Answer
$$V = \frac{{81}}{2}\pi $$
Work Step by Step
$$\eqalign{
& x = {y^2},{\text{ }}x = 0,{\text{ and }}y = 3 \cr
& {\text{Use the Shell method about the }}x{\text{ - axis}} \cr
& V = \int_c^d {2\pi y\left[ {p\left( y \right) - q\left( y \right)} \right]} dy \cr
& {\text{From the graph we can see that}} \cr
& {y^2} > 0{\text{ on the interval }}\left[ {0,3} \right] \cr
& {\text{Therefore}}{\text{,}} \cr
& V = \int_0^3 {2\pi y\left( {{y^2} - 0} \right)} dy \cr
& V = 2\pi \int_0^3 {{y^3}} dy \cr
& {\text{Integrating}} \cr
& V = 2\pi \left[ {\frac{1}{4}{y^4}} \right]_0^3 \cr
& V = \frac{\pi }{2}\left[ {{y^4}} \right]_0^3 \cr
& {\text{Simplifying}} \cr
& V = \frac{\pi }{2}\left[ {{{\left( 3 \right)}^4} - {{\left( 0 \right)}^4}} \right] \cr
& V = \frac{{81}}{2}\pi \cr} $$
