One of the standard applications of Green's theorem is the following:
Let $\mathbf{F} : \mathbb{R}^2\setminus\{0,0\} \to \mathbb{R}^2$ be a vector field such that $\nabla\times\mathbf{F}=0$ on all of its domain. Then, if $C_1$ and $C_2$ are two concentric, positively-oriented loops containing $\{0,0\}$ such that one is entirely contained in the other, then: $$ \oint_{C_1} \mathbf{F}\cdot d\mathbf{r} = \oint_{C_2} \mathbf{F}\cdot d\mathbf{r} $$ The proof of this is a straightforward consequence of Green's theorem on multiple boundaries. In other words, the proof goes as follows. $$ \int_{C_1} \mathbf{F}\cdot d\mathbf{r} - \int_{C_2} \mathbf{F}\cdot d\mathbf{r} = \iint_D (\nabla\times \mathbf{F})\cdot \mathbf{k}\ dA = 0 $$ Here, $D$ denotes the region between $C_1$ and $C_2$.
Now, I'm wondering about the behaviour when there are multiple singularities. Let's say there are two singularities $\mathbf{a}$ and $\mathbf{b}$. Suppose that $C_1$ and $C_2$ are again concentric, positively-oriented loops such that one entirely contains the other. However, this time $C_1$ goes around both $\mathbf{a}$ and $\mathbf{b}$, whereas $C_2$ wraps around $\mathbf{a}$ only.
My intuition tells me that the line integrals around $C_1$ and $C_2$ do not have to agree in this case, because somehow we are not "protecting" the singularity $\mathbf{b}$ as it is in our region of integration $D$.
But I don't see how proving the situation in the second diagram would be any different from proving the situation in the first diagram. Since, after all, we know that $\nabla\times\mathbf{F}=0$ everywhere that $\mathbf{F}$ is defined.
So my question is the following:
- For the situation in the second diagram, are the line integrals around $C_1$ and $C_2$ always equal?
- If not, then what is the condition that I would be violating (but which is not violated for the situation in the first diagram) if I were to attempt a proof of the situation in the second diagram?
EDIT: Assume the integrals in question always converges.
