Answer
\[\frac{2}{15}\]
Work Step by Step
\[\begin{align}
& \text{From the graph we can define the region }R\text{ as:} \\
& R=\left\{ \left( x,y \right):0\le x\le \sqrt{1-{{y}^{2}}},\text{ }-1\le y\le 1\text{ } \right\} \\
& \text{Then,} \\
& \iint_{R}{f\left( x,y \right)}=\int_{-1}^{1}{\int_{0}^{\sqrt{1-{{y}^{2}}}}{x{{y}^{2}}}dxdy} \\
& \text{Integrating} \\
& =\int_{-1}^{1}{\left[ \frac{{{x}^{2}}{{y}^{2}}}{2} \right]_{0}^{\sqrt{1-{{y}^{2}}}}dy} \\
& =\int_{-1}^{1}{\left[ \frac{{{\left( \sqrt{1-{{y}^{2}}} \right)}^{2}}{{y}^{2}}}{2}-\frac{{{\left( 0 \right)}^{2}}{{y}^{2}}}{2} \right]dy} \\
& =\int_{-1}^{1}{\left[ \frac{\left( 1-{{y}^{2}} \right){{y}^{2}}}{2} \right]dy} \\
& =\int_{-1}^{1}{\left( \frac{1}{2}{{y}^{2}}-\frac{1}{2}{{y}^{4}} \right)dy} \\
& =\left[ \frac{1}{6}{{y}^{3}}-\frac{1}{10}{{y}^{5}} \right]_{-1}^{1} \\
& =\left[ \frac{1}{6}{{\left( 1 \right)}^{3}}-\frac{1}{10}{{\left( 1 \right)}^{5}} \right]-\left[ \frac{1}{6}{{\left( -1 \right)}^{3}}-\frac{1}{10}{{\left( -1 \right)}^{5}} \right] \\
& =\frac{1}{15}+\frac{1}{15} \\
& =\frac{2}{15} \\
\end{align}\]
