Answer
\[\frac{152}{3}\]
Work Step by Step
\[\begin{align}
& \iint_{R}{\left( x+y \right)}dA \\
& \text{The region }R\text{ is represented in the graph shown below} \\
& \text{Then,} \\
& R=\left\{ \left( x,y \right):{{x}^{2}}\le y\le 8-{{x}^{2}},\text{ }0\le x\le 2 \right\} \\
& \iint_{R}{xy}dA=\int_{0}^{2}{\int_{{{x}^{2}}}^{8-{{x}^{2}}}{\left( x+y \right)}}dydx \\
& \text{Integrate} \\
& =\int_{0}^{2}{\left[ xy+\frac{1}{2}{{y}^{2}} \right]_{{{x}^{2}}}^{8-{{x}^{2}}}}dx \\
& =\int_{0}^{2}{\left[ x\left( 8-{{x}^{2}} \right)+\frac{1}{2}{{\left( 8-{{x}^{2}} \right)}^{2}}-x\left( {{x}^{2}} \right)-\frac{1}{2}{{\left( {{x}^{2}} \right)}^{2}} \right]}dx \\
& =\int_{0}^{2}{\left( 8x-{{x}^{3}}+32-8{{x}^{2}}+\frac{1}{2}{{x}^{4}}-{{x}^{3}}-\frac{1}{2}{{x}^{4}} \right)}dx \\
& =\int_{0}^{2}{\left( 8x+32-8{{x}^{2}}-2{{x}^{3}} \right)}dx \\
& =\left[ 4{{x}^{2}}+32x-\frac{8}{3}{{x}^{3}}-\frac{1}{2}{{x}^{4}} \right]_{0}^{2} \\
& =\left[ 4{{\left( 2 \right)}^{2}}+32\left( 2 \right)-\frac{8}{3}{{\left( 2 \right)}^{3}}-\frac{1}{2}{{\left( 2 \right)}^{4}} \right]-\left[ 0 \right] \\
& =\frac{152}{3} \\
\end{align}\]
