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?

  • $\begingroup$ It is certainly true for a piecewise linear centerline, if at each bend we just extrapolate the cylinders of each meeting piece linearly until the symmetry plane -- the amount of surface that is lost on the inside of the bend corresponds exactly to the extra surface on the outside. A smooth flexible tube can be approximated arbitrarily well by piecewise linear ones, and there seems to be no good reason why the equivalence wouldn't continue holding in the limit. (Assuming that the smooth centerline doesn't bend with a radius of curvature less than the radius of the tube). $\endgroup$ – hmakholm left over Monica Nov 16 '11 at 20:51
  • $\begingroup$ @Henning I think I see, at least in the case of constant $r$... each cylindrical extension is cut in half diagonally by the adjacent segment. What if $r$ varies? I played around with the idea of approximating the tube as piecewise tori, but couldn't convince myself the surface area of such approximations converged to that of the tube... but maybe using cylinders (or truncated cones?) as you suggest is easier. $\endgroup$ – user7530 Nov 16 '11 at 20:58
  • 3
    $\begingroup$ We can't use cylinders for a variable radius; it would have to be infinitesimal truncated cones. And then it's not immediately obvious to me that if we put two truncated cones (even with the same aperture) together at an angle, they would intersect in a planar curve, much less one where the area surplus and deficit would cancel out. $\endgroup$ – hmakholm left over Monica Nov 16 '11 at 21:05

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.

  • $\begingroup$ This approach will give the right answer, of course, but carrying out this integration doesn't seem particularly pleasant. I'm hoping there's a simpler solution, such as TonyK's idea below of transforming the tube into a surface of revolution. $\endgroup$ – user7530 Nov 19 '11 at 10:13

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.

  • $\begingroup$ Interesting! It's not obvious to me that as you increase the number of slices, the error in surface area due to "ripples" along the surface (whose frequency increases as the number of slices increases) vanishes, but it's certainly plausible that the details work out. $\endgroup$ – user7530 Nov 17 '11 at 15:59

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