Functions of Random variables as integrals

Question

Suppose I have

and we are asked to find the cdf of U=Y1-Y2

Then we must integrate f(y1,y2) over the dark shaded area for any given u.

according to the text the means by which we find the area of the triangle is:

However, focusing on the double integral, I feel like if I integrated these regions, I would get the 3D volume over this rectangular area instead of the triangle.

I don't understand what about that integral restricts the region to that triangle rather than the rectangle of the same vertices.

Answer 1

The inner integral is over the range $0 \le y_2 \le y_1 - u$, and the RHS implies $y_1 - y_2 \ge u$ as desired.