If $\vec{F}=\frac{y\hat i-x\hat j}{x^2+y^2}$ then evaluate $\int_c \vec{F}\cdot\vec{r} $ where C is the circle $x^2+y^2=1$ traversed in counter clockwise direction.
My trial: using Green's theorem over XY-plane $$\int_c\vec{F}\cdot d\vec{r}=\int_c\left(\frac{y\hat i-x\hat j}{x^2+y^2}\right)\cdot d\vec{r}$$ $$=\int_c\left(\frac{y\hat i-x\hat j}{x^2+y^2}\right)\cdot (\hat idx+\hat jdy)$$ $$=\int_c\left(\frac{y}{x^2+y^2}\right)dx-\left(\frac{x}{x^2+y^2}\right)dy$$ $$=\int\int_R\left(\frac{\partial}{\partial x}\left(\frac{-x}{x^2+y^2}\right)-\frac{\partial}{\partial y}\left(\frac{y}{x^2+y^2}\right)\right)dxdy$$ $$=\int\int_R\left(\left(\frac{x^2-y^2}{(x^2+y^2)^2}\right)-\left(\frac{x^2-y^2}{(x^2+y^2)^2}\right)\right)dxdy$$ $$=0$$
my book suggests that the answer is $-2\pi$ but i don't know how. Am I wrong if yes then please explain or help solve this problem. thanks.