Density of states of one classical harmonic oscillator I have to determine the density of states of one tridimensional harmonic oscillator. I have to prove that the expression is the following $D(E) = aE^2$, a is a constant.
I know this is a 6-dimensional phase space $(x,y,z)$ and $(p_x,p_y,p_z)$ and the hamiltonian depends on the position and momentum. So when performing the integral I have to determine the area of a hypersphere, but I don't get why is $E^2$, I can't get that result. Can someone explain how do I determine the density in this situation? I added a picture with the my solution and where is my doubt.The constants are correct according to the solutions I have, my problem is really the E^(5/2). 

 A: For harmonic oscillator,
$$E=\sum_{x,y,z}\left(n_i+\frac{1}{2}\right)\hbar \omega_i$$
If I define $G(E)$ as the total number of state that lies between $0$ to $E$ then the density of state can be found using
$$g(E)=\frac{dG}{dE}$$
$$G(E)=\frac{1}{\hbar^3\omega_x\omega_y\omega_z}\int_0^E\int_0^{E-E_x}\int_0^{E-E_x-E_y}dE_zdE_ydE_x$$
If you perform the trivial integral, you will find
$$G(E)=\frac{1}{(\hbar\omega_0)^3}\frac{E^3}{6}$$
where we have taken $\omega_i=\omega_0$.
$$\boxed{g(E)=\frac{dG}{dE}=\frac{1}{2}\frac{E^2}{(\hbar\omega_0)^3}}$$

Edit: For classical case,
$$H=\sum_i^N[P^2_i+Q^2_i]=\frac{2E}{\omega}$$
where $Q_i=\sqrt{m\omega} q_i$ and $P_i=p_i/\sqrt{m\omega}$.
The volume of the available phase space is thus given by
$$\Omega (E,V,N)=V^{2N}\frac{2\pi^{3N}}{\Gamma(3N)}\left(\frac{2E}{\omega}\right)^{(6N-1)/2}\Delta_R$$
Note $P^2=2E/\omega\Rightarrow dP=dE/\sqrt{2\omega E} $ so $\Delta_R=dE/\sqrt{2\omega E}$.
$$\Omega (E,V,N)=V^{2N}\frac{2\pi^{3N}}{\Gamma(3N)}\left(\frac{2E}{\omega}\right)^{(6N-1)/2}\frac{1}{\sqrt{2\omega E}}dE$$
The density of state then
$$D(E)=V^{2N}\frac{2\pi^{3N}}{\Gamma(3N)}\left(\frac{2E}{\omega}\right)^{(6N-1)/2}\frac{1}{\sqrt{2\omega E}}$$
Putting  $N=1$
$$\boxed{D(E)\propto E^2}$$
