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Im not good in geometric interpretations... any help is very welcome.

Consider the unitary disc $$D=\{(x,y,0)\in\mathbb{R}^3, x^2+y^2\leq1\},$$

parameterized by $$\varphi(r,\theta)=(r\cos\theta,r\sin\theta), (r,\theta)\in[0,1]\times[0,2\pi].$$ Let $\Omega(x,y,z)$ be the solid angle of $\varphi$, viewed from $(x,y,z)$. Consider a closed curve $\gamma:[a,b]\rightarrow\mathbb{R}^3\backslash S$ of class $C^1$, with $$S=\{(x,y,0)\in\mathbb{R}^3, x^2+y^2=1\}.$$ Let $p$ be the number of times that $\gamma$ cuts $D$, coming from $z>0$ to $z<0$, and $q$ the number of times that $\gamma$ cuts $D$, coming from $z<0$ to $z>0$. Use geometric arguments to conclude that $$\int_\gamma d\Omega=4\pi(p-q).$$

PS: if someone wants to know about Solid Angle, take a look at http://en.wikipedia.org/wiki/Solid_angle

or

http://mathworld.wolfram.com/SolidAngle.html

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  • $\begingroup$ Maybe a stupid question. If $\gamma$ is closed, don't we have $\gamma(a) = \gamma(b)$? $\endgroup$
    – newbie
    Commented Jul 2, 2013 at 10:58
  • $\begingroup$ I think you are right, im gonna take out thos part. $\endgroup$
    – diff_math
    Commented Jul 14, 2013 at 0:50
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    $\begingroup$ Can you explain the definition of $\Omega$ a little more clearly? What do you mean by the solid angle of $\varphi$? The general gist of this problem is that this $\Omega$ in some sense relates to "going around" the edge of the disk, but I don't understand its definition. $\endgroup$
    – dfeuer
    Commented Jul 14, 2013 at 1:46
  • $\begingroup$ This is suposed to be familiar to mathematicians. My teacher words. en.wikipedia.org/wiki/Solid_angle $\endgroup$
    – diff_math
    Commented Jul 15, 2013 at 16:08
  • $\begingroup$ In this case, the "point of view" is not the origin but (x,y,z), so it is necessary to do a translation before, then work with the classic solid angle. $\endgroup$
    – diff_math
    Commented Jul 15, 2013 at 16:11

1 Answer 1

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Hint: Prove that $\int_\gamma d\Omega = \int_{\gamma'} d\Omega$ if $\gamma$ and $\gamma'$ are homotopic in $\mathbb{R}^3\backslash S$ (for some regular enough homotopy, probably you will need $C^2$). Use this to reduce (through deforming and "cutting" the path $\gamma$) the problem to the case where $\gamma$ is for example a circle (or some similarly easy to work with form) cutting $D$ exactly once (through deforming and separating the path $\gamma$).

Here's a related problem (the one I mentioned in the comments) with solution:

Problem (problem 2)

Solution

Ich hoffe, dass du Deutsch lesen kannst ;)

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  • $\begingroup$ I know how to prove that if $\gamma$ and $\gamma '$ are homotopic in $\mathbb{R^3}\backslash S$ then $\int_{\gamma}d\Omega =\int_{\gamma '}d\Omega$. But why are $\gamma$ and $\gamma '$ homotopic? $\endgroup$
    – diff_math
    Commented Jul 15, 2013 at 22:36
  • $\begingroup$ If you can show that $\int_{\gamma'}d\Omega = \int_{\gamma}d\Omega$ for homotopic paths, then you can reduce the problem to easier cases. For example, consider a path crossing $D$ exactly twice and both times from $z<0$ to $z>0$. Then you can deform it to a circle which you run trough twice, and thus you are reduce to compute the integral only for the simple case of a circle crossing $D$ once! $\endgroup$
    – Daniel Robert-Nicoud
    Commented Jul 16, 2013 at 9:20
  • $\begingroup$ I forgot to add @diff_math to notify you of my previous comment. Anyway, I wanted to add that this is a standard technique to solve similar problems. Also, I think that you can show that $\int_\gamma d\Omega = \int_\gamma\int_S\frac{(d\vec{s_1}\wedge\vec{ds_2})\cdot \vec{r}}{r^3}$, where $s_1$, $s_2$ run on $\gamma$ and $S$ and $\vec{r}$ is the difference of position. If that is the case, you can then use I think Faraday's equation (from ED) to solve the problem. $\endgroup$
    – Daniel Robert-Nicoud
    Commented Jul 16, 2013 at 9:51

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