The original problem is given as thus
Find $$\iint_Dx\,dxdy $$
where $D$ is a triangle with vertices $(0,2),(2,0),(3,3)$.
Green's theorem says that $$\iint_D(G_x-F_y)dxdy = \int_{\partial D}Fdx+Gdy $$
I could parametrize the individual sides of the triangle as such:
$$L_1 = (0,2)\to(2,0) : \begin{cases}x = t\\y=2-t\end{cases}0\leq t\leq 2 $$
$$L_2 = (2,0)\to (3,3) : \begin{cases}x = t+2\\y=3t\end{cases}0\leq t\leq 1$$
$$L_3 = (3,3)\to (0,2) : \begin{cases}x = 3-t\\y=3-\frac{t}{3}\end{cases}0\leq t\leq 3 $$
Is this parametrization correct?
The confusing part is utilizing the theorem. I have to pick $F,G$ such that $G_x,F_y$ are continous in $D$, but $x = x-0 = 2x – x = (y+x)-y = \ldots$ there are endless possibilities. Is the theorem implying no matter which representation of $x$ I choose, the result will always be the same?
For instance: $x = x-0 \Longrightarrow G = \frac{x^2}{2}\quad F = x\qquad$ or $\qquad x = (y+x)-y\Longrightarrow G=yx+\frac{x^2}{2}\quad F = \frac{y^2}{2}$
Best Answer
Your parametrizations look good. You can see that they actually are parametrizations of lines since the derivative of each is a constant vector, and you can verify that they start & stop where intended by plugging in the min & max values of $t$.
Now, as long as you pick $F$ and $G$ (continuous in $D$) such that $\displaystyle \frac{ \partial G}{\partial x} - \frac{ \partial F}{\partial y} = x$, then you're good to go. There are often many choices for $F$ and $G$, and the integral will be invariant under different choices for $F$ and $G$ satisfying those criteria, similar to the fact that contour integrals are invariant under the choice of parametrization for the contour.
The discussions here and here (ctrl-F "many functions") note that multiple choices for $F$ and $G$ are valid when applying Green's theorem.