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?