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I'm not sure if this question already exists, because I'm not sure how to call it. I want to integrate over the line of a square. In a simplified problem, I'd like to solve:

$$ l=\oint\limits_c dl $$

Therefore I figured I could take one line and multiply the result by 4:

$$ l=4\oint\limits_{c'}dl $$

And then I wanted to use an angle ($d\alpha$) to represent $dl$ - however, that's where I failed. In the following figure I illustrated which values I can easily use to get $dl$ ($R$ is known and $r=\frac{R}{\cos\alpha})$: illustration With the help of the sine theorem I got the following: $$\frac{dl}{\sin(d\alpha)}=\frac{r}{\sin(180^\circ-d\alpha-(90^\circ-\alpha))}=\frac{r}{\sin(90^\circ+\alpha-d\alpha)}=\frac{r}{\cos(d\alpha-\alpha)}$$ and with the addition theorem: $$dl=\frac{r\sin(d\alpha)}{\cos(d\alpha-\alpha)}=\frac{r\sin(d\alpha)}{\sin(d\alpha)\sin\alpha+\cos(d\alpha)\cos\alpha}\simeq\frac{rd\alpha}{d\alpha\sin\alpha+\cos\alpha}$$ But I have no idea how to calculate the resulting integral: $$l=4\oint\limits_{c'}\frac{rd\alpha}{d\alpha\sin\alpha+\cos\alpha}$$ as $d\alpha$ is in the denominator.

Am I missing some trigonometric rule here?


Edit if I use $r=\frac{R}{cos\alpha}$, I'd have: $$dl\simeq\frac{Rd\alpha}{\sin\alpha\cos\alpha d\alpha+\cos^2\alpha}$$ which (as wolframalpha tells me) is $$dl\simeq\frac{2Rd\alpha}{\sin(2\alpha)d\alpha+\cos(2\alpha)+1}$$ However, that doesn't help me much, as $d\alpha$ still is in the denominator.

More with the help of guesswork then through my own thoughts, I found a $dl$ that would work: $$dl=|2\sin(2\alpha)|d\alpha$$ For a square with $R=1$ for example, I'd have: $$l=4\oint\limits_{c'}|2\sin(2\alpha)|d\alpha=4\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}|2\sin(2\alpha)|d\alpha=4\cdot2=8$$ so this behaves as expected - however, I have no idea how this could be trigonometrically (or mathematically, for that matter) explained. Furthermore, this seems to contrast my previous thoughts about $dl$, as illustrated in the following graph from geogebra (with $d\alpha=0.1$ - green is my first attempt, blue is my guessed attempt): graph

I also did some more research and got here, where the line element in polar coordinates is given by: $$dl^2=dr^2+r^2d\alpha^2$$ this seems to use the pythagorean theorem, in the sense that $d\alpha$ is straigt for small values, which makes sense. Follwing that, one might say: $$\tan(d\alpha)=\frac{dr}{d\alpha}$$ implying the following for the line element: $$dl=\sqrt{d\alpha^2\tan^2(d\alpha)+d\alpha^2}=d\alpha\sqrt{\tan^2(d\alpha)+1}$$ however, I don't know how to use this, as $d\alpha$ also is in the Tangens, is squared, is a summand and is taken the root of. I guess there's something wrong with this solution, as the term $\alpha$ doesn't even appear in it, but even if it wasn't I could not use it any more than my first solution.

I have the feeling that I'm looking at this problem in a way to complicated manner, especially as I can't seem to find useful information in the internet. But if this is true, what is the obvious answer here?

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$$l=R\tan\alpha\implies dl=\frac{R\,d\alpha}{\cos^2\alpha}.$$

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  • $\begingroup$ Ok that actually makes sense. Could you nevertheless elaborate on that? $\endgroup$ – MetaColon Jun 26 at 4:29
  • $\begingroup$ @MetaColon What kind of elaboration do you mean? I thought your only problem was to express $dl$ via $d\alpha$... Of course the expression is valid only for one side of the square ($-\pi/4<\alpha<\pi/4$) but it is straightforward to write down similar expressions for 3 other sides as well. $\endgroup$ – user Jun 26 at 9:14
  • $\begingroup$ In my understanding it is a rather unusual way to express the variable to integrate over ($dl$) as a differential and hence applying the rules for differentiating on it. At least it's a method I've never seen in practice - maybe you could elaborate on why exactly this works? And also how it fits to my attempts, if it does at all. $\endgroup$ – MetaColon Jun 26 at 11:00
  • $\begingroup$ This is the usual way. In one-dimensional case one introduces a new variable $t$, expresses it as function of the old one $t=f(x)$, and finds the differential as $dt=f'(x)dx$. Quite oppositely I do not understand the background of your attempts. Please give a reference to description of this procedure and - if possible - to its successful application. $\endgroup$ – user Jun 26 at 11:24
  • $\begingroup$ To be honest I'm a physics student, hence we learned a kind of intuitive way to understanding things like $dl$. This answer for example shows how we explained volume elements in spherical coordinates. I tried to apply similar reasoning to get the line element in polar coordinates. $\endgroup$ – MetaColon Jun 26 at 14:05

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