How to understand Density of States with dispersion relation

I am having trouble understanding the Density of states concept. As I currently understand it, for the density of states $g(k)$ it is the number of microstates with wave number in the range $[k,k+\mathrm{d}k]$.

However, I'm not sure how I actually go about calculating this in practice. The only example I can find is one which takes the partition function of an ideal monatomic gas and just rewrites the integral:

\begin{align*}Z_{1}&=\frac{V}{(2\pi)^{3}}\int_{0}^{\infty}4\pi k^{2} \exp\left(-\frac{\beta \hbar^{2} k^{2}}{2m}\right)\:\mathrm{d}k \\ &= \int_{0}^{\infty}g(k)\exp\left(-\frac{\beta \hbar^{2} k^{2}}{2m}\right)\:\mathrm{d}k \implies g(k)=\frac{V k^{2}}{2\pi^{2}}\end{align*}

However, I was then given a question:

Given the dispersion relationship for a gas of non-interacting, massless spin-$0$ bosons: $$\omega = \alpha k^{2}$$ Show that in $3$-dimensions, the density of states is $$g(\omega) \propto \omega^{1/2}$$

However, I don't understand why the density of states would differ, as we are dealing with a non-interacting gas again, i.e. we'd have:

$$g(k)=\frac{Vk^{2}}{2\pi^{2}}\implies g(\omega)=\frac{V \omega}{2\alpha\pi^{2}}$$

But this is wrong. What have I misunderstood?

You have to express $dk$ in terms of $\omega$ and $d\omega$ as well. This is just a matter of using the method of u-substitution: Just reexpress the integral in terms of $\omega$.