Answer
$$V = \frac{{608}}{3}\pi $$
Work Step by Step
$$\eqalign{
& {\text{Let the functions }}y = x,{\text{ }}y = 2x + 2,{\text{ and the lines }}x = 2,{\text{ }}x = 6 \cr
& {\text{revolved about the }}y{\text{ - axis}} \cr
& {\text{Using the shell method about the }}y{\text{ - axis}} \cr
& V = \int_a^b {2\pi x\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr
& {\text{From the graph of the region shown below}} \cr
& f\left( x \right) = 2x + 2{\text{ and }}g\left( x \right) = x{\text{ on the interval }}\left[ {2,6} \right],{\text{ then}} \cr
& V = \int_2^6 {2\pi x\left[ {\left( {2x + 2} \right) - x} \right]} dx \cr
& V = 2\pi \int_2^6 {x\left( {x + 2} \right)} dx \cr
& V = 2\pi \int_2^6 {\left( {{x^2} + 2x} \right)} dx \cr
& {\text{Integrating}} \cr
& V = 2\pi \left[ {\frac{1}{3}{x^3} + {x^2}} \right]_2^6 \cr
& V = 2\pi \left[ {\frac{1}{3}{{\left( 6 \right)}^3} + {{\left( 6 \right)}^2}} \right] - 2\pi \left[ {\frac{1}{3}{{\left( 2 \right)}^3} + {{\left( 2 \right)}^2}} \right] \cr
& V = 2\pi \left( {108} \right) - 2\pi \left( {\frac{{20}}{3}} \right) \cr
& V = \frac{{608}}{3}\pi \cr} $$