$\displaystyle \int{\int_D{max(x^2,y)}}dxdy$
D is the square with corners in (0,0) and (1,1).
I know how to solve double integrals in general, but this one is a bit tricky.
Draw the curve y = x^2 inside the square. This divides the square into two regions. In the upper region, y > x^2 so max(x^2, y) = y. In the lower region, y < x^2 so max(x^2, y) = x^2. So your integral becomes:
$\displaystyle \int_{0}^{1} \int_{y = x^2}^{y = 1} y \, dy \, dx + \int_{0}^{1} \int_{y = 0}^{y = x^2} x^2 \, dy \, dx$.