# Geometric representation of domain of $z=x+\sqrt{y}$

I have to find the domain of this function and represent geometrically the domain in a three dimensional plan. The function is $z=x+\sqrt{y}$.

Now,the domain is $y\geq0$. How do I represent this? How about the domain of $z=\frac{1}{\sqrt{x^2+y^2-1}}$?

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If I understood correctly, this is a function on two variables, so its domain is in a two-dimensional plane, not three-dimensional. Now draw the $xy-$plane, and think which points have non-negative $y$ coordinates. – Pedro M. Mar 12 '13 at 16:29
No,I have to represent it geometrically. – gfg Mar 12 '13 at 16:30
@PedroMilet has given you a way to represent it geometrically, as a subset of the $xy$ plane. That is very much geometric. – Dylan Yott Mar 12 '13 at 17:18
Can you draw the domain in the default, two-dimensional setting (as Pedro Milet suggested)? Now redraw it as if it was lying flat on (infinitesimally thin, colorless) carpet in a room. In other words, add third dimension, so it looks more 'realistic'. – Kuba Helsztyński Mar 12 '13 at 17:19

Thanks Babak: yes, helping is sweeter than all the rest. That's why I'm here. $\ddot\smile$ – amWhy Mar 13 '13 at 5:36