Answer
$$\frac{2}{3}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\int_0^{2 - y} y dxdy} \cr
& \int_0^1 {\left[ {\int_0^{2 - y} y dx} \right]dy} \cr
& \int_0^1 {y\left[ {\int_0^{2 - y} {dx} } \right]dy} \cr
& {\text{Evaluate inner integral}} \cr
& \int_0^{2 - y} {dx} = \left[ x \right]_{x = 0}^{x = 2 - y} \cr
& = 2 - y \cr
& {\text{Therefore,}} \cr
& \int_0^1 {y\left[ {\int_0^{2 - y} {dx} } \right]dy} = \int_0^1 {y\left( {2 - y} \right)dy} \cr
& = \int_0^1 {\left( {2y - {y^2}} \right)dy} \cr
& {\text{Integrating}} \cr
& = \left[ {{y^2} - \frac{1}{3}{y^3}} \right]_0^1 \cr
& = \left[ {{{\left( 1 \right)}^2} - \frac{1}{3}{{\left( 1 \right)}^3}} \right] - \left[ {{{\left( 0 \right)}^2} - \frac{1}{3}{{\left( 0 \right)}^3}} \right] \cr
& = 1 - \frac{1}{3} \cr
& = \frac{2}{3} \cr} $$
