# Surface area of a torus

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?

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

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

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

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

• 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. – sequence Jun 7 '15 at 6:38
• 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. ${}\qquad{}$ – Michael Hardy Sep 15 '15 at 19:35
• 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. – Rogelio Molina Sep 16 '15 at 1:25

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

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

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.

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 =

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