Answer
\[\frac{2}{3}\]
Work Step by Step
\[\begin{align}
& y=-x+1\Rightarrow x=1-y \\
& y=x-1\Rightarrow x=y+1 \\
& \text{From the graph, the region }R\text{ is} \\
& R=\left\{ \left( x,y \right):1-y\le x\le y+1,\text{ 0}\le y\le 1\text{ } \right\} \\
& \text{Then,} \\
& \iint_{R}{{{y}^{2}}}dA=\int_{0}^{1}{\int_{1-y}^{y+1}{{{y}^{2}}dx}dy} \\
& \text{Integrating} \\
& =\int_{0}^{1}{\left[ x{{y}^{2}} \right]_{1-y}^{y+1}dy} \\
& =\int_{0}^{1}{\left[ \left( y+1 \right){{y}^{2}}-\left( y-1 \right){{y}^{2}} \right]dy} \\
& =\int_{0}^{1}{\left( {{y}^{3}}+{{y}^{2}}-{{y}^{3}}+{{y}^{2}} \right)dy} \\
& =\int_{0}^{1}{2{{y}^{2}}dy} \\
& =\left[ \frac{2{{y}^{3}}}{3} \right]_{0}^{1} \\
& =\frac{2{{\left( 1 \right)}^{3}}}{3}-\frac{2{{\left( 0 \right)}^{3}}}{3} \\
& =\frac{2}{3} \\
\end{align}\]
