An integration problem

**wslgx1024** · July 11th 2010, 10:13 AM

Here is the problem.
$ \begin{displaymath} \int_{0}^{2\pi} \frac{1}{\sqrt{a^{2}+1-2a\cos\theta}} \mathrm{d} \theta \end{displaymath} $

**parkhid** · July 11th 2010, 11:37 AM

Hi , solve by this way :

$ \int \frac{d\theta}{\sqrt{(a-cos\theta)^2-sin^2\theta}} $

now you can use the this substitute :

$ sin\theta = (a-cos\theta)sinx $

**CreaturesOfPrometheus** · July 11th 2010, 08:01 PM

Originally Posted by parkhid

Hi , solve by this way :

$ \int \frac{d\theta}{\sqrt{(a-cos\theta)^2-sin^2\theta}} $

now you can use the this substitute :

$ sin\theta = (a-cos\theta)sinx $

But $(a-\cos\theta)^2-\sin^2\theta} \ne a^2-2a\cos{\theta}+1$ , unless I'm missing something.

**mr fantastic** · July 12th 2010, 05:07 AM

Originally Posted by wslgx1024

Here is the problem.
$ \begin{displaymath} \int_{0}^{2\pi} \frac{1}{\sqrt{a^{2}+1-2a\cos\theta}} \mathrm{d} \theta \end{displaymath} $

Where has the integral come from?

Read this: Elliptic Integral of the First Kind -- from Wolfram MathWorld

**wslgx1024** · July 12th 2010, 07:42 AM

Originally Posted by mr fantastic

Where has the integral come from?

Read this: Elliptic Integral of the First Kind -- from Wolfram MathWorld

The original problem is

In electrostatics, a charge distribution over a curve $\mathcal{C}\subset \textbf{R}^{3}$ induces an electric potential in $$\textbf{R}^{3}\backslash\mathcal{C}$. $
The charge distribution is given by the charge density $\rho: \mathcal{C} \rightarrow \textbf{R}$ , and the induced electric potential at $p \in \textbf{R}^{3}\backslash\mathcal{C}$
is given by
$ \begin{center} $V(p)=\frac{1}{4\pi\varepsilon}\int_{\mathcal{C}} \frac{\rho}{\mathrm{dist(p,\cdot)}}$ \end{center} $
where $\mathrm{dist(p,\cdot)} = \|p-x\|$ denotes the distance between $p\in \textbf{R}^{3}\backslash\mathcal{C}$ and $x\in \mathcal{C}$ .
Consider a uniform charge density $\rho$ on the set
$ \begin{center} $\mathcal{C}=\{(x,y,z)\in \textbf{R}^{3}:x^{2}+y^{2}=1,z=0\}$. \end{center} $
Find the potential at points on the x-axis, |x|>1.

And by using the parametrization $x=\cos(\theta)$ and $$y=\sin(\theta)$

I got the previous integral.
And how could the integral in the link you gave solve this integral?

**mr fantastic** · July 12th 2010, 06:05 PM

Originally Posted by wslgx1024

The original problem is

In electrostatics, a charge distribution over a curve $\mathcal{C}\subset \textbf{R}^{3}$ induces an electric potential in $$\textbf{R}^{3}\backslash\mathcal{C}$. $
The charge distribution is given by the charge density $\rho: \mathcal{C} \rightarrow \textbf{R}$ , and the induced electric potential at $p \in \textbf{R}^{3}\backslash\mathcal{C}$
is given by
$ \begin{center} $V(p)=\frac{1}{4\pi\varepsilon}\int_{\mathcal{C}} \frac{\rho}{\mathrm{dist(p,\cdot)}}$ \end{center} $
where $\mathrm{dist(p,\cdot)} = \|p-x\|$ denotes the distance between $p\in \textbf{R}^{3}\backslash\mathcal{C}$ and $x\in \mathcal{C}$ .
Consider a uniform charge density $\rho$ on the set
$ \begin{center} $\mathcal{C}=\{(x,y,z)\in \textbf{R}^{3}:x^{2}+y^{2}=1,z=0\}$. \end{center} $
Find the potential at points on the x-axis, |x|>1.

And by using the parametrization $x=\cos(\theta)$ and $$y=\sin(\theta)$

I got the previous integral.
And how could the integral in the link you gave solve this integral?

I thought that might be the case. These sorts of integrals often occur in those contexts. You need to re-arrange your integral to recognise the elliptic-integral form you have. Evaluate your integral here: http://www.wolframalpha.com/ to see where you need to head.

**wslgx1024** · July 13th 2010, 01:26 AM

Originally Posted by mr fantastic

I thought that might be the case. These sorts of integrals often occur in those contexts. You need to re-arrange your integral to recognise the elliptic-integral form you have. Evaluate your integral here: Wolfram|Alpha—Computational Knowledge Engine to see where you need to head.

Thanks a lot for help. BTW, the website is great!

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