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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?

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    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).2011-11-16
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    @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.2011-11-16
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    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.2011-11-16

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