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$$ \frac{\sqrt{x^2-y}}{\ln(1-x^2-y^2)} $$ 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 Badia Nov 13 '12 at 17:50
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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

1 Answer 1

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

3) is the easiest of them all... it's all of the plane except the origin!


The last step would be to look at what the intersection of all three regions would be. Thta is, all three regions determined by your constraints.

The first one gives you the region under a certain parabola, the second one gives you the interior of a certain circle, and the last one we have discussed.

It's a strange shape... try to sketch it!

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