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:

enter image description here

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:

enter image description here 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??????

  • $\begingroup$ Did you try to calculate $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$ directly? $\endgroup$ – user99914 Jan 21 '16 at 3:44
  • $\begingroup$ @JohnMa I did now, it gives $0$ :) $\endgroup$ – Guerlando OCs Jan 21 '16 at 3:56
  • $\begingroup$ So do you understand the calculation of the "figure 8" now? that's because the Green's theorem tells you that the $\gamma_2$ part integrates to zero. $\endgroup$ – user99914 Jan 21 '16 at 4:00
  • $\begingroup$ @JohnMa yes, but what about the $\gamma_1$? $\endgroup$ – Guerlando OCs Jan 21 '16 at 4:01
  • $\begingroup$ For that $\gamma_1$, just like what you did in the first example, draw a small circle $\gamma(t) = (r\cos t, r\sin t)$, $t\in [0, 2\pi]$ ($r$ is small so that the circle is "inside" $\gamma_1$) and then use Green's theorem. $\endgroup$ – user99914 Jan 21 '16 at 4:05

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