Triple Integration: Transform Equation to Spherical Coordinates

[big_gie]

Homework Statement

Transform the equation from cartesians coordinates to spherical coordinates.

Homework Equations

$$\int_\infty\int_\infty\int_\infty<br /> exp\left\{<br /> \frac{-\left| \vec{x'}-\vec{x}_0 \right|^2}{2 \sigma}<br /> \right\}<br /> \frac{\left( \vec{x} - \vec{x'} \right)}{\left| \vec{x} - \vec{x'} \right|^{3}} d^3x'$$

The Attempt at a Solution

I'm confused by the $$\vec{x}$$, $$\vec{x'}$$ and $$\vec{x}_0$$... I know it can be done: nothing depends on the angle here, so I should just get something depending on $$\vec{r}$$.

Thank you for any hints...

 

[Defennder]

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?

 

[big_gie]

Hi Defennder, thanks for your reply.

Yes, $$d^3x'$$ is indeed $$dx'~dy'~dz'$$: a volume element.

The integration is over infinity.

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

Maybe I'll explain more the problem...

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

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

I've done a variable change for $$\vec{x} - \vec{x}'$$ 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...)