My book first tells me to take:
$$\oint_\gamma \frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy$$
over $\gamma$ which is a closed path, simple, such that its edge is a set $B$, with interior that contains the circle $x^2+y^2\le 1$
I've seen the resolution of this exercise, and it does it in this way:
The parametric plot of the circle is:
$$x=\cos(t)\\y=\sin(t)\\0\le t\le 2\pi$$
Then, the book separates the regions like this:
It says that $K_1$ is the region delimited by $\gamma_1$ and $\gamma$, so it's the outer part of the circle that is delimited by the outer path. So $K$ is the region of the circle (that's what I understood). Then it does:
$$\oint_\gamma Pdx + Qdy-\oint_{\gamma_1} Pdx+Qdy = \int\int_{K_1}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)$$
and then:
$$\oint_{\gamma_1} Pdx + Qdy = \int_0^{2\pi} (\sin^2(t)+\cos^2(t))dt = 2\pi$$
Well, ok, this I understood. You cannot use Green's theorem here but you can directly use line integration to come up with the answer: $2\pi$. But then, for the other area it does:
$$\int\int_K\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy = 0$$
First of all, shouldn't it be $K_1$? Because $K$ is the circle. I think there's a typo here. Also, why $0$? It's a strange looking path, it's not even defined by a parametrization, how can I know that this is $0$?
The book ends says that $$\oint Pdx + Qdy=2\pi$$
Then, the book considers another even crazier path $\gamma$, for the same integrand:
It says that for $K_1$ with edge $\gamma_1$ we have by the previous exercise:
$$\oint_\gamma \frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy = 2\pi$$
WHAT??? It didn't even say how the path is parametrized, but it used the result that's supposed to chose a circle as a path, but $K_1$ does not look like a circle! And the exercise don't mention a relation to the previous exercise...
Then, for $\gamma_2$ it says that Green's theorem applies and we have:
$$\int\int_K\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy = 0$$
so it follows that $$\oint_{\gamma_2}Pdx+ Qdy=0$$
so the result over the entire path is $2\pi$
Can someone please explain to me what's going on??????