A Question about evaluating surface integrals Vector calculus is relatively new to me so I have a little trouble understanding double integrals intuitively. The question is esssentially this:

Calculate the surface integral of $\vec v=x^2 \hat j $ over the triangle formed by the points $(1,0,0),(0,2,0)$ and $(0,0,1)$.

The plane is $2x+y+2z=2$ and I am aware of the algorithm to solve this problem(projecting the infinitesimal area element onto the $xy$ plane, taking its dot with the given vector field etc. ) The double integral that we are supposed to end up with is $\int_0^1 \int_0^{2-2x}\frac{x^2}{2}dxdy$. 
I am having trouble with understanding the limits that have been substituted. I understand that a projection of the area element on the $xy$ plane means that $z=0$ so the limits will be a result of that. 
But I don't see why the same procedure can't be done on a plane parallel to the $xy$ plane,for instance $z=3$. I do not think anything in the beginning would change-  Even the projected area of the infinitesimal element would remain the same since both $z=0$ and $z=3$ have $\hat k$ as their unit normal. But how can one explain the limits in this case?
Hope I've been able to explain my problem properly!Thanks in advance.
 A: The limits for the double integral don't come from the intersection of the triangle's plane with the $xy$-plane. They come from the triangle itself. (In this problem, it just happens to be the case that one of the triangle's sides is in the $xy$-plane.)
Projecting the triangle with vertices $(1,0,0), (0,2,0), (0,0,1)$ onto the $z=0$ plane gives the triangle with vertices $(1,0,0), (0,2,0), (0,0,0)$. To calculate in plane geometry, we can just ignore $z$ and consider the plane points $(1,0), (0,2), (0,0)$. This is a right triangle in the first quadrant bounded by the lines $x=0$, $y=0$, $2x+y=2$. That last equation is found as the line in the 2-D plane containing $(1,0)$ and $(0,2)$.
Projecting the triangle with vertices $(1,0,0), (0,2,0), (0,0,1)$ onto the $z=3$ plane gives the triangle with vertices $(1,0,3), (0,2,3), (0,0,3)$. This is another right triangle directly above the one projected onto the $z=0$ plane with exactly the same sides and angles, and once we ignore $z$ to use plane geometry techniques, it actually is the same triangle.
So no matter which parallel plane you project onto, you get an identical copy of the projected region of integration. In general, the limits of integration on the projected region are the curves which come from projecting the curves at the boundary ("edges") of the original 3-D surface.
