Answer
$${\text{Volumes are equal}}{\text{}}$$
Work Step by Step
$$\eqalign{
& {\text{Let the graphs }}y = 2x{\text{ and }}x = 5 \cr
& \cr
& *{\text{ Revolving the region about the }}x{\text{ - axis:}} \cr
& {\text{Using the Washer Method about the }}x{\text{ - axis:}} \cr
& V = \int_a^b {\pi \left[ {f{{\left( x \right)}^2} - g{{\left( x \right)}^2}} \right]} dx \cr
& {\text{From the graph the interval of integration }}\left[ {a,b} \right]{\text{ is }}\left[ {0,5} \right] \cr
& 2x \geqslant 0,{\text{ for }}\left[ {0,5} \right] \to f\left( x \right) = 2x{\text{ and }}g\left( x \right) = 0{\text{ }} \cr
& {\text{We can represent the volume as:}} \cr
& V = \int_0^5 {\pi \left[ {{{\left( {2x} \right)}^2} - {{\left( 0 \right)}^2}} \right]} dx \cr
& V = \int_0^5 {\pi \left( {4{x^2} - 0} \right)} dx \cr
& {\text{Integrating we obtain}} \cr
& V = \frac{{500}}{3}\pi \cr
& \cr
& *{\text{ Revolving the region about the }}y{\text{ - axis:}} \cr
& {\text{Using the Washer Method about the }}y{\text{ - axis:}} \cr
& V = \int_c^d {\pi \left[ {p{{\left( y \right)}^2} - g{{\left( y \right)}^2}} \right]} dy \cr
& {\text{From the graph the interval of integration }}\left[ {c,d} \right]{\text{ is }}\left[ {0,10} \right] \cr
& y = 2x \to x = \frac{1}{2}y{\text{ and }}x = 5 \cr
& 5 \geqslant \frac{1}{2}y{\text{ for }}\left[ {0,10} \right] \to p\left( y \right) = 5{\text{ and }}q\left( y \right) = \frac{1}{2}y \cr
& {\text{We can represent the volume as:}} \cr
& V = \int_0^{10} {\pi \left[ {{{\left( 5 \right)}^2} - {{\left( {\frac{1}{2}y} \right)}^2}} \right]} dy \cr
& V = \int_0^{10} {\pi \left( {25 - \frac{1}{4}{y^2}} \right)} dy \cr
& {\text{Integrating we obtain}} \cr
& V = \frac{{500}}{3}\pi \cr
& \cr
& {\text{Therefore}}{\text{, the volumes are equal}}{\text{.}} \cr
& \cr
& {\text{Graph}} \cr} $$