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Problem:

Let $P={-y \over x^2+y^2}$ and $Q={x \over x^2+y^2}$ for $(x,y)\ne(0,0)$.

Show that $\oint_{\partial \Omega}(Pdx + Qdy)=2\pi$ if $\Omega$ is any open set containing $(0,0)$ and with a sufficiently regular boundary.

Working:

Clearly, we cannot immediately apply Green's Theorem, because $P$ and $Q$ are not continuous at $(0,0)$. So, we can create a new region $\Omega_\epsilon$ which is $\Omega$ with a disc of radius $\epsilon$ centered at the origin excised from it.

We then note ${\partial Q \over \partial x} - {\partial P \over \partial y} = 0$ and apply Green's Theorem over $\Omega_\epsilon$. Furthermore, $\oint_C(Pdx + Qdy)=2\pi$ if $C$ is any positively oriented circle centered at the origin.

I get the general scheme of how to approach this problem, however I am unsure of how to argue it in a rigorous manner.

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By the definition of open set, since $(0,0) \in \Omega$, there is an $r>0$ such that the disk of radius $r$ centred at $(0,0)$, $B_r(0,0)$ is contained in $\Omega$. Hence you can define $\Omega_r = \Omega \setminus B_r(0,0)$, which has two distinct boundaries, $\partial \Omega_r$ and $\partial B_r(0,0)$.

Now, as you say, we can use Green's Theorem on $\Omega_r$, so $$ 0 = \iint_{\Omega_r} (\partial_x Q - \partial_y P)\, dx\,dy = \int_{\partial\Omega_r} P \, dx + Q \, dy = \left( \int_{\partial\Omega} -\int_{x^2+y^2=r^2} \right) (P \, dx + Q \, dy), $$ because the boundary of $B_r(0,0)$ is traversed in the opposite direction because we have to keep the interior of $\Omega_r$ on the left.

Therefore we have to compute $\int_{x^2+y^2=r^2} (P \, dx + Q \, dy)$. Since this is just a circle, we parametrise it by $x=-r\cos{t}$, $y=r\sin{t}$, $dx=r\cos{t} \, dt$, $ dy=r\sin{t} \, dt $: $$\int_{x^2+y^2=r^2} (P \, dx + Q \, dy) = \int_0^{2\pi} \frac{-r\sin{t}}{r^2(\cos^2{t}+\sin^2{t})} (-r\sin{t}) \, dt + \frac{r\cos{t}}{r^2(\cos^2{t}+\sin^2{t})} (r\cos{t}) \, dt \\ = \int_0^{2\pi} 1 \, dt = 2\pi. $$

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If your boundary is smooth, then you may parametrize it by $x(t) = r(t) \cos t$ and $y(t) = r(t) \sin t$, with $t \in [0,2\pi]$. Then your differential form $P \Bbb d x + Q \Bbb d y$ becomes

$$\frac {-r(t) \sin t \Big (r'(t) \cos t - r(t) \sin t \Big) + r(t) \cos t \Big( r'(t) \sin t + r(t) \cos t \Big)} {r^2 (t)} = 1$$

which, integrated from $0$ to $2\pi$, will give $2 \pi$.

Please note that the application of Green's theorem on $\Omega _\varepsilon$ will tell you that the integral on $C$ is equal to the integral on the small circle of radius $\varepsilon$ surrounding the origin, but won't calculate things for you. You'll still have to parametrize the circle and do an explicit integral.

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  • $\begingroup$ What if there is no bijection from the angle variable to the boundary? I.e. you can easily have radial segments or even retrograde parts. You need to say something about those cases if you do it this way. $\endgroup$ – Chappers Jun 1 '15 at 12:22
  • $\begingroup$ @Chappers: Agreed, my argument is not a general one. I assumed the domain contained within to be convex. The simplest proof, then, is the one indicated in my last paragraph - integrating on the small circle surrounding the origin. $\endgroup$ – Alex M. Jun 1 '15 at 12:24

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