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Math Help - An integration problem

  1. #1
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    An integration problem

    Here is the problem.
    <br />
\begin{displaymath} <br />
\int_{0}^{2\pi} \frac{1}{\sqrt{a^{2}+1-2a\cos\theta}} \mathrm{d} \theta<br />
\end{displaymath}<br />
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  2. #2
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    Easy way

    Hi , solve by this way :

    <br />
\int \frac{d\theta}{\sqrt{(a-cos\theta)^2-sin^2\theta}}<br />

    now you can use the this substitute :

    <br />
sin\theta = (a-cos\theta)sinx<br />
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  3. #3
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    Quote Originally Posted by parkhid View Post
    Hi , solve by this way :

    <br />
\int \frac{d\theta}{\sqrt{(a-cos\theta)^2-sin^2\theta}}<br />

    now you can use the this substitute :

    <br />
sin\theta = (a-cos\theta)sinx<br />
    But (a-\cos\theta)^2-\sin^2\theta} \ne a^2-2a\cos{\theta}+1, unless I'm missing something.
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  4. #4
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    Quote Originally Posted by wslgx1024 View Post
    Here is the problem.
    <br />
\begin{displaymath} <br />
\int_{0}^{2\pi} \frac{1}{\sqrt{a^{2}+1-2a\cos\theta}} \mathrm{d} \theta<br />
\end{displaymath}<br />
    Where has the integral come from?

    Read this: Elliptic Integral of the First Kind -- from Wolfram MathWorld
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    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 $\textbf{R}^{3}\backslash\mathcal{C}$.<br />
    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
    <br />
\begin{center}<br />
$V(p)=\frac{1}{4\pi\varepsilon}\int_{\mathcal{C}} \frac{\rho}{\mathrm{dist(p,\cdot)}}$<br />
\end{center}<br />
    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
    <br />
\begin{center}<br />
$\mathcal{C}=\{(x,y,z)\in \textbf{R}^{3}:x^{2}+y^{2}=1,z=0\}$.<br />
\end{center}<br />
    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?
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  6. #6
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    Quote Originally Posted by wslgx1024 View Post
    The original problem is

    In electrostatics, a charge distribution over a curve \mathcal{C}\subset \textbf{R}^{3} induces an electric potential in $\textbf{R}^{3}\backslash\mathcal{C}$.<br />
    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
    <br />
\begin{center}<br />
$V(p)=\frac{1}{4\pi\varepsilon}\int_{\mathcal{C}} \frac{\rho}{\mathrm{dist(p,\cdot)}}$<br />
\end{center}<br />
    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
    <br />
\begin{center}<br />
$\mathcal{C}=\{(x,y,z)\in \textbf{R}^{3}:x^{2}+y^{2}=1,z=0\}$.<br />
\end{center}<br />
    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.
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    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!
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