Domain of a function in two variables

**Robb** · May 24th 2009, 03:18 AM

Hello MHF,

Have this question, and I am really unsure about function in multiple variables..

Find the domain of each of the following functions and sketch it, showing clearly whether any parts of the boundary are included in the domain
(a) $f(x,y) = \sqrt{\frac{1}{y}-x^2}$
(b) $g(x,y) = \ln(1-xy)$

So for equation (a) obviously $\frac{1}{y}-x^2\geq 0 \Rightarrow D=\{(x,y)|\frac{1}{y}-x^2\geq 0, y \neq 0\}$ which I have reduced to $y \leq \frac{1}{x^2}$ so that i can graph it, and the boundry of the graph is included in the domain.

For part (b) it is similiar because $\ln 0$ isn't defined, so $1-xy > 0 \Rightarrow D=\{(x,y)|1-xy>0\}$ , so this can be reduced to $y<\frac{1}{x}$ so that it can be graphed, and the boundry of the graph isn't in the domain?

**Min.Lu** · May 24th 2009, 03:40 AM

The answer to the part (a) is right . However, as to part(b), you miss some parts . The domain of the function g is $D=\left\{\left(x,y\right)|xy<1\right\}$ , which can be dived into three parts:

$D_1=\left\{{\left(x,y\right)|x<\frac{1}{y},y>0}\ri ght\}$

$D_2=\left\{{\left(x,y\right)|x \in \mathcal{R}, y=0}\right\}$

$D_3=\left\{{\left(x,y\right)|x>\frac{1}{y},y<0}\ri ght\}$

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