<s>[INST] Find the domain of this function in the x,y,z graph Find the domain of this function in set notation:
$f(x,y)=\dfrac{\sqrt{x^2+y^2+4}}v$
$v=y-x$
Could I say that $x^2+y^2+4 \geq 0$ and $y-x \geq 0$ and based off those statements put that into set notation? and get:
$D= \{(x,y)|x^2+y^2+4 \neq 0$ and $y-x \neq 0\}$
 [/INST] Since $v \neq 0$ by definition, then $y-x \neq 0$ ---> $y \neq x$. Therefore, the line y= x is excluded from the domain of f. Also, since $\sqrt{x^2+y^2+4}\geq 0$ and $y \neq x$,  $\sqrt{x^2+y^2+4}$. So the domain is $ D = \{(x,y)| \sqrt{x^2+y^2+4}\geq 0$ where $y\neq x$}  Surb gives an excellent sketch of the domain above in his/her reply. 
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