There is the following graph in my notes:
and then there is the formula:
$$I=\oint _C \frac{xdy-ydx}{x^2+y^2}=2 \pi$$
So, at the formula of the Green theorem: $\displaystyle{M=\frac{-y}{x^2+y^2}} \text{ and } N=\frac{x}{x^2+y^2}$
Does this mean,that whenever we have a circle, $M$ and $N$ are given from the formulas above? Or is it just a specific case?
Also,according to my notes:
$$dx=dr \cos \theta -r \sin \theta d \theta$$ $$dy=dr \sin \theta+r \cos \theta d \theta$$
Why is it like that??
I thought that it would be:
$$dx=-r \sin \theta d \theta$$ $$dy=r \cos \theta d \theta$$
Let $C$ be any circle counter clockwise oriented and centered at the origin then set $C:h(t)=(rcos(t),rsin(t))$ for $t\in [0,2\pi]$ then
$$\int_C F dr=\int_0^{2\pi}F. h'(t) dt$$
$$\int_0^{2\pi}\dfrac {(-rsin(t),rcos(t))}{r^2cos^2(t)+r^2sin^2(t)}(-rsin(t),rcos(t))dt$$ $$\int_0^{2\pi} \dfrac{r^2}{r^2}dt=2\pi$$
So, it is independent of the radius of the circle $r$.
Note: In our case, you can easily see that $M_y=N_x$ but you can not directly use the Green theorem since $M_y,N_x$ is not defined on the origin. But if $C$ be any closed path which does not enclose the origin, then $$\oint_C F dr=\int\int_DN_x-M_y dA=0$$ by Green theorem.