Integral of 1-form $\omega=\dfrac{-y \,dx + x \,dy}{x^2 +y^2}$ over a triangle.

Question

I'm trying to evaluate the integral of the $1$-form

$$\omega=\dfrac{-y \,dx +x\,dy}{x^2 +y^2}$$

through the corners of a triangle with the vertices $A= (-5,-2)$, $B=(5,-2)$, $C=(0,3)$.

I've tried to use Green's theorem but it didn't work because the 1-form isn't differentiable on (0,0) so I had to parametrize the paths from each point and got three line integrals but it got so complicated at the end, so it doesn't seem to be the right solution or at least the one I'm supposed to calculate. Is there anyway else to evaluate the integral?

Answer 1

We know that $$\omega=\dfrac{-y \,dx +x\,dy}{x^2 +y^2}=d\arctan\dfrac{y}{x}$$ is an exact differential and the area $\Delta$ includes the origin, since the $\arctan\dfrac{y}{x}$ winds around the origin one time then $$\int_\Delta \omega=2\pi$$

Answer 2

This is essentially a long comment to the other answer, but worth making an answer of its own, since it highlights a useful trick which also works for Stoke's Theorem, as well as explains the $2 \pi n$ answer. It's true that you can't apply Green's theorem to this problem directly, since the function contains a pole at the origin, but this doesn't stop you from applying Green's theorem on other regions that may also be useful.

Observe that your triangle $T$ does not pass through the origin. This means that you can find a small disk inside the triangle containing the origin $D$. Now consider the region $T - D$. This has two boundary components with opposite orientations, and we can consider the integral over $T - D$.

This gives us:

$$\int_{\text{interior of }T - D} d \omega = \int_T \omega - \int_D \omega$$ by Green's Theorem (or the Stoke's Theorem for differential forms, if you prefer to call it this). You can check that on this region, the left integral is $0$, as $d \omega$ is $0$ everywhere here. This means that you can replace the integral of $T$ with the integral over any nice circle you like encompassing the origin, since the left side is $0$.

Now recognize that this integral is super friendly to polar coordinates, and convert to these. You will get the correct answer.