## Calculus 8th Edition

The domain is $$\mathcal{D}=\left\{(x,y)\left|\frac{x^2}{3^2}+\frac{y^2}{1^2}<1\right.\right\},$$ and it is sketched in the figure bellow
The argument of the logarithm has to be greater than zero i.e. we need that $9-x^2-9y^2>0$. Rearranging terms this gives $$x^2+9y^2<9.$$ Dividing by $9$ we get $$\frac{x^2}{9}+y^2<1.$$ Rewrite previous inequality as $$\frac{x^2}{3^2}+\frac{y^2}{1^2}<1.$$ Knowing that $$\frac{x^2}{a^2}+\frac{y^2}{b^2}<1$$ is equation of a centered ellipse with major semiaxis equal to $a$ and minor semiaxis equal to $b$, we see that the domain is the region bounded by the ellipse with major semiaxis equal to $3$ and minor semiaxis equal to $1$ (the region does not include points from the ellipse because we have strictly less sign in the inequality). We can write for the domain $$\mathcal{D}=\left\{(x,y)\left|\frac{x^2}{3^2}+\frac{y^2}{1^2}<1\right.\right\},$$ and it is shown in the figure below