you can replace x y instead of u v
i cant think of a way to find the are described here
?
1. To be honest: The only thing my computer did was drawing the coordinate system.
2. Consider $\displaystyle D=\{(u,v) | max(|u|, v) \le 2~\wedge~v\ge 0\}$
Then you'll get a rectangle from u = -2 to u = 2 and from v = 0 to v = 2.
3. If you take the additional condition $\displaystyle max(|u|, v) \ge 1$ into account you have to exclude all those points whose both coordinates u and v start with 0.
To give you an example:
P(-1.2 , 0.3) belongs to the area because $\displaystyle |u| \ge 1$
Q(0.7 , 0.2) does not belong to the area because both coordinate values are smaller than 1.
4. All points R(|0.###| , 0.***) are situated in a rectangle from u = -1 to u = 1 and from v = 0 to v = 1
As earboth says, that hole is because of the condition $\displaystyle max(|u|,v)\ge 1$.
If, in the u,v plane, you draw the lines v= 1, u= 1, u= -1, notice that any point inside those lines has v< 1 and -1< u< 1 or |u|< 1. Saying "$\displaystyle max(|u|, v)\ge 1$" means neither of those can happen and so we are outside or on those lines. Similarly, the lines v= 2, |u|= 2 (or u= -2 and u= 2) form the outer "box", from $\displaystyle max(|u|,v)\le 2$. The lower boundary is at v= 0, of course, because of the condition $\displaystyle v\ge 0$.
If the fact that these are u and v, rather than x and y is bothering you, just go ahead and replace with u with x and v with y- it is exactly the same thing in the xy-plane. If fact, I notice that earboth, in his last post, has accidently replace the condition $\displaystyle v\ge 0$ with $\displaystyle y\ge 0$!