Answer
$$V = \frac{{280}}{3}\pi $$
Work Step by Step
$$\eqalign{
& {\text{Let }}y = 8,{\text{ }}y = 2x + 2,{\text{ }}x = 0{\text{ and }}x = 2 \cr
& y = 2x + 2 \to x = \frac{1}{2}y - 1 \cr
& {\text{Use the Shell method about the }}x{\text{ - axis}} \cr
& V = \int_c^d {2\pi y\left[ {p\left( y \right) - q\left( y \right)} \right]} dy \cr
& x = \frac{1}{2}y - 1 > x = 0{\text{ on the interval }}\left[ {2,6} \right] \cr
& {\text{Therefore}}{\text{,}} \cr
& V = \int_2^6 {2\pi y\left( {\frac{1}{2}y - 1 - 0} \right)} dy + \int_6^8 {2\pi y\left( {2 - 0} \right)} dy \cr
& V = 2\pi \int_2^6 {\left( {\frac{1}{2}{y^2} - y} \right)} dy + 4\pi \int_6^8 y dy \cr
& {\text{Integrating}} \cr
& V = 2\pi \left[ {\frac{1}{6}{y^3} - \frac{1}{2}{y^2}} \right]_2^6 + 4\pi \left[ {\frac{1}{2}{y^2}} \right]_6^8 \cr
& V = 2\pi \left[ {\frac{1}{6}{{\left( 6 \right)}^3} - \frac{1}{2}{{\left( 6 \right)}^2}} \right] - 2\pi \left[ {\frac{1}{6}{{\left( 2 \right)}^3} - \frac{1}{2}{{\left( 2 \right)}^2}} \right] + 2\pi \left[ {{8^2} - {6^2}} \right] \cr
& {\text{Simplifying}} \cr
& V = 2\pi \left( {18} \right) - 2\pi \left( { - \frac{2}{3}} \right) + 2\pi \left( {28} \right) \cr
& V = \frac{{280}}{3}\pi \cr} $$
