I am trying to work a seemingly simple practice problem but I am having some confusion. The question asked to verify the divergence theorem in the plane for the vector field $$F=(xy)i+(2x-y)j,$$ where $C$ is the triangle with vertices $(0,0)$, $(1,0)$, $(0,1)$.
I.e., I want to verify that
$$\int_{C} \bar F \cdot n\, dS = \iint_{Rxy} \nabla \cdot \bar F\, dA$$
The right hand side I computed the gradient to be $$y-1$$ and integrated over the triangle to be $$\frac{-1}{3}$$
But my issue now is with the other integral. I am confused about what a normal is in only the plane.
For example, I could split it into $C_1$, $C_2$, and $C_3$ maybe with $n$ being $-j$ for$ C_1$ and $-i$ for $C_3$ and for $C_2$ it would be perpendicular to $y=1-x$.
But then I don't know about $dS$.
Overall my question is: How to make sense of the left integral in the plane, when it involves element of surface area, etc.
Thanks
In the plane, you should use $ds$ instead of $dS$. It is not infinitesimal surface area, but instead infinitesimal arc-length. So $ds = \sqrt{x'(t)^2 + y'(t)^2} \, dt$, where $\vec{r}(t)=\left<x(t),y(t)\right>$ is a parameterization of your curve $C$.