Finding probability of a region I have a question that says:
Given $f_{X,Y} (x, y) = 1/17(x\cdot y + y^2) , 0 < x < 3 , 0 < y < 2$.
Set up the integral that gives $P(X+Y>2)$.
They set up the integral like this:
$$1/17\cdot \int^2_0\int^3_{2-y}xy+y^2 \, dx\,dy$$ 
But I set up my integral differently... and I get a different answer. I don't know why I'm wrong, can someone help?
This is how I set up my integral:
$$1/17\cdot \int^3_0\int^2_{2-x}xy+y^2 \, dy\,dx$$
It just seems like they solved for $P(Y>-X+2)$ and I solved for $P(X>-Y+2)$. Why do we get different answers? Shouldnt they be the same? If not, how should I know that I should solve for the latter and not the former Probability?
Thanks

Sketch a diagram of the domain. Observe that the region of interest is a trapezoid. In particular, if we slice the region vertically, then there are two types of "lower" boundaries: the diagonal at $y = 2 - x$ for $x \in (0, 2)$, and the horizontal at $y = 0$ for $x \in (2, 3)$. Thus, if we want to change the order of integration, then we'd have to split it into two integrals as follows:
$$
\int_0^2 \int_{2 - x}^2 f_{XY}(x, y) \, dy \, dx
+ \int_2^3 \int_{0}^2 f_{XY}(x, y) \, dy \, dx
$$
