$$ \frac{\sqrt{x^2-y}}{\ln(1-x^2-y^2)} $$ 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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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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