Density of states for arbitrary dispersion relation If I have a 3D dispersion relation 
$E=E(k_x, k_y, k_z)$ I have an equation for the density of states, which is 
$D(E)=\frac{1}{\nabla_k E}\int\frac{dS}{(2\pi)^3}$
1) I am confused about the integral. What am I integrating over, exactly? It must have units of the $k$ vector, so it's a k-space integral, but what is the surface that is relevant?
3) Where does that equation come from, how is it derived?
 A: deriving the desired equation for the density of states we are reminded that the surface which we are integrating over is a surface of constant energy in the reciprocal space which is denoted by $S(e)$, where $e$ is the 3-dimensional dispersion relation
we know that the number of states between the surfaces $S(e)$ and $S(e + de)$ is given by the integral. $$D(e)\, \mathrm{d}e = \int_{S(e)} \frac{\mathrm{d}S}{(2\pi )^3} \mathrm{d}q_\perp(\boldsymbol q)$$ if we linearly expand the dispersion relation $e$ as $e+\mathrm{d}e = e + |\nabla_\boldsymbol q e(\boldsymbol q)|\mathrm{d}q_\perp(\boldsymbol q)$ where $\boldsymbol q$ is a vector, $\mathrm{d}q_\perp (\boldsymbol q)$ is the perpendicular distance between the surfaces and $\mathrm{d}q_\perp(\boldsymbol q)= \frac {\mathrm{d}e}{|\nabla_\boldsymbol q e(\boldsymbol q)|}$. this leads to $$D(e) = \frac{1}{(2\pi)^3} \int_{S(e)} \frac {\mathrm{d}S}{|\nabla_\boldsymbol q e(\boldsymbol q)|}$$ 
which is the desired equation for the density of states. 
