Integral of a differential form over a circumference.

Question

I'm asked to find $$\int_{\gamma}\omega$$ where $\gamma$ is the circumference of center $C(2,0)$ and radius $1$ and $$\omega = \frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy$$ We can parametrize the curve as $$\gamma \colon t \mapsto (2+\cos t,\sin t)$$ and so, by applying theorems about integration of differential forms over a curve we have $$\int_{\gamma} \omega = \int_0^{2\pi}\left(\frac{-\sin t}{5+\cos t}(-\sin t)+\frac{2+\cos t}{5+4\cos t}\cos t\right)dt=\int_0^{2\pi}\frac{1+2\cos t}{5+4\cos t}dt$$ which I cannot solve. I tried different sostitutions but they lead me to nothing. How can I do it?

Answer 1

Multiply top and bottom by the conjugate of the denominator to get

$$I = \int_0^{2\pi}\frac{5+6\cos t - 8\cos^2 t}{25-16\cos^2t}\:dt = \int_0^{2\pi}\frac{1}{2} + \frac{6\cos t}{9+16\sin^2t} - \frac{15}{2}\frac{1}{9\cos^2t+25\sin^2t}\:dt$$

$$= \pi + 0 -\frac{3}{10}\int_0^{2\pi}\frac{\sec^2t}{\frac{9}{25}+\tan^2t}\:dt$$

$$ = \pi - \frac{1}{2}\Biggr[\tan^{-1}\left(\frac{5}{3}\tan t\right)\Biggr|_0^{\frac{\pi}{2}^-}+\tan^{-1}\left(\frac{5}{3}\tan t\right)\Biggr|_{\frac{\pi}{2}^+}^{\frac{3\pi}{2}^-}+\tan^{-1}\left(\frac{5}{3}\tan t\right)\Biggr|_{\frac{3\pi}{2}^+}^{2\pi}\Biggr]$$

$$= \pi - \frac{1}{2}\Biggr[\frac{\pi}{2} + \pi + \frac{\pi}{2}\Biggr] = 0$$

Answer 2

Maybe a simpler way could be noticing the fact that $\omega = Im \hspace{0.1cm}(\frac{dz}{z})$ .

Since $\overline{B((2,0),1)} \subseteq \mathbb{C} - \left\lbrace 0 \right\rbrace$ it follows that $\gamma$ is homotopically trivial.

So $\int_{\gamma} \frac{dz}{z} = 0$ which means $\int_{\gamma}\omega = 0$

Answer 3

Here's an alternative way to do this. Note that on the region $x>0$ we have $$\omega=d\big(\arctan(y/x)\big).$$ Note that $\gamma$ is a subset of this half-plane. By the Fundamental Theorem of Calculus (or baby Stokes's Theorem, if you prefer), since $\gamma$ is a closed curve, the integral of an exact form around $\gamma$ is $0$.

(This particular $1$-form $\omega$ is quite famous for being closed, and hence locally exact, but not exact in all of $\Bbb R^2-\{0\}$.)