TITLE: Prove that sphere is the only surface which can be generated by rotation in more than one way
QUESTION [7 upvotes]: In Hilbert's book Geometry and the imagination, he said that 

sphere is the only surface which can be generated by rotation in more
  than one way.

It is quite intuitive, but I can't give a rigorous proof.
How to prove it?
PS: Here rotation means rotating a closed curve with respect to the axis of symmetry of it that is in the same plane.

REPLY [5 votes]: Let $S$ be a nonempty subset of $\mathbb R^3$ that can be generated by rotating a closed curve $C$ around $a$ and is also invariant under rotation around $b$.
Then $S$ is compact and connected because $C$ is compact and connected.
If $a$ and $b$ do not intersect, let $c$ be a line intersecting both $a$ and $b$ perpendicuularly.
Then the rotation by $\pi$ around $a$ followed by rotation by $\pi$ about $b$ leave $c$ fixed but translate it by twice the distance between $a$ and $b$. Thus this is the same as a screw operation along $c$ and causes $S$ to be unbounded, contradicting compactness. 
If $a$ and $b$ intersect (wlog. in the origin), their rotations generate all of $SO(3)$, as joriki says. Therefore a single point $x\in S$ has as orbit a sphere aroound the origin (or consists of $x$ alone if $x=0$). We conclude that $S$ is the union of concentric spheres.
Since $S$ is compact and connected, this leaves only the possibilities 
$$\tag1 S=\{0\} $$
$$\tag2S=\{x\in\mathbb R^3 \colon |x|=r\}\text{ for some }r>0$$
$$\tag3S=\{x\in\mathbb R^3 \colon |x|\le r\}\text{ for some }r>0$$
$$\tag4S=\{x\in\mathbb R^3 \colon r_1\le |x|\le r_2\}\text{ for some }0<r_1<r_2.$$
The only case that actually leads to a surface is indeed $(2)$. The other cases may be obtained with suitable exotic curves.