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Text book says:

The area form $dS$ of a surface $S\subseteq \mathbb R ^3$ is defined as for any positively oriented orthonormal frame $\{E_1, E_2\}$, $dS(E_1,E_2)=1$.

Then given parametrization $ x:D\rightarrow M$ where $M$ is a surface in $\mathbb R^3$, what is area form on $M$ in general?

In particular, given parametrization of unit sphere centered at the origin $x:(u,v)\mapsto(\cos u\cos v,\cos u\sin v, \sin u)$, the area form is $$\int \int _D ||x_u\times x_v||dudv=\int\int_D(\cos^4u+4\sin^2u\cos^2u)dudv$$

Then how about torus?

Given parametrization $x:[0,2\pi]\times[0,2\pi]\rightarrow \mathbb R^3$ as $$(u,v)\mapsto((R+r\cos u)\cos v,(R+r\cos u)\sin v,r\sin u)$$ where $r<R$.

Or any other surface? If problem ask me given parametrization $x$, write area form in terms of basis $du$ and $dv$ where $du$ and $dv$ are the dual basis of $\{x_u,x_v\}$, then is it okay just to write $$\int\int _D ||x_u\times x_v||dudv$$?

Darae-Uri
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    Your formula for the sphere is not correct. – Ted Shifrin Jun 17 '16 at 16:34
  • @TedShifrin Oh, seems to be a mistake. Is it correct if I replace it to $$\sqrt {\cos^2u(\cos^2u+1)}$$? – Darae-Uri Jun 17 '16 at 17:11
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    Nope, still not right. – Ted Shifrin Jun 17 '16 at 23:30
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    Generally, an area form for a surface is a gadget that, when fed an ordered pair of tangent vectors to a surface, returns the oriented area of the parallelogram spanned by those vectors. If $(u, v)$ denotes Cartesian coordinates, for example, the area form in the Euclidean plane is $du, dv$ (or $du \wedge dv$, depending on your notation). If $U = (U_{1}, U_{2})$ and $V = (V_{1}, V_{2})$ are vectors, then$$(du, dv)(U, V) = du(U) dv(V) - du(V) dv(U) = U_{1}V_{2} - U_{2} V_{1}.$$ – Andrew D. Hwang Jun 18 '16 at 00:58

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