>>1521034If you want area, you want a positive number. So just pick the one that gives positive values. Since the integrand is positive, you'll want the smaller bound on the bottom and the larger bound on the top.
For y between 0 and 1, sqrt(y) > y, so y goes on the bottom, sqrt(y) on the top.
In general, when you're talking about integrals, you specify an orientation. If you use a transform, the Jacobi determinant manages any changes in orientation (as well as stretching and squishing). This is probably too advanced for now.
(By the way, this particular example can be solved with a single integral.)
