In this exercise my initial thought was to split up the integral into a part for the transverse and parallel polarizations. When I checked the answers I noticed that the DOS is combined and a single \omega_D is used as upper bound to compute the energy. This however seems a bit strange to me.
If we consider the two dispersion relations, we find in that case that v_{||}k_1= v_\perp k_2 = \omega_D. Since v_{||} > v_\perp then k_1 < k_2. In other words, the perpendicular modes of vibration can have shorter wavelengths than the longitudinal ones.
What seemed logical to me is two different $\omega_D$’s such that the maximum k vectors for both polarizations are equal. This way we find that \omega_{D||} > \omega_{D\perp}, so two different Debye frequencies, that tell you that the transverse modes will be occupied faster than the longitudinal ones and that the longitudinal modes can support higher frequencies (which maybe sounds reasonable given their higher speed of sound?).
If we can use a single \omega_D, then how does that make sense? How can the minimum wavelength be shorter in one mode of propagation than the other?