Answer
$$V = 288\pi $$
Work Step by Step
$$\eqalign{
& {\text{Let the functions }} \cr
& y = 2{\text{ and }}y = 2x + 2,{\text{ }}x = 6,{\text{ revolved about }}y{\text{ - axis}} \cr
& y = 2x + 2 \to x = \frac{1}{2}y - 1 \cr
& {\text{From the graph of the region shown below}} \cr
& 6 > \frac{1}{2}y - 1{\text{ on the interval }}2 \leqslant y \leqslant 14 \cr
& {\text{Using the washer method about the }}y{\text{ - axis}} \cr
& V = \int_c^d {\pi \left[ {p{{\left( y \right)}^2} - q{{\left( y \right)}^2}} \right]} dy \cr
& {\text{Let }}p\left( y \right) = 6{\text{ and }}q\left( y \right) = \frac{1}{2}y - 1 \cr
& V = \int_2^{14} {\pi \left[ {{{\left( 6 \right)}^2} - {{\left( {\frac{1}{2}y - 1} \right)}^2}} \right]} dy \cr
& V = \pi \int_2^{14} {\left( {36 - \frac{1}{4}{y^2} + y - 1} \right)} dy \cr
& V = \pi \int_2^{14} {\left( {35 - \frac{1}{4}{y^2} + y} \right)} dy \cr
& {\text{Integrating}} \cr
& V = \pi \left[ {35y - \frac{1}{{12}}{y^3} + \frac{1}{2}{y^2}} \right]_2^{14} \cr
& V = \pi \left[ {35\left( {14} \right) - \frac{1}{{12}}{{\left( {14} \right)}^3} + \frac{1}{2}{{\left( {14} \right)}^2}} \right] - \pi \left[ {35\left( 2 \right) - \frac{1}{{12}}{{\left( 2 \right)}^3} + \frac{1}{2}{{\left( 2 \right)}^2}} \right] \cr
& {\text{Simplifying}} \cr
& V = \pi \left( {\frac{{1078}}{3}} \right) - \pi \left( {\frac{{214}}{3}} \right) \cr
& V = 288\pi \cr} $$
