Answer
$$V = \frac{2}{3}\pi $$
Work Step by Step
$$\eqalign{
& y = x,{\text{ }}y = 2 - x,{\text{ and }}y = 0 \cr
& y = 2 - x \to x = 2 - y \cr
& {\text{Use the Shell method about the }}x{\text{ - axis}} \cr
& V = \int_c^d {2\pi y\left[ {p\left( y \right) - q\left( y \right)} \right]} dy \cr
& {\text{From the graph we can see that}} \cr
& 2 - y > y{\text{ on the interval }}\left[ {0,1} \right] \cr
& {\text{Therefore}}{\text{,}} \cr
& V = \int_0^1 {2\pi y\left( {2 - y - y} \right)} dy \cr
& V = 2\pi \int_0^1 {\left( {2y - 2{y^2}} \right)} dy \cr
& V = 4\pi \int_0^1 {\left( {y - {y^2}} \right)} dy \cr
& {\text{Integrating}} \cr
& V = 4\pi \left[ {\frac{1}{2}{y^2} - \frac{1}{3}{y^3}} \right]_0^1 \cr
& V = 4\pi \left[ {\frac{1}{2}{{\left( 1 \right)}^2} - \frac{1}{3}{{\left( 1 \right)}^3}} \right] - 4\pi \left[ {\frac{1}{2}{{\left( 0 \right)}^2} - \frac{1}{3}{{\left( 0 \right)}^3}} \right] \cr
& {\text{Simplifying}} \cr
& V = 4\pi \left( {\frac{1}{6}} \right) \cr
& V = \frac{2}{3}\pi \cr} $$
