= Solution
By reflection symmetry, $G$ is the midpoint of the <cylinder> axis. With position $r$ measured from $G$, define $J^G_{ij}=\int(|r|^2\delta_{ij}-r_ir_j)\,dm$. Symmetry makes the off-diagonal entries zero. Uniform <probability density function>[density] gives $\langle x^2\rangle=\langle y^2\rangle=a^2/4$ and $\langle z^2\rangle=b^2/3$, by integrating over a disk and the <interval> $[-b,b]$. Hence
$$
\boxed{J^G=M\operatorname{diag}(a^2/4+b^2/3,\ a^2/4+b^2/3,\ a^2/2).}
$$
For parallel axes based at $P$, the tensor form of the parallel-axis theorem is $J^P=J^G+M(|d|^2I-dd^T)$, where $d=\overrightarrow{PG}$. This follows by substituting $r+d$ in the defining integral and using $\int r\,dm=0$. Here $d$ is an axial vector of length $b$, so
$$
\boxed{A=M(a^2/4+4b^2/3),\quad C=Ma^2/2,\qquad J^P=\operatorname{diag}(A,A,C).}
$$
The <Euler angles> $\psi,\phi$ are <cyclic coordinates>. Their conserved conjugate momenta are $p_\psi=C(\dot\psi+\dot\phi\cos\theta)=Cn$ and $p_\phi=A\dot\phi\sin^2\theta+Cn\cos\theta=h$. The <Lagrangian> has no explicit time dependence, so the conserved energy is kinetic plus gravitational potential, namely the expression given with constant $E$.
Eliminate the angular velocities using $\dot\phi=(h-Cn\cos\theta)/(A\sin^2\theta)$ and $\dot\psi+\dot\phi\cos\theta=n$. The energy equation becomes
$$
A\dot\theta^2=2E-Cn^2+2Mgb\cos\theta-\frac{(h-Cn\cos\theta)^2}{A\sin^2\theta}.
$$
Thus
$$
\boxed{F(\theta)=A\sin^2\theta(2E-Cn^2+2Mgb\cos\theta)-(h-Cn\cos\theta)^2.}
$$
The elimination is valid away from the coordinate poles; the multiplied equation extends by <continuity> to any pole reached by a regular motion.
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