Given $f(x,y) = x\sqrt{3 - x^2 - y^2}$ find and sketch the domain? I have been trying to do this but I am having a hard time. All I have is that $\sqrt{3-x^2-y^2}\geq0$
The next step I want to do is the following:
\begin{aligned}(\sqrt{3-x^2-y^2})^2\geq0^2\end{aligned}
\begin{aligned}3 -x^2-y^2\geq0\end{aligned}
\begin{aligned}(3 -x^2-y^2)-3\geq0-3\end{aligned}
\begin{aligned}-x^2-y^2\geq-3\end{aligned}
\begin{aligned} -1*(-x^2- y^2)\geq-3*-1\end{aligned}
\begin{aligned}  x^2 + y^2 \leq3\end{aligned}
I would just assume that the domain is the circle inequality above and graph the circle in the $(x,y)$ plane including the radius. Maybe I am wrong. What do you guys think?

The domain of the function is the set of points $(x,y)$ on which the function is well defined. Now, you know that you can only take the square root of a non-negative number
Hence,
$$ 3-x^2-y^2 \geq 0$$
$$\implies x^2+y^2 \leq 3$$
Only if $(x,y)$ satisfies this can you get a real value for $f(x,y)$
As for sketching, this would be a disc centered at origin of radius $\sqrt{3}$
