Cross-sectional area of a Gaussian beam of particles On page 47 of the seventh edition of Particles and Nuclei, the authors write the following:

For a Gaussian distribution of the beam particles around the beam centre (with horizontal and vertical standard deviations $\sigma_x$ and $\sigma_y$ respectively), $A$ [the cross-sectional area] is given by \begin{equation}
A=4\pi\sigma_x\sigma_y.
\end{equation}

This formula isn't proven or even motivated, however. Could anyone help me to understand how this result comes to be?
 A: I read your question again. 
I can't say why a particle beam would have the same Gaussian beam shape as a laser. But in lasers, this is the most common and desirable beam. If nothing else, it has the most collimated beam because other modes have larger diffraction effects. Diffraction is often important for lasers. 
The intensity profile across the center of a Gaussian beam has is a bell shaped curve. Because of this, there is no obvious beam diameter. The beam fades away as you get farther from the center, without any definite edge. The beam diameter is a matter of convention. 
For a bell shaped curve, the standard deviation, $\sigma$, is an obvious way to speak of beam radius. And this is one of the usual conventions for beam radius. At this radius, the E field is $1/e$ of the central value, and the intensity has dropped to $1/e^2$. Another convention for beam diameter is Full Width at Half Max (FWHM).
There is plenty of intensity outside these diameters. Typically lenses used to focus the beam must have an aperture 1.5 times the beam $\sigma$ based diameter. At that distance from the center, the intensity is about 1% of the central intensity. This is low enough that diffraction effects from truncating the beam are usually negligible. 
Not all Gaussian beams are circular. Some are elliptical. These have two different beam radii, $\sigma_x$ and $\sigma_y$. 
Given this, a convention for the area would reasonably be 
$$A = \pi \sigma_x \sigma_y$$
It appears this author is using $2\sigma$ as his convention. So he gets 
$$A = 4\pi \sigma_x \sigma_y$$
