$$D=\{(x,y)| |x-1|+|y-1|\leq \frac{1}{2}\}$$
I know how $y=|x-1|$ looks, and $|y-1|$ but how do I find this? The problem I have is to find $$\iint_{D}|\ln(xy)|dxdy$$. I think I can do this, just need to find how this set looks. And how this is generaly done.
You have $$|y-1|+|x-1|\le \frac{1}{2}\Rightarrow |y-1|\le -|x-1|+\frac{1}{2}$$
This means that
$$y-1\le -|x-1|+\frac{1}{2}\Rightarrow y\le-|x-1|+\frac{3}{2}$$ and $$y-1\ge |x-1|-\frac{1}{2}\Rightarrow y\ge |x-1|+\frac{1}{2}$$
The bound will just be the equalities
$$y=-|x-1|+\frac{3}{2}, y= |x-1|+\frac{1}{2}$$
The first equality you will shade below because your original inequality symbol was 'less than or equal to'. The second equality you will shade above because your second inequality was 'greater than or equal to'. The solution set is the intersection of the two graphs. See my original comment for a graph from Wolfram Alpha.
If you are not sure how to graph this, just recall that the general form of the absolute value equation is $$y=a|x-h|+k$$ where the ordered pair $(h,k)$ is the vertex.