{Integrating}} \cr
& V = \frac{{4\pi }}{3}\left[ {{{\left( 3 \right)}^3} - {{\left( 2 \right)}^3}} \right] \cr
& V = \frac{{4\pi }}{3}\left( {27 - 8} \right) \cr
& V = \frac{{76}}{3}\pi \cr} $$
