Answer
\[\overline{f}=0\]
Work Step by Step
\[\begin{align}
& \text{From the graph we can define the region }R\text{ as:} \\
& R=\left\{ \left( x,y \right):0\le x\le 1-{{y}^{2}},\text{ }-1\le y\le 1\text{ } \right\} \\
& \text{The area of the region is } \\
& A=\int_{-1}^{1}{\left( 1-{{y}^{2}} \right)}dy \\
& A=\frac{4}{3} \\
& \text{The average of }f(x,y)\text{ on the region }R\text{ is} \\
& \overline{f}=\frac{1}{A}\iint_{R}{f\left( x,y \right)}dA \\
& \overline{f}=\frac{3}{4}\int_{-1}^{1}{\int_{0}^{1-{{y}^{2}}}{y}dx}dy \\
& \text{Integrate} \\
& \overline{f}=\frac{3}{4}\int_{-1}^{1}{\left[ xy \right]_{0}^{1-{{y}^{2}}}}dy \\
& \overline{f}=\frac{3}{4}\int_{-1}^{1}{\left[ \left( 1-{{y}^{2}} \right)y \right]}dy \\
& \overline{f}=\frac{3}{4}\int_{-1}^{1}{\left( y-{{y}^{3}} \right)}dy \\
& \overline{f}=\frac{3}{4}\left[ \frac{1}{2}{{y}^{2}}-\frac{1}{4}{{y}^{4}} \right]_{-1}^{1} \\
& \overline{f}=\frac{3}{4}\left[ \frac{1}{2}{{\left( 1 \right)}^{2}}-\frac{1}{4}{{\left( 1 \right)}^{4}} \right]-\frac{3}{4}\left[ \frac{1}{2}{{\left( -1 \right)}^{2}}-\frac{1}{4}{{\left( -1 \right)}^{4}} \right] \\
& \overline{f}=\frac{3}{4}\left( \frac{3}{16} \right)-\frac{3}{4}\left( \frac{3}{16} \right) \\
& \overline{f}=0 \\
\end{align}\]
