Answer
\[\frac{4}{3}\]
Work Step by Step
\[\begin{align}
& \text{From the graph we can define the region }R\text{ in the quadrant I} \\
& \text{as:} \\
& R=\left\{ \left( x,y \right):-\frac{1}{2}x+1\le y\le -x+2,\text{ }0\le x\le 2\text{ } \right\} \\
& \text{Then, by symmetry of }z={{x}^{2}} \\
& \iint_{R}{f\left( x,y \right)}=2\int_{0}^{2}{\int_{-\frac{1}{2}x+1}^{-x+2}{{{x}^{2}}}dydx} \\
& \text{Integrating} \\
& =2\int_{0}^{2}{\left[ {{x}^{2}}y \right]_{1-\frac{1}{2}x}^{2-x}dx} \\
& =2\int_{0}^{2}{\left[ {{x}^{2}}\left( 2-x \right)-{{x}^{2}}\left( 1-\frac{1}{2}x \right) \right]dx} \\
& =2\int_{0}^{2}{\left[ 2{{x}^{2}}-{{x}^{3}}-{{x}^{2}}+\frac{1}{2}{{x}^{3}} \right]dx} \\
& =2\int_{0}^{2}{\left( {{x}^{2}}-\frac{1}{2}{{x}^{3}} \right)dx} \\
& =2\left[ \frac{1}{3}{{x}^{3}}-\frac{1}{8}{{x}^{4}} \right]_{0}^{2} \\
& =2\left[ \frac{1}{3}{{\left( 2 \right)}^{3}}-\frac{1}{8}{{\left( 2 \right)}^{4}} \right]-2\left[ \frac{1}{3}{{\left( 0 \right)}^{3}}-\frac{1}{8}{{\left( 0 \right)}^{4}} \right] \\
& =\frac{4}{3} \\
\end{align}\]
