Integrating triangle in a 2D plane I am interested in integrating $(x^2y+y^2x)$ on the following loop: $(x=1,y=2)\rightarrow(x=2,y=1)\rightarrow(x=3,y=3)\rightarrow(x=1,y=2)$. 
I know this loop forms a triangle with all three sides changing with respect to the $x$ and $y$ axis. I understand how to do the integration part, but unfortunately I am really stuck on how to set up the integration itself. Should I do a double integral for each side of the triangle and then add them?
If anyone can help me set up the integration for this problem it would be greatly appreciated. Thank you! 

Yes, it can be done using two double integrals. Here you are dividing the triangle into two parts by the vertical line $x=2$.
The equation of the line joining $(1,2)$ and $(2,1)$ is $x+y=3$, and that of the line joining $(1,2)$ to $(3,3)$ is $x-2y+3=0$.
You can set $x=1$ to $x=2$, and then you will have the range of $y$ as $y=3-x$ to $y=\frac{x+3}{2}$.
The equation of the line joining $(2,1)$ to $(3,3)$ is $2x-y=3$.
Then you can set $x=2$ to $x=3$, and you will get the range of $y$ as $y=2x-3$ to $y=\frac{x+3}{2}$.
EDIT:
The question meant to ask integrating along the loop, not over the region bounded by it.
For each line, the curve can be parameterized. Example procedure for $y=3-x$.
Let $x=t$. Then, $y=3-t$. $t$ varies from $1$ to $2$.
The function $x^2y+y^2x$ then becomes $t^2(1-t)+t(1-t)^2$.
$$\oint {(x^2y+y^2x).dr}= \int_{1}^{2} {(t^2(1-t)+t(1-t)^2)|v(t)|dt}+otherterms$$.
where $$|v(t)|=\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}=\sqrt{2}$$
