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?
 A: 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 
$$
A = 4\pi^2 ab
$$
A: 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.
A: 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!
A: 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 
$$
S= ab \int_0^{2\pi}du \int_0^{2\pi} dv = 4\pi^2 ab
$$
A: $$\frac{\partial\vec{g}}{\partial u}=(-a\sin u\cos v, -a\sin u\sin v, a \cos u),$$
$$\frac{\partial\vec{g}}{\partial v}=(-(b+a\cos u)\sin v, (b+a\cos u)\cos v, 0).$$
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.
