Answer
$$V = \frac{{32\pi }}{3}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}y = 4 - x,{\text{ }}y = 2,{\text{ }}x = 0{\text{ }} \cr
& y = 4 - x \to x = 4 - y \cr
& {\text{Use the Shell method about the }}x{\text{ - axis}} \cr
& V = \int_c^d {2\pi y\left[ {p\left( y \right) - q\left( y \right)} \right]} dy \cr
& 4 - y > 0{\text{ on the interval }}\left[ {2,4} \right] \cr
& {\text{Therefore}}{\text{,}} \cr
& V = \int_2^4 {2\pi y\left( {4 - y} \right)} dy \cr
& V = 2\pi \int_2^4 {\left( {4y - {y^2}} \right)} dy \cr
& {\text{Integrating}} \cr
& V = 2\pi \left[ {2{y^2} - \frac{1}{3}{y^3}} \right]_2^4 \cr
& V = 2\pi \left[ {2{{\left( 4 \right)}^2} - \frac{1}{3}{{\left( 4 \right)}^3}} \right] - 2\pi \left[ {2{{\left( 2 \right)}^2} - \frac{1}{3}{{\left( 2 \right)}^3}} \right] \cr
& {\text{Simplifying}} \cr
& V = 2\pi \left( {\frac{{32}}{3}} \right) - 2\pi \left( {\frac{{16}}{3}} \right) \cr
& V = \frac{{32\pi }}{3} \cr} $$
