Is This the Correct Way to Graph the Domain of a Function in Calc 3?

[gokugreene]

Hello guys I am not sure if I am doing this right. If you could offer any advice or point out my mistakes I would appreciate it.

Problem: Sketch the graph of the domain of f(x,y,z) = \frac{1}{x^2+y^2-z}

Domain: (x,y,z) \in R^3 \mid x^2+y^2does not equal z

Graph: I set f(x,y,z) = 1 and solved for z
I got z=x^2+y^2-1 <<-- Would this be the proper graph of the domain?

Thanks

 

[CarlB]

No, setting f(x,y,z)=1 has nothing to do with the domain. You were right when you said that the domain is the points in R^3 where x^2+y^2 isn't equal to z. But from there you jumped in the wrong direction.

Carl