1. ## surface of rev.

Hello,

I try to solve this problem:

$h(t)=(h_1(t),h_3(t))$ some smooth path in $\mathbb{R}^2$, $t \in (0,1), |h'(t)|=1, h_1(t)>0$ and $g(t,\alpha)=(h_1(t)*cos\alpha, h_1(t)*sin\alpha, h_3(t)), \alpha \in [0,2\pi)$ the corresponding surface of revolution.
$h_3$ is a diffeomorphism onto its image.

Find $U \subset \mathbb{R}^3$ and a function f:U-> $\mathbb{R}$,s.t. 0 is a regular value of f and Im(g)= $f^{-1}(0)$

I have try to define some function f on a set U, s.t. the Points on the surface are mapped to 0. But i couldn't find it.
My results are quite little, we know that the length of the curve h is 1. Therefore we can find a open set U, which contains the surface.

I try it with some sinus and cosinus relations, like $sin^2+cos^2=1$, but it doesn't help. because in equations like f(x,y,z)=x^2+y^2-... we get $c_1^2$

How can i define f reasonable?

Regards

2. Need only to find a function H of R^2 so that $Im(h)=H^{-1}(0)$. One of the ways to define such a function is as follows:

Let n(t) be the unit normal vector field on h(t). For t in [0,1] and small values of s, a parametrization of the plane near Im(h) is given by r(t,s)=h(t)+s*n(t). Since Im(h) is compact, the parametrization is well defined on [0,1]*[-d,d], where d is a small positive value. Define H on R^2 as H(t,s)=s, then $Im(h)=H^{-1}(0)$.

Then rotate the function H to define the function f on R^3 as $f(t,s,\alpha)=H(t,s)$.