Answer
\[14\]
Work Step by Step
\[\begin{align}
& x=4-{{y}^{2}} \\
& y=-x+2\Rightarrow x=2-y \\
& \text{From the graph, the region }R\text{ is} \\
& R=\left\{ \left( x,y \right):2-y\le x\le 4-{{y}^{2}},\text{ 0}\le y\le 2\text{ } \right\} \\
& \text{Then,} \\
& \iint_{R}{{{y}^{2}}}dA=\int_{0}^{2}{\int_{2-y}^{4-{{y}^{2}}}{3xydx}dy} \\
& \text{Integrating} \\
& =\int_{0}^{2}{\left[ \frac{3{{x}^{2}}y}{2} \right]_{2-y}^{4-{{y}^{2}}}dy} \\
& =\frac{3}{2}\int_{0}^{2}{\left[ {{x}^{2}}y \right]_{2-y}^{4-{{y}^{2}}}dy} \\
& =\frac{3}{2}\int_{0}^{2}{\left[ {{\left( 4-{{y}^{2}} \right)}^{2}}y-{{\left( 2-y \right)}^{2}}y \right]dy} \\
& =\frac{3}{2}\int_{0}^{2}{\left( 16y-8{{y}^{3}}+{{y}^{5}}-4y+4{{y}^{2}}-{{y}^{3}} \right)dy} \\
& =\frac{3}{2}\int_{0}^{2}{\left( 12y-9{{y}^{3}}+{{y}^{5}}+4{{y}^{2}} \right)dy} \\
& =\frac{3}{2}\left[ 6{{y}^{2}}-\frac{9}{4}{{y}^{4}}+\frac{1}{6}{{y}^{6}}+\frac{4}{3}{{y}^{3}} \right]_{0}^{2} \\
& =\frac{3}{2}\left[ 6{{\left( 2 \right)}^{2}}-\frac{9}{4}{{\left( 2 \right)}^{4}}+\frac{1}{6}{{\left( 2 \right)}^{6}}+\frac{4}{3}{{\left( 2 \right)}^{3}} \right] \\
& =\frac{3}{2}\left[ \frac{28}{3} \right] \\
& =14 \\
\end{align}\]
