The fact that surface of spherical segment only depends on its height follows from symplectic geometry It is quite quite well known that the surface of the piece of a sphere with $z_0<z<z_1$ for some values of $z_0,z_1$ is given by 
$ S = 2\pi R (z_1-z_0) $. So this surface area only depends on the height of the spherical segment. 
My teacher mentioned that this fact follows from some general theorem in symplectic geometry. He mentioned this right after giving the action of $U(1)$ on $S^2$ by rotation as an example of a hamiltonian Lie group action. So I guess that it should follow somehow from this action. (Whose moment map is essentially the z-coordinate of the sphere.)
What could be the general theorem my teacher was talking about?
 A: Consider a hamiltonian $S^1$-action on a symplectic manifold $(M, \omega)$. Denote by $\mu : M \to \mathbb{R}$ the associated moment map. Since $Lie(S^1) \cong T_0S^1 \cong \mathbb{R}$ is generated by $\frac{\partial}{\partial \theta}$, the moment map is determined by the hamiltonian function $H : M \to \mathbb{R} : m \mapsto \mu(m)(\frac{\partial}{\partial \theta})$.
Given any 'invariant cylinder' $C' \subset M$, we show that its $\omega$-area is completely determined by the values of $\mu$ on $\partial C'$.
Indeed, consider the cylinder $C = S^1 \times [0,1]$. It has a distinguished area-form, that is $d\theta \wedge dt$. Consider a map $\phi : C \to M$ such that for each $t \in [0,1]$, the map $\phi_t = \phi(-,t) : S^1 \to M$ is an orbit of the $S^1$-action on $M$. In particular, the 'velocity vector' $X = (\phi_t)_{\ast} \frac{\partial}{\partial \theta}$ coincides with the hamiltonian vector field $X_H$ implicitly given by $dH = X_H \lrcorner \,  \omega$. Notice that $H$ is constant on any circle $\phi_t(S^1)$ and that, moreover, the quantity $dH \left( \phi_{\ast}\frac{\partial}{\partial t} \right)$ is independent of $\theta$.
We compute
$$ \begin{align}
\int_C \phi^{\ast}\omega &= \int_C \omega \left( \phi_{\ast}\frac{\partial}{\partial \theta}, \phi_{\ast}\frac{\partial}{\partial t} \right) \, d\theta \wedge dt = \int_C dH \left( \phi_{\ast}\frac{\partial}{\partial t} \right) \, d\theta \wedge dt \\ 
&= 2\pi \int_{[0,1]} dH \left( \phi_{\ast}\frac{\partial}{\partial t} \right) \, dt = 2\pi \int_{[0,1]} \phi^{\ast}dH \\
&=  2\pi \int_{[0,1]} d(\phi^{\ast}H) = 2\pi \int_{\{0,1\}} \phi^{\ast}H \\
&= 2\pi (\left. H \right|_{\phi_1(S^1)} -   \left. H \right|_{\phi_0(S^1)}) \, .
\end{align} $$
A: It follows from the Duistermaat-Heckman theorem as described here 
