Describing multivariable functions So I am presented with the following question:
Describe and sketch the largest region in the $xy$-plane that corresponds to the domain of the function:
$$g(x,y) = \sqrt{4 - x^2 - y^2} \ln(x-y).$$
Now to be I can find different restrictions like $4 - x^2 - y^2 \geq 0$... but I'm honestly not even sure where to begin this question! Any help?
 A: So, you need two things:
$$ 4 - x^2 - y^2 \geq 0 $$
to get make the square root work, and also
$$ x-y > 0 $$
to make the logarithm work.
You will be graphing two regions in the $xy$-plane, and your answer will be the area which is in both regions.
A good technique for graphing a region given by an inequality is to first replace the inequality by an equality.  For the first region this means
$$ 4 - x^2 - y^2 = 0$$
$$ 4 = x^2 + y^2 $$
Therefore, we're talking about the circle of radius two centered at the origin.  The next question to answer: do we want the inside or outside of that circle?  To determine that, we use a test point: pick any point not on the circle and plug it into the inequality.  I'll choose $(x,y) = (5,0)$.  Note that this point is on the outside of the circle.
$$ 4 - x^2 - y^2 \geq 0 $$
$$ 4 - 5^2 - 0^2 \geq 0 $$
That's clearly false, so we do not want the outside of the circle.  Our region is the inside of the circle.  Shade that lightly on your drawing.
Now, do the line by the same algorithm.
A: When is $g(x,y)$ defined and takes a real value?  Well, the square root has to be of a positive number, so $x^2+y^2\le4$.  Also, when you are taking $\log(x-y)$, for the logarithm to be defined, $x-y$ must be greater than $0$.  Thus, the domain of $g(x,y)$ consists of all points $(x,y)$ such that
(1) $x^2+y^2\le4$ and
(2) $y<x$
Sketch (1) and (2) on an $xy$ plane.  (1) is a filled-in circle with radius $2$ centered around the origin (all points on the boundary are points in the domain).  (2) is all points strictly below the line $y=x$ (all points on the boundary are not points in the domain).  For $g(x,y)$ to be defined, both (1) and (2) must be satisfied, so we must take the intersection of these two figures.
