Answer
$$V = 8\pi $$
Work Step by Step
$$\eqalign{
& x = {y^2},{\text{ }}x = 4,{\text{ and }}y = 0 \cr
& {\text{Use the Shell method about the }}x{\text{ - axis}} \cr
& V = \int_c^d {2\pi y\left[ {p\left( y \right) - q\left( y \right)} \right]} dy \cr
& {\text{From the graph we can see that}} \cr
& 4 > {y^2}{\text{ on the interval }}\left[ {0,2} \right] \cr
& {\text{Therefore}}{\text{,}} \cr
& V = \int_0^2 {2\pi y\left( {4 - {y^2}} \right)} dy \cr
& V = 2\pi \int_0^2 {\left( {4y - {y^3}} \right)} dy \cr
& {\text{Integrating}} \cr
& V = 2\pi \left[ {2{y^2} - \frac{1}{4}{y^4}} \right]_0^2 \cr
& V = 2\pi \left[ {2{{\left( 2 \right)}^2} - \frac{1}{4}{{\left( 2 \right)}^4}} \right] - 2\pi \left[ {2{{\left( 0 \right)}^2} - \frac{1}{4}{{\left( 0 \right)}^4}} \right] \cr
& {\text{Simplifying}} \cr
& V = 2\pi \left[ 4 \right] - 2\pi \left[ 0 \right] \cr
& V = 8\pi \cr} $$
