Triple Integration: Transform Equation to Spherical Coordinates

AI Thread Summary
The discussion focuses on transforming a complex integral from Cartesian to spherical coordinates, specifically for calculating the electric field due to a Gaussian charge distribution. Participants clarify the meaning of variables such as \vec{x}, \vec{x'}, and \vec{x}_0, with \vec{x'} representing the integration variable as a position vector. The integral's limits extend to infinity, and the volume element is confirmed as d^3x' = dx' dy' dz'. One user suggests using the electric potential for easier integration, leading to a potential expression involving the same Gaussian charge density. The conversation highlights the challenges of handling the Gaussian function within the integral transformation.
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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&#039;}-\vec{x}_0 \right|^2}{2 \sigma}<br /> \right\}<br /> \frac{\left( \vec{x} - \vec{x&#039;} \right)}{\left| \vec{x} - \vec{x&#039;} \right|^{3}} d^3x&#039;<br />


The Attempt at a Solution


I'm confused by the \vec{x}, \vec{x&#039;} 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...
 
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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?
 
Hi Defennder, thanks for your reply.

Yes, d^3x&#039; is indeed dx&#039;~dy&#039;~dz&#039;: a volume element.

The integration is over infinity.

\vec{x&#039;} is the integration variable. It is a position vector (x&#039;,y&#039;,z&#039;).

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&#039;=-\infty}^{\infty} \int_{y&#039;=-\infty}^{\infty} \int_{z&#039;=-\infty}^{\infty}<br /> \rho\left(\vec{x&#039;}\right) \frac{\vec{x} - \vec{x&#039;}}{\left| \vec{x} - \vec{x}&#039;\right|^3} ~dx&#039;~dy&#039;~dz&#039;
<br /> \rho\left(\vec{x}\right) &amp; = &amp; \rho_0 \exp\left(<br /> -\frac{\left(\vec{x} - \vec{x_0}\right)^2}{2 \sigma^2}<br /> \right)<br />
where:
\vec{x} is the position where the field is wanted;
\vec{x&#039;} 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)
<br /> \phi\left(\vec{x}\right) = <br /> k \int_{x&#039;=-\infty}^{\infty} \int_{y&#039;=-\infty}^{\infty} \int_{z&#039;=-\infty}^{\infty}<br /> \rho\left(\vec{x&#039;}\right) \frac{1}{\left| \vec{x} - \vec{x}&#039;\right|} ~dx&#039;~dy&#039;~dz&#039;<br />

I've done a variable change for \vec{x} - \vec{x}&#039; 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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