Answer
\[\frac{2}{3}\]
Work Step by Step
\[\begin{align}
& \text{From the graph we can define the region }R\text{ in the quadrant I} \\
& \text{as:} \\
& R=\left\{ \left( x,y \right):0\le y\le 1-x,\text{ }0\le x\le 1\text{ } \right\} \\
& \text{Then, by symmetry of }z={{x}^{2}}+{{y}^{2}} \\
& \iint_{R}{f\left( x,y \right)}=4\int_{0}^{1}{\int_{0}^{1-x}{\left( {{x}^{2}}+{{y}^{2}} \right)}dydx} \\
& \text{Integrating} \\
& =4\int_{0}^{1}{\left[ {{x}^{2}}y+\frac{1}{3}{{y}^{3}} \right]_{0}^{1-x}dx} \\
& =4\int_{0}^{1}{\left[ {{x}^{2}}\left( 1-x \right)+\frac{1}{3}{{\left( 1-x \right)}^{3}} \right]dx} \\
& =4\int_{0}^{1}{\left[ {{x}^{2}}-{{x}^{3}}-\frac{1}{3}{{\left( 1-x \right)}^{3}}\left( -1 \right) \right]dx} \\
& =4\left[ \frac{1}{3}{{x}^{3}}-\frac{1}{4}{{x}^{4}}-\frac{1}{12}{{\left( 1-x \right)}^{4}} \right]_{0}^{1} \\
& =4\left[ \frac{1}{3}{{\left( 1 \right)}^{3}}-\frac{1}{4}{{\left( 1 \right)}^{4}}-\frac{1}{12}{{\left( 1-1 \right)}^{4}} \right]-4\left[ 0-0-\frac{1}{12}{{\left( 1-0 \right)}^{4}} \right] \\
& =\frac{1}{3}+\frac{1}{3} \\
& =\frac{2}{3} \\
\end{align}\]
