The Hamiltonian for a particle is given by:
$$H_1 = \frac{P^2}{2m} + \frac{p^2_\theta}{2I} + \frac{p^2_\phi}{2Isin^2 \theta}$$
$I$ is the moment of inertia
$$I = \frac{mR^2}{4}$$
To get the partition function for one particle $Z_1(P, Q)$ the following integral has to be solved:
$$Z_1(P, Q, p_ \theta, \theta, p_\phi, \phi) = \int_ {-\infty}^{+\infty}\frac{dPdQ}{\lambda^3} dp_\theta d\theta dp_\phi d\phi e^{-\beta H_1}$$
Where $\lambda$ is the thermal wavelength:
$$\lambda = h \sqrt{\frac{\beta}{2 \pi m}}$$
My issue here is that I do not know how to solve this 6 dimensional integral.
I am asked for the average energy < E >
$$\langle E \rangle = -\frac{ \partial \log(Z_1)}{\partial \beta}$$
Answer:
$$\langle E \rangle = \frac{5}{2\beta}$$
With $$H = \frac{P^2}{2m} + \frac{P_{\theta}^{2}}{2 m r} + \frac{P_{\phi}^{2}}{2 m r \sin^2\theta}$$ then $$I = \int e^{- \beta H} \, dP \, dP_{\theta} \, dP_{\phi} \, dr \, d\theta \, d\phi$$ becomes: \begin{align} I &= \int d^{3}q \, \int e^{-\beta P^{2}/2m} \, dP \cdot \int e^{- \beta P_{\theta}^{2}/(2 m r)} \, dP_{\theta} \, \cdot \int e^{- \beta P_{\phi}^{2}/(2 m r \sin \theta)} \, dP_{\phi} \\ &= \int d^{3}q \, \left(\frac{2 m \pi}{\beta}\right)^{1/2} \, \left(\frac{2 m \pi r}{\beta}\right)^{1/2} \, \left(\frac{2 m \pi r \sin^2\theta}{\beta}\right)^{1/2} \\ &= \left( \frac{2 \pi m}{\beta}\right)^{3/2} \, \int r^2 \sin\theta \, dr \, d\theta \, d\phi \\ &= 4 \pi \left( \frac{2 \pi m}{\beta}\right)^{3/2}. \end{align}
This is how the integral is computed. The remainder is to add in the various constants and compute the average energy.