double integral over D 2 dydx
where the set D satisfies your 4 inequalities. Turning this double integral into an iterated integral to evaluate has to go back to the picture. From that picture I get the iterated integral will have two parts depending on whether x/2 < 1 - x or not:
int (0,2/3) int (0,x/2) 2 dydx + int (2/3,1) int (0,1-x) 2 dydx =
int (0,2/3) 2(x/2) dx + int (2/3,1) 2(1-x) dx.
This will equal 1/3 as you say.