Surface area of a flexible tube

Question

Consider the hollow tube formed by sweeping a circle of radius $r(t)$ along a curve $\gamma(t)$ in $\mathbb{R}^3$; in other words, the set of points

$$S=\{\gamma(t) + r(t) \hat{n}\quad \vert\quad \|\hat n\| = 1, \hat{n}\cdot \gamma'(t) = 0, t \in [a,b]\}.$$

What is the surface area of this tube?

There are some easy special cases: when $\gamma$ is a straight line, $S$ is a surface of revolution. When $r$ is constant and $\gamma$ is a circle, $S$ is a torus -- and, surprisingly, the surface area of this torus is the same as if we "straightened" the centerline of the tube, turning the torus into a cylinder! Does a similar result hold generally?

Answer 1

In the first place you need a parametric representation of $S$. Note that nobody would guess from the equation $x^2+y^2=1$ that this curve has length $2\pi$. So your equation $n\cdot \gamma'(t)=0$ is of no help.

Given the "soul" $t\mapsto \gamma(t)$ of $S$ and assuming that the curvature $\kappa$ of $\gamma$ is nonzero there is a well-defined orthonormal Frenet frame $\bigl(e_1(t),e_2(t),e_3(t)\bigr)$ along $\gamma$ with $e_1=\gamma'$ (assuming $t$ is arc length), $e_2$ in the osculating plane, and $e_3:=e_1\times e_2$.

In terms of this frame a parametric representation of $S$ is given by

$$(t,\phi)\ \mapsto\ {\bf r}(t,\phi):= \gamma(t)+r(t)\cos\phi\ e_2(t) +r(t)\sin\phi\ e_3(t)\qquad(a\leq t\leq b,\ 0\leq\phi\leq 2\pi) .$$

The area $\omega(S)$ of this surface is then obtained as follows:

$$\omega(S)\ =\ \int_a^b\int_0^{2\pi}\ |{\rm r}_t\times {\rm r}_\phi|\ d\phi\ dt\ .$$

Because of the so-called Frenet equations for the $e_i'$ it turns out that we can compute the integrand $|{\rm r}_t\times {\rm r}_\phi|$ without actually computing the $e_i$. The result is

$$|{\rm r}_t\times {\rm r}_\phi|^2=r^2\bigl(r'^2 +(\kappa r\cos\phi-1)^2\bigr)\ .$$

Unless $r'(t)\equiv0$ (a tube of constant radius) or $\kappa(t)\equiv0$ (the "soul" is a straight line) the "inner" integration $\int_0^{2\pi}\ldots\ d\phi$ will result in a complete elliptic integral, which cannot be done in elementary terms.

Answer 2

I think my answer here to the torus problem can be applied to this problem too.

Given a short stretch of curve, of length $\delta$ say, consider the section of tube generated by that stretch. It will be approximately a slice of a torus. But if we take the whole of the curve after the midpoint of the section, and rotate it by $180^\circ$ about the tangent at that point, then it becomes approximately a cylindrical slice, with surface area $2\pi r \delta$.

You can partition the whole of the curve into stretches of length $\delta$, and perform this straightening operation successively from the first stretch to the last stretch. You end up with a nearly straight curve, whose enclosing tube is nearly a cylinder. In the limit as $\delta \rightarrow 0$, the curve becomes straight, and the tube becomes a cylinder.

Edited in response to user's comment: The statement of the problem presupposes that $\gamma(t)$ is differentiable. Also, we need to assume that its curvature is $\le r$, otherwise the tube folds back in on itself, and we get negative surface areas. So the ripples can't be too ripply.