The number $\Omega$ of microstates accessible to the particle system is defined as :
$$
\Omega=\omega(E)\Delta
$$
where $\omega(E)$ is the density of state. To determine this guy, you need first to calculate the number $\chi(E)$ of µ-states corresponding to an energy inferior or equal to $E$.
Let us call the associated phase space region $\Sigma_{[0,E]}\equiv\lbrace\Gamma,0\leqslant\mathcal{H}(\Gamma)\leqslant E\rbrace$.
Then, it basically reads :
$$
\chi(E)=\frac{1}{h^{3N}}\int_{\Sigma_{[0,E]}}\mathrm{d}\Gamma \quad \text{with} \quad \mathrm{d}\Gamma=\prod^N_{i=1}\mathrm{d}\textbf{q}_i\,\mathrm{d}\textbf{p}_i
$$
$h^{3N}$ is here the element phase space volume corresponding to one µ-state.
It is quite straighforward to compute :
$$
\chi(E)=\frac{1}{h^{3N}}\left[\prod^N_{i=1}\int_{\Sigma_{[E,E+\Delta]}}\mathrm{d}\textbf{q}_i\right]\left[\prod^N_{i=1}\int_{\Sigma_{[0,E]}}\mathrm{d}\textbf{p}_i\right]=\frac{1}{h^{3N}}\,V^N\times V_{3N}(\sqrt{2mE})
$$
where $V_{3N}(\text{r})$ is the volume of a $3N$ dimension hyperball with a $\text{r}$ radius, and $\Sigma_{[E,E+\Delta]}$ is the phase space region $\lbrace\Gamma,E\leqslant\mathcal{H}(\Gamma)\leqslant E+\Delta \rbrace$.
For a 3D gas, we have for a given particle $i$ :
$$
\int_{\Sigma_{[E,E+\Delta]}}\mathrm{d}q_{i,x}\mathrm{d}q_{i,y}\mathrm{d}q_{i,z}=V\quad\text{volume of the gas}
$$
and for $N$ particles :
$$
\int_{\Sigma_{[0,E]}}\prod_{i=1}^N\mathrm{d}p_{i,x}\mathrm{d}p_{i,y}\mathrm{d}p_{i,z}=V_{3N}(\sqrt{2mE})
$$
by integrating over all possible direction of the total momentum $\sum_i\textbf{p}_i$ and all magnitudes below the energy shell $\lbrace\Gamma,\mathcal{H}(\Gamma)=E \rbrace$ so that :
$$
\left|\sum_i\textbf{p}_i\right|=\sqrt{\sum_i\textbf{p}_i^2}=\sqrt{2mE}\quad\text{with}\quad E=\frac{1}{2m}\sum_i\textbf{p}_i^2
$$
Since all direction are possible, the integration is performed over a $3N$-ball.
Then, it follows :
$$
\omega(E)=\frac{\mathrm{d}\chi}{\mathrm{d}E}(E)=\frac{1}{h^{3N}}\,V^N\times S_{3N}(\sqrt{2mE})
$$