# Thread: An integration problem

1. ## An integration problem

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

2. ## Easy way

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
$

3. 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.

4. 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

5. 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?

6. 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.

7. 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&mdash;Computational Knowledge Engine to see where you need to head.
Thanks a lot for help. BTW, the website is great!