On surfaces of revolution

There are two ways to define a surface of revolution in Euclidean space ${\mathbb R}^3$.

1. A surface $S$ is a surface of revolution with respect to the line $L$ is $\phi(S)=S$ for any rotation of axis $L$ (type I)

2. A surface of revolution is a surface constructed as follows. Fix $L$ a straight-line and let $P$ be a plane containing $L$. Consider a curve $C$ contained in $P$. Then the surface of revolution generated by $C$ is the set of points obtained when we rotate $C$ about the axis $L$ (type II). We denote this surface as $S(C)$

It is clear that any surface of type II is a surface of type I by the definition given in 2. It is less clear if a surface of type I is of type II, that is, if $S$ satisfies I, is there exists a curve $C$ in a plane $P$ containing $L$ such that $S=S(C)$? After a rigid motion of ${\mathbb R}^3$, we suppose that $L$ is the $z$-axis. 

Take $P$ a plane containing $L$. Then at any point   $p\in S\cap P$, the surfaces $S$ and $P$ are transversal, that is, $T_pS\not=T_pP$. Indeed, if $p=(x,y,z)\in S$ with $x^2+y^2\not=0$ (we are assuming that $S$ does not intersect the axis $L$), the rotation about the $z$ axis is the curve $$\alpha(\theta)=  \left(\cos\theta x-\sin\theta y,\sin\theta x+\cos\theta y,z \right).$$ Since $\alpha(0)=p$, then $\alpha'(0)\in T_pS$, that is, $(-y,x,0)$. The plane $P$ containing $p$ and the $z$ axis is the plane orthogonal to $(-y,x,0)$ through $p$. This proves $T_pS\not=T_pP$.

As a consequence $S\cap P$ defines a regular curve $C_p$ around $p$. Since $C_p\subset S$, then $\phi_\theta(C_p)\subset S$ for any rotation $\phi_\theta$ about the $z$-axis. In particular, the surface of revolution $S(C_p)\subset S$. 

Therefore we have prove that if $C=S\cap P$ is the intersection curve (with possible many components), then $S(C)\subset S$.

For the other inclusion, if $(x,y,z)\in S$, it is immediate that the circle $\alpha(\theta)$ defined previously intersects $P$. For example, if $P$ is the $xz$-plane, we are asking if there exists $\theta$ such that   $$\sin\theta x+\cos\theta y=0.$$ It suffices by taking $\theta$ such that $\tan\theta=-y/x$ if $x\not=0$ and $\theta=\pi/2$ if $x=0$ (it is not possible $x=y=0$). If $q=\alpha(\theta)$, then it is immediate that a suitable rotation of $q$ (exactly that rotation with angle $-\theta$) gives $p$.