Answer
$$V = \int_0^4 {\pi \left( {y - \frac{1}{4}{y^2}} \right)} dy$$
Work Step by Step
$$\eqalign{
& {\text{Let the graphs }}f\left( x \right) = 2x{\text{ and }}g\left( x \right) = {x^2} \cr
& {\text{Solving for }}y \cr
& y = 2x \to x = \frac{1}{2}y,{\text{ }}q\left( y \right) = \frac{1}{2}y \cr
& y = {x^2} \to x = \sqrt y ,{\text{ }}p\left( y \right) = \sqrt y \cr
& {\text{The graph of the region }}R{\text{ is shown below}}{\text{.}} \cr
& {\text{From the graph we can note that the interval of integration}} \cr
& {\text{is from }}\underbrace {\left[ {0,4} \right]}_{\left[ {c,d} \right]} \cr
& {\text{Revolving the region about the }}y{\text{ - axis}}{\text{, using the}} \cr
& {\text{Washer Method about the }}y{\text{ - axis }} \cr
& V = \int_a^b {\pi \left[ {p{{\left( y \right)}^2} - q{{\left( y \right)}^2}} \right]} dy \cr
& {\text{We can represent the volume as:}} \cr
& V = \int_0^4 {\pi \left[ {{{\left( {\sqrt y } \right)}^2} - {{\left( {\frac{1}{2}y} \right)}^2}} \right]} dy \cr
& V = \int_0^4 {\pi \left( {y - \frac{1}{4}{y^2}} \right)} dy \cr} $$
