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Thread: surface of rev.

  1. #1
    Oct 2010

    surface of rev.


    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?

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  2. #2
    Senior Member
    Mar 2010
    Beijing, China
    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).
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