Answer
$$32$$
Work Step by Step
Using the rules of integrals, we obtain:
$\begin{aligned} \text { Area }=& \iint_{R} d A=\int_{-3}^{3} \int_{1-y^{2} / 9}^{9-y^{2}} d x d y \\ &=\int_{-3}^{3}[x]_{1-y^{2} / 9}^{9-y^{2}} d y \\=& \int_{-3}^{3} 9-y^{2}-1+\frac{y^{2}}{9} d y \\ &=\int_{-3}^{3} -\frac{8 y^{2}}{9} +8 d y \\ &=\frac{8}{9} \int_{-3}^{3} 9-y^{2} d y \\ &=\frac{8}{9}\left[9 y-\frac{y^{3}}{3}\right]_{-3}^{3} \\=& \left[9 \cdot 3-\frac{3^{2}}{3}\right]\frac{16}{9} =(-1+3)16=32 \end{aligned}$
