Consider a simple closed curve $C$ in the plane (positively oriented) with $p$ and $q$ as two points inside $C$. Let $f$ be a given function that is differentiable in $\mathbb{R}^2 \setminus \{p, q\}$ and $\Delta f = 0$ in $\mathbb{R}^2 \setminus \{p, q\}$. Furthermore, assume that $$ \ \int_{C_p} (-f_y) \, dx + f_x \, dy = A, \quad \int_{C_q} (-f_y) \, dx + f_x \, dy = 2A$$
where $A$ is a real number, and $C_p$, $C_q$ are circles centered at $p$ and $q$, respectively, with arbitrary radii (positively oriented). Note that we have assumed $A$ is independent of the radius.
Determine the line integral
$$\int_C -f_y \, dx + f_x \, dy $$
expressed in terms of the number $A$.
The answer: $I = \int_C = \int_{C+C^{-}_p} + \int_{C^-_q} - \int_C = I_1 - I_2 - I_3$, where we have chosen the negative orientation for $C^{-}_{p}$ and $C^{-}_{q}$ to have a positively oriented region $D$ bounded by these curves.
We can now apply Green's theorem and the fact that $\Delta f = 0$ to obtain $I_1 = 0$. We have $I_2 = \int_{C^-_p} = -\int_{C_p} = -A$, and $I_3 = -2A$.
Thus, we have $I = 3A$.
Okay my question is in what way did they use green theorem? Isn't green's theorem only for vector fields? Also why did they choose the negative orientation for the circular curves around p and q?
If $\vec{F}=(-f_y,f_x)$ and $d\vec{r}=(dx,dy)$, your integral simply reads $\int_{C}\vec{F}\cdot d \vec{r}$. Let $r_1,r_2>0$ be arbitrary numbers and $B(p,r_1), B(q,r_2)$ the disks centered at $p,q$ with radius $r_1,r_2$ respectively. If $D$ is the region bounded by the curve $C$, then you apply Green's theorem to $E:= D \setminus\{B(p,r_1)\cup B(q,r_2)\}$, since $\vec{F}$ is differentiable on $E$ by assumption. Thus,
$$\int_{C}\vec{F}\cdot d\vec{r} = \iint_{E}\partial_{x}(f_x)-\partial_y(-f_y)dA + \int_{C_p^{+}}\vec{F}\cdot d\vec{r}+\int_{C_q^{+}}\vec{F}\cdot d\vec{r}=\int_{C_p^{+}}\vec{F}\cdot d\vec{r}+\int_{C_q^{+}}\vec{F}\cdot d\vec{r}=A+2A=3A$$ since $\partial_x(f_x)-\partial_y(-f_y)=f_{xx}+f_{yy}=0$.