TITLE: Surface area of a torus
QUESTION [15 upvotes]: One can generate a torus as follows: $\vec{g}=((b+a\cos u)\cos v, (b+a\cos u)\sin v, a \sin u)$. To find its area, we can use a surface integral of the form $S=\iint_{D_{uv}} {\lVert \frac{∂g}{∂u} \times \frac{∂g}{∂v} \rVert \, du \, dv}$. However, in the case of torus the integral becomes seemingly unnecessarily tedious to evaluate. Are there nicer approaches to evaluating the surface area using a surface integral?

REPLY [3 votes]: In fact this problem necessitates only a simple integral, you just have to integrate (b+cos(s))2pia ds from 0 to 2pi. It is a matter of adding an infinity of circles of infinitesimal width ads. The radius of those circles is in average b, but it can be as much as b+a (outer ridge of the torus) and as little as b-a (inner ridge of the torus). For any s, the radius of the circle will be b+cos s, if we choose s so that s=0 corresponds to the position where the radius is maximal. I was very troubled at first to realize that the cosine term becomes zero when we integrate, so that the formula boils down to the absurdly simple 2pia times 2pi*b, big perimeter times small perimeter. We understand why it is so simple if we consider any horizontal slice of the torus. It will be made of two circles, of radius b+cos(s) and b-cos(s), where s is the angle corresponding to one of the two circles (any will do, just keep the same s for both circles). The cosine terms will always cancel! Magic!