Answer
$$V = 8\pi $$
Work Step by Step
$$\eqalign{
& y = 3\left( {2 - x} \right),{\text{ }}y = 0,{\text{ }}x = 0 \cr
& y = 6 - 3x \cr
& x = 2 - \frac{1}{3}y \cr
& {\text{Let }}R\left( y \right) = 2 - \frac{1}{3}y \cr
& {\text{Apply the disk method}} \cr
& V = \pi \int_c^d {{{\left[ {R\left( y \right)} \right]}^2}} dy \cr
& {\text{So}},{\text{ the volume of the solid of revolution is}} \cr
& V = \pi \int_0^6 {{{\left( {2 - \frac{1}{3}y} \right)}^2}} dy \cr
& V = \pi \int_0^6 {\left( {4 - \frac{4}{3}y + \frac{1}{9}{y^2}} \right)} dy \cr
& {\text{Integrate}} \cr
& V = \pi \left[ {4y - \frac{2}{3}{y^2} + \frac{1}{{27}}{y^3}} \right]_0^6 \cr
& V = \pi \left[ {4\left( 6 \right) - \frac{2}{3}{{\left( 6 \right)}^2} + \frac{1}{{27}}{{\left( 6 \right)}^3}} \right] - \pi \left[ 0 \right] \cr
& V = 8\pi \cr} $$
