# How to draw the domain of this function.

$$\frac{\sqrt{x^2-y}}{\ln(1-x^2-y^2)}$$ I see, that the domain is real, but: $$1) x^2\geq y$$ and $$2) x^2+y^2<1$$ and $$3) x^2+y^2 \ne0$$ I can draw 1 and 2, but how to draw 3) ?

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If the sum of two non-negative numbers is zero, what does that say about those nunbers? – Javier Nov 13 '12 at 17:50
do you mean, that I should exclude the point (0,0)? – TomDavies92 Nov 13 '12 at 17:52
@amWhy Surely you don't mean to say the domain is the whole plane without the origin... there are the other two constraints, of course... – rschwieb Nov 13 '12 at 17:58
@rschwieb It gets confusing when discussing the domain in contrast to the Image of f(x,y). And I wasn't looking at the original function! (at the top)! – amWhy Nov 13 '12 at 18:01
Normally, we say a domain is "real" if the domain is a subset of $\mathbb R$. In this case, that isn't true, so the domain isn't "real" under any definition that I've heard of. Rather, the domain is a subset of the "real plane." – Thomas Andrews Nov 13 '12 at 18:05

Well, the domain is real in the sense that it's the real plane $\mathbb{R}^2$.