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Consider the $1$-form $$\alpha=\frac{1}{x^2+y^2}(-ydx+xdy)\in \Omega^1(\mathbb{R}^2\setminus \{(0,0)\})$$ If $\gamma:[0,1]\to \mathbb{R}^2\setminus \{(0,0)\}$ is such that $\gamma(t)=(x(t),y(t))$ where $y(t)>0$ for all $t\in [0,1]$ and $\gamma(0)=\gamma(1)$, find $\int_{\gamma}\alpha$.

I've been stuck on this for a while and am not sure how to get started. Going through this post, the two answers proposed, to be honest, are confusing me (which is mainly because I'm quite new to differential forms). Can someone please explain, without using Stokes' theorem (since this has not been covered yet in my class) or contraction, how I am supposed to use the assumption that $y(t)>0$ for all $t\in [0,1]$ to evaluate the integral?

Liebster Jugendtraum
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  • Complex integration works. – Ѕᴀᴀᴅ Apr 09 '18 at 23:49
  • @AlexFrancisco Hi Alex. Thanks for the suggestion. I know this is somewhat related to winding number but is there an alternative way in which real integration works? (This is from my differential geometry course) – Liebster Jugendtraum Apr 09 '18 at 23:53
  • Since the exact expression of $γ(t)$ is not given, it's probably true that the only method using real integration is to use Stoke's theorem. – Ѕᴀᴀᴅ Apr 09 '18 at 23:59
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    If $y(t) > 0$ for all $t \in [0,1]$ and $\gamma(0) = \gamma(1)$, then $\gamma$ does not enclose the origin. Then $\gamma$ is null-homotopic (in $\Bbb R^2 \setminus {0,0}$) and so $\int_\gamma \alpha = 0$ since $\alpha$ is exact. – Ivo Terek Apr 10 '18 at 00:03
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    But $\alpha$ is not exact on its entire domain. So be careful. It's certainly exact on the upper half-plane, however. – Ted Shifrin Apr 10 '18 at 03:48

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