= Solution
For an autonomous <Lagrangian>, the <Euler-Lagrange equations> imply
$$
\frac d{dt}\left(\sum_{q=\theta,\phi}\dot q\frac{\partial L}{\partial\dot q}-L\right)=-\frac{\partial L}{\partial t}=0.
$$
Indeed, differentiating the expression cancels the acceleration terms, leaving $\sum_q\dot q(dL_{\dot q}/dt-L_q)-L_t$, which vanishes. Here $T$ is quadratic in the velocities, so $\sum\dot qL_{\dot q}=2T$. Thus the conserved <energy> is
$$
\boxed{H=T+V=E.}
$$
The azimuth $\phi$ is a <cyclic coordinate>. Its conjugate <momentum> is constant:
$$
p_\phi=\frac{\partial L}{\partial\dot\phi}=\frac{Ma^2}{3}\sin^2\theta\,\dot\phi.
$$
Therefore $\boxed{\sin^2\theta\,\dot\phi=h}$ is constant as well. Physically, $p_\phi$ is the vertical component of <angular momentum>.
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