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Homework Help: Triple integration

  1. Aug 5, 2008 #1
    1. The problem statement, all variables and given/known data
    Transform the equation from cartesians coordinates to spherical coordinates.


    2. Relevant equations
    [tex]\int_\infty\int_\infty\int_\infty
    exp\left\{
    \frac{-\left| \vec{x'}-\vec{x}_0 \right|^2}{2 \sigma}
    \right\}
    \frac{\left( \vec{x} - \vec{x'} \right)}{\left| \vec{x} - \vec{x'} \right|^{3}} d^3x'
    [/tex]


    3. The attempt at a solution
    I'm confused by the [tex]\vec{x}[/tex], [tex]\vec{x'}[/tex] and [tex]\vec{x}_0[/tex]... I know it can be done: nothing depends on the angle here, so I should just get something depending on [tex]\vec{r}[/tex].

    Thank you for any hints...
     
  2. jcsd
  3. Aug 5, 2008 #2

    Defennder

    User Avatar
    Homework Helper

    I'm equally confused by your integrand expression. It's supposed to be dxdydz isn't it? And what are the limits of your integration? And what does x' mean as both a scalar variable or vector function variable?
     
  4. Aug 5, 2008 #3
    Hi Defennder, thanx for your reply.

    Yes, [tex]d^3x'[/tex] is indeed [tex]dx'~dy'~dz'[/tex]: a volume element.

    The integration is over infinity.

    [tex]\vec{x'}[/tex] is the integration variable. It is a position vector [tex](x',y',z')[/tex].

    Maybe I'll explain more the problem...

    The integral is the electric field at position [tex]\vec{x}[/tex], caused by a charge distribution of gaussian shape:
    [tex]\vec{E}\left(\vec{x}\right) =
    k \int_{x'=-\infty}^{\infty} \int_{y'=-\infty}^{\infty} \int_{z'=-\infty}^{\infty}
    \rho\left(\vec{x'}\right) \frac{\vec{x} - \vec{x'}}{\left| \vec{x} - \vec{x}'\right|^3} ~dx'~dy'~dz'[/tex]
    [tex]
    \rho\left(\vec{x}\right) & = & \rho_0 \exp\left(
    -\frac{\left(\vec{x} - \vec{x_0}\right)^2}{2 \sigma^2}
    \right)
    [/tex]
    where:
    [tex]\vec{x}[/tex] is the position where the field is wanted;
    [tex]\vec{x'}[/tex] is the integration variable;
    [tex]\vec{x_0}[/tex] is the particle center;
    [tex]\sigma[/tex] is the particle width.

    I think I'll use the potential instead, for an easier integration:
    [tex]\vec{E}\left(\vec{x}\right) = - \nabla \phi\left(\vec{x}\right)[/tex]
    [tex]
    \phi\left(\vec{x}\right) =
    k \int_{x'=-\infty}^{\infty} \int_{y'=-\infty}^{\infty} \int_{z'=-\infty}^{\infty}
    \rho\left(\vec{x'}\right) \frac{1}{\left| \vec{x} - \vec{x}'\right|} ~dx'~dy'~dz'
    [/tex]

    I've done a variable change for [tex]\vec{x} - \vec{x}'[/tex] but then I'm stuck in the gaussian...

    Thanx for any hints.

    (Sorry if any mistakes have slipped, I'm writting this from memory and it's getting late...)
     
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