Show $\oint_C (Pdx+Qdy) \,=2\pi$ if $P$ and $Q$ aren't $C^1$

Question

Define $P=\frac{-y}{x^2+y^2}$ and $Q=\frac{x}{x^2+y^2}$ for $(x,y)\neq(0,0)$

I wish to show that $$\oint_C (Pdx+Qdy) \,=2\pi \\$$ where C is any circle with centre $(0,0)$ orientated counterclockwise.

I'm unsure how to solve this problem, as I believe Green's Theorem does not apply in this instance as $P$ and $Q$ are not $C^1$. I am not looking for an explicit answer, rather hints on how to proceed.

Answer 1

To parameterize a circle of indeterminate radius

$x = R\cos t\\ y = R\sin t$

$\int_0^{2\pi} \frac {-R\sin t}{R^2} (-R\sin t) + \frac{R\cos t}{R^2} (R\cos t)\ dt$

Answer 2

Your field $${\bf A}(x,y):=\left({-y\over x^2+y^2},{x\over x^2+y^2}\right)\qquad\bigl((x,y)\ne(0,0)\bigr)$$ (which is $C^\infty$ througout its domain of definition) is the gradient field of the polar angle ${\rm arg}$ (even though the latter is only determined up to $2\pi$). It follows that integrating this field along any arc not passing through $(0,0)$ gives the total increment of the polar angle along this arc.

Green's theorem applies to a nice region $\Omega$ and its boundary $\partial\Omega$. An essential assumption is that the field in question is $C^1$ throughout $\bar\Omega$. Therefore you cannot apply Green's theorem with ${\bf A}$and a region $\Omega$ containing $(0,0)$ and its boundary $\partial\Omega$.