# Thickness of a $3N$ dimensional spherical shell (entropy of classical gas)

I'm brushing up on statistical mechanics and calculating the entropy of a classical gas (i.e. particles in a box).

Working through the calculation, we end up with an integral of the form: $$\int^{'} d^3\mathbf{r}_1...d^3\mathbf{r}_N$$

where N is the number of particles and $\mathbf{r}_i \equiv \sqrt{\hbar^2/2m} (\pi/L) \mathbf{n}_i$, and $\mathbf{n}_i = (n_i^x, n_i^y, n_i^z)$ is just the set of quantum numbers of the $i^{\text{th}}$ particle. Clearly $|\mathbf{r}_i|$ has dimensions of $\sqrt{\text{energy}}$

The prime on the integral denotes integration over the points in 3N dimensional space satisfying:

$$E - \delta E/2 \leq \sum_{i=1}^N |\mathbf{r}_i|^2 \leq E + \delta E/2$$.

Where $E$ is the fixed total energy of the system, and $\delta E$ is a small uncertainty in the energy.

This restriction is clearly a spherical shell, and my professor's notes say the radius is $\sqrt{E}$ (makes sense) and the thickness $\frac{1}{2} \delta E/\sqrt{E}$. He then proceeds as usual by calculating the integral etc.

My question is, why is the thickness $\frac{1}{2} \delta E/\sqrt{E}$?

I know this question is borderline absurd, but I'm not sure what's going on here.

If the radius is determined by $\sqrt{E}$, then the length after a small increase is $\sqrt{E + \delta E}$
But that’s not really in a useful form, so we expand using the fact that $\delta E$ is infinitesimal:
$\sqrt{E + \delta E} = \sqrt{E} \sqrt{1 + \delta E/ E}$
$= \sqrt{E} (1 + \delta E /(2E))$
$= \sqrt{E} + \delta E/(2\sqrt{E})$