Im studing integration over manifolds. I want to compute the surface area of the torus.
Im given the usual parametrization
$f(u,v)= ((a+b\cos(v))\cos(u), (a+b\cos(v))\sin(u), b\sin(v))$ for $0\leq u,v \leq 2\pi$.
I know that the area can be computed as
$\int_T \omega$ where $T$ denotes the torus and $\omega$ denotes the volume form of the manifold.
I was able to compute the volume form of the 3-dimensional manifold "the solid torus". But I dont get how to compute the volume form of the boundary of this manifold (which is $T$).
I appreciate any idea.
Best Answer
Since you have a representation of the Torus by an embedding $f:\mathbb{R}^2\rightarrow \mathbb{R}^3$ and the volume (area) is something you get from the target space (it is there, where you do the measurement) you need to 'pull back' the metric to the domain of defintion $U$. The resulting formula is, in differential geometry, usually given by $$\int_U \sqrt{|g|} du\wedge dv$$ where $g$ is the determinant of the so called metric tensor $(g_{ij})$. In the two -dimensional case you simply have $$g_{11} = \langle f_u, f_u\rangle, \, g_{12}= g_{21} = \langle f_u, f_v\rangle, \, g_{22} = \langle f_v, f_v\rangle $$
(In general it is
$$g_{ij} = \langle f_{x_i}, f_{x_i}\rangle$$ with $x_i$ the coordinate functions).