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