# Double integral area of domain

I am revisiting some of the double integral problems, but I am having trouble conceptualize the area of the domain to integrate.

Ex: $f(x,y)$ of $(x,y) = 1/2$

where $0 < x < y < 2$

How do I know the area of interests is the upper or lower triangle? This is a very basic question. But as soon as the question changes between $x < y$ or $y < x$, I will get confused about the area I am suppose to integrate from.

Breaking up that inequality will help. $$0 < x < y < 2 \quad\implies\quad \begin{cases} 0 < x < 2 \\ 0 < y < 2 \\ x < y \end{cases}$$
It is the area such that both $x$ and $y$ are non-negative but less than $2$, and where $x < y$. This last inequality tells us that it must be the upper triangle. You can tell by plugging in a point that resides in either triangle and seeing whether it makes $x < y$ true or false.
So if we pick the point $(1,1\frac12)$ in the upper triangle, since this points yields true for $x < y$, the upper triangle is the region of interest. Otherwise if we had chosen a point in the lower triangle instead, say the point $(1\frac23,\frac13)$, then since the inequality $x < y$ is false for this point, the lower triangle is not the region of interest.