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Thread: An integration problem

  1. #1
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    An integration problem

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

    Hi , solve by this way :

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

    now you can use the this substitute :

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

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

    now you can use the this substitute :

    $\displaystyle
    sin\theta = (a-cos\theta)sinx
    $
    But $\displaystyle (a-\cos\theta)^2-\sin^2\theta} \ne a^2-2a\cos{\theta}+1$, 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.
    $\displaystyle
    \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
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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 $\displaystyle \mathcal{C}\subset \textbf{R}^{3}$ induces an electric potential in $\displaystyle $\textbf{R}^{3}\backslash\mathcal{C}$.
    $
    The charge distribution is given by the charge density $\displaystyle $\rho: \mathcal{C} \rightarrow \textbf{R}$$, and the induced electric potential at $\displaystyle $p \in \textbf{R}^{3}\backslash\mathcal{C}$ $
    is given by
    $\displaystyle
    \begin{center}
    $V(p)=\frac{1}{4\pi\varepsilon}\int_{\mathcal{C}} \frac{\rho}{\mathrm{dist(p,\cdot)}}$
    \end{center}
    $
    where $\displaystyle $\mathrm{dist(p,\cdot)} = \|p-x\|$$ denotes the distance between $\displaystyle $p\in \textbf{R}^{3}\backslash\mathcal{C}$$ and $\displaystyle $x\in \mathcal{C}$$.
    Consider a uniform charge density $\displaystyle $\rho$$ on the set
    $\displaystyle
    \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 $\displaystyle $x=\cos(\theta)$$ and $\displaystyle $y=\sin(\theta)$

    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 $\displaystyle \mathcal{C}\subset \textbf{R}^{3}$ induces an electric potential in $\displaystyle $\textbf{R}^{3}\backslash\mathcal{C}$.
    $
    The charge distribution is given by the charge density $\displaystyle $\rho: \mathcal{C} \rightarrow \textbf{R}$$, and the induced electric potential at $\displaystyle $p \in \textbf{R}^{3}\backslash\mathcal{C}$ $
    is given by
    $\displaystyle
    \begin{center}
    $V(p)=\frac{1}{4\pi\varepsilon}\int_{\mathcal{C}} \frac{\rho}{\mathrm{dist(p,\cdot)}}$
    \end{center}
    $
    where $\displaystyle $\mathrm{dist(p,\cdot)} = \|p-x\|$$ denotes the distance between $\displaystyle $p\in \textbf{R}^{3}\backslash\mathcal{C}$$ and $\displaystyle $x\in \mathcal{C}$$.
    Consider a uniform charge density $\displaystyle $\rho$$ on the set
    $\displaystyle
    \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 $\displaystyle $x=\cos(\theta)$$ and $\displaystyle $y=\sin(\theta)$

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