I'm currently reviewing some problems in Statistical Mechanics and I have come across a question that I'm struggling to resolve. Specifically, in certain parts of the study of ideal gases, the concept of $n$-dimensional spheres is used, and two important quantities related to them are their area $S(n)$ and their volume $V(n)$. One method for finding solutions to these quantities involves integrating Gaussian $\varphi:\mathbb{R}^{n}\to\mathbb{R}$ (see Gaussian Integrals here).
So, my question is: why do we specifically use this scalar field $\varphi$? And how is it related to spheres in higher dimensions?
I've searched through several articles and texts, but I haven't found anything that could assist me in comprehending this question.
Suppose you have a positive radial function $\phi:\Bbb{R}^n\to\Bbb{R}$, meaning there is a function $f:[0,\infty)\to\Bbb{R}$ such that $\phi(x)=f(\|x\|)$. Due to the radial nature, by changing to polar coordinates, we can evaluate the integral of $\phi$ as follows: \begin{align} \int_{\Bbb{R}^n}\phi(x)\,dx=\int_0^{\infty}f(r) A_{n-1}r^{n-1}\,dr,\tag{$*$} \end{align} and hence we have (by positivity of the integrand, we can divide) \begin{align} A_{n-1}&=\frac{\int_{\Bbb{R}^n}\phi(x)\,dx}{\int_0^{\infty}f(r)r^{n-1}\,dr}.\tag{$**$} \end{align} This gives us an equation for the surface area of the unit sphere $S^{n-1}\subset\Bbb{R}^n$. The question now becomes whether there is a convenient choice of $\phi$ which simplifies computations of the two integrals above. In the case where you choose $\phi$ to be a Gaussian, you can evaluate the top using Fubini (and the fundamental property of exponentials $e^{a+b}=e^ae^b$), and the bottom by definition of the Gamma function (which is essentially an integral of exponential times a power), hence we get after some algebra the nice formula $A_{n-1}=\frac{2\pi^{n/2}}{\Gamma(n/2)}$.
I will admit that coming up with the formula $(*)$, and rearranging it to $(**)$ is not complicated at all and with some practice it is a pretty natural/common thing to do as well. But, the choice of exponentials is indeed very clever which requires some insight, and I can't motivate it a-priori. I can only say in hindsight that this is a very clever choice. The most naive, but cumbersome, approach to surface areas and volumes of spheres/balls is by doing things recursively with Fubini, as shown in the Wikipedia pages (but like I said, this is a little annoying).
The relationship between the volume $V_n$ and the surface area $A_{n-1}$, being $V_n=\frac{A_{n-1}}{n}$ is very easy to deduce using $(*)$ above. Let $B$ denote the unit ball in $\Bbb{R}^n$. Then, \begin{align} V_n&=\int_{B}1\,dx=\int_0^1A_{n-1}r^{n-1}\,dr=\frac{A_{n-1}}{n}. \end{align}
Another way to prove this, is by a clever observation and invoking the divergence theorem: \begin{align} V_n&=\int_{B}1\,dx=\int_B\frac{1}{n}\text{div}(F)\,dx=\frac{1}{n}\int_{\partial B}\langle F,\nu\rangle\,dA=\frac{1}{n}\cdot A_{n-1}, \end{align} where $F(x)=x$ is the radial vector field, and $\nu$ is the unit outward normal of the unit ball $B$, so $\nu(x)=x$ on $\partial B$, hence $\langle F,\nu\rangle=1$.