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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?

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Another way to proceed would be by writing out the surface integral using differential forms. To this end we need to set up a chart and a coordinate system on the Torus. Fortunately we need only one chart and the coordinate system can be made global (using the coordinates already introduced in the question $u,v$), the surface integral will be

$$ S=\int_{M} ab~ du \wedge dv $$ The integration runs over the coordinate range of the open cover $M=T^2$, that is $u \in [0, 2\pi], ~v \in [0, 2\pi]$ leading to

$$ S= ab \int_0^{2\pi}du \int_0^{2\pi} dv = 4\pi^2 ab $$

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$$\frac{\partial\vec{g}}{\partial u}=(-a\sin u\cos v, -a\sin u\sin v, a \cos u),$$

$$\frac{\partial\vec{g}}{\partial v}=(-(b+a\cos u)\sin v, (b+a\cos u)\cos v, 0).$$

Without surprise, these two vectors are orthogonal, and the norm of their cross product is the product of their norms, $a(b+a\cos u)$.

Integrating on $u,v$ both in range $[0,2\pi]$, noticing that the average value of the cosine is zero, will yield $$(2\pi)^2ab.$$ Nothing really tedious.

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Use Pappus theorem. If the radius of the transversal section of the torus is $r$ then its perimeter is $2\pi r$ and Pappus theorem states that the surface of the torus (it is a revolution surface) equals $A=2\pi r \cdot 2 \pi R$ where $R$ is the radius of rotation that generates the torus. In your case this is

$$ A = 4\pi^2 ab $$

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  • $\begingroup$ Thanks, this is very useful. But I'm wondering if there's a better way to find the area using a surface integral. Sorry for not being precise. $\endgroup$
    – sequence
    Jun 7 '15 at 6:38
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    $\begingroup$ But how to prove Pappus's theorem would seem closer to the meaning of the original question. There are of course ways of doing it other than the methods that we now know as integral calculus; otherwise Pappus couldn't have done it. ${}\qquad{}$ $\endgroup$ Sep 15 '15 at 19:35
  • $\begingroup$ Indeed, actually using Pappus theorem was my first answer, but the OP asked the use of a surface integral. I believe that Pappus did not have integrals at his disposal as we know them but relied on Cavalieri's principle and the use of the indivisibles. $\endgroup$ Sep 16 '15 at 1:25
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Choosing a toroidal coordinate system the computation of area is I believe most easy. However, a hybrid between spherical and cylindrical coordinate system can be adopted.

Let radius at crown be $b$, $R$ tube radius, $r$ variable radius of cyl coordinate system, and $\phi$ the latitude of spherical system. Area of torus =

$$ \int_{b-R}^{b+R} \int_{-\pi}^{ \pi} R\, d\phi \cdot 2 \pi\, dr = 2 \pi R \int_{-\pi} ^{ \pi} ( b- R \cos \phi) d \phi = 2 \pi R \cdot 2 \pi b. $$

This is essentially surface area generation by rotation using Pappu's theorem.

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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!

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