TITLE: Surface area of a flexible tube
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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.