I'm having trouble graphing the domain of the function $\displaystyle f(x,y)=ln(9-x^2-9y^2)$. I don't need help drawing the graphg itself, I just need assurance that I'm on the right track.

The domain should be all $\displaystyle x,y$ such that $\displaystyle 9-x^2-9y^2 \geq 0$

$\displaystyle -9y^2 \geq x^2-9$

$\displaystyle y^2 \leq 1-\frac{1}{9}x^2$

$\displaystyle y\leq\pm\sqrt{(1-\frac{1}{3}x)(1+\frac{1}{3}x)}$

So do I just need to sketch the region below the curves $\displaystyle y=\pm\sqrt{(1-\frac{1}{3}x)(1+\frac{1}{3}x)}$

? It seems like there must be an easier way.

Originally Posted by adkinsjr

I'm having trouble graphing the domain of the function $\displaystyle f(x,y)=ln(9-x^2-9y^2)$. I don't need help drawing the graphg itself, I just need assurance that I'm on the right track.

The domain should be all $\displaystyle x,y$ such that $\displaystyle 9-x^2-9y^2 \geq 0$

$\displaystyle -9y^2 \geq x^2-9$

$\displaystyle y^2 \leq 1-\frac{1}{9}x^2$

$\displaystyle y\leq\pm\sqrt{(1-\frac{1}{3}x)(1+\frac{1}{3}x)}$

So do I just need to sketch the region below the curves $\displaystyle y=\pm\sqrt{(1-\frac{1}{3}x)(1+\frac{1}{3}x)}$

? It seems like there must be an easier way.

Do you know what the ellipse $\displaystyle \frac{x^2}{9} + y^2 \leq 1$ looks like?

Originally Posted by Defunkt

Do you know what the ellipse $\displaystyle \frac{x^2}{9} + y^2 \leq 1$ looks like?

yeah, I see it now. I need to shade in the ellipse with center at origin, vertices at $\displaystyle (0,\pm 1)$ and $\displaystyle (\pm 3,0)$.