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I'm trying to find out the Area of the following surface:

Let $C$ be the curve associated to a regular, simple path $\theta:[0,l]\rightarrow \Bbb R^2 $; also assume that $((x'(s))^2+((y'(s))^2=b^2$ and let $S$ be the surface generated by the circles of radius $b$, orthogonal to, and centered in points of the curve $\rho(s)=(\theta(s),0) $.

With help from this source. I concluded that a convenient parametrization for $S$ is given by:

$$ H(s, t) = ( x(s), y(s), 0) + \sin(t) (0, 0, b) + \cos(t) (b\ y'(s), -b\ x'(s), 0). $$

I intended to use the fact that the area of $S$ is given by the surface integral:

$$ \int_S ||T_s \times T_t|| \ dv $$

However,by this approach, the terms of $||T_s \times T_t||$ become really awful. Am I doing something wrong here? What would you recommend?

I found this approach, wich uses the Divergence Theorem. However in this case I´m not sure I can use that since I don´t have a vector field.

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Weierstraß Ramirez
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    I asked a similar question a long time ago (http://math.stackexchange.com/questions/82804/surface-area-of-a-flexible-tube) but never got a full answer. Still interested in the solution! – user7530 Dec 14 '14 at 18:53
  • Lets hope we get something helpful! – Weierstraß Ramirez Dec 14 '14 at 19:01
  • A gifted student asked me this once (actually his question was about a surface created by revolving one curve about another). My excuse for not calculating it was that the "surface" could be really awful. Say, if $\Theta$ is a circle with diameter $b$. Then $S$ will intersect pretty badly given all the points of $\theta$ will also lie on $S$. – Jyrki Lahtonen Dec 14 '14 at 21:54
  • @JyrkiLahtonen I'd be interested in a solution even if it requires assuming that the curvature of $\theta$ is sufficiently large. For instance is it obvious that the answer isn't always just $2\pi bl$? – user7530 Dec 28 '14 at 18:53

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