Answer
\[5\]
Work Step by Step
\[\begin{align}
& y=\sqrt{x}\Rightarrow x={{y}^{2}} \\
& y=-x+2\Rightarrow x=2-y \\
& \text{From the graph, the region }R\text{ is} \\
& R=\left\{ \left( x,y \right):{{y}^{2}}\le x\le 2-y,\text{ 0}\le y\le 2\text{ } \right\} \\
& \text{Then,} \\
& \iint_{R}{12y}dA=\int_{0}^{2}{\int_{{{y}^{2}}}^{2-y}{12ydx}dy} \\
& \text{Integrating} \\
& =\int_{0}^{1}{\left[ 12xy \right]_{{{y}^{2}}}^{2-y}dy} \\
& =\int_{0}^{1}{\left[ 12\left( 2-y \right)y-12\left( {{y}^{2}} \right)y \right]dy} \\
& =\int_{0}^{1}{\left( 24y-12{{y}^{2}}-12{{y}^{3}} \right)dy} \\
& =\left[ 12{{y}^{2}}-4{{y}^{3}}-3{{y}^{4}} \right]_{0}^{1} \\
& =\left[ 12{{\left( 1 \right)}^{2}}-4{{\left( 1 \right)}^{3}}-3{{\left( 1 \right)}^{4}} \right]-\left[ 12{{\left( 0 \right)}^{2}}-4{{\left( 0 \right)}^{3}}-3{{\left( 0 \right)}^{4}} \right] \\
& =5 \\
\end{align}\]
