I am trying to use differential forms to determine the surface area element for a sphere. For a sphere of radius $r=1$. I think I am loosing something in the algebra (tried to check symbolic calculations on computer, still don't know how to proceed)
In terms of Cartesian coordinates the surface of the sphere is: $x^2+y^2+z^2=1$. The spherical coordinates relate to Cartesian coordinates in the standard way:
$$ \begin{align} x=&\sin\theta\cos\phi \\ y=&\sin\theta\sin\phi \\ z=&\cos\theta \\ \end{align} $$
The area element in Cartesian coordinates is: $$ d^2S = dx \wedge dy - dx \wedge dz + dy \wedge dz $$
Computing the equivalents in spherical coordinates (not quite equivalent since radius is fixed):
$$ \begin{align} dx=& \cos\theta\cos\phi\,d\theta - \sin\theta\sin\phi\,d\phi \\ dy=& \cos\theta\sin\phi\,d\theta + \sin\theta\cos\phi\,d\phi \\ dz=& -\sin\theta\,d\theta \\ \end{align} $$
Therefore:
$$ \begin{align} d^2 S =\quad&\left(\cos\theta\sin\theta\cos^2\phi+\cos\theta\sin\theta\sin^2\phi\right)d\theta\wedge d\phi-\\ -&\left(-\sin^2\theta\sin\phi\right)d\theta\wedge d\phi+ \\ +&\left(\sin^2\theta\cos\phi\right)d\theta\wedge d\phi \\ \\ d^2 S =&\sin\theta\cdot\left(\cos\theta + \sin\theta\cdot\left(\cos\phi+\sin\phi\right)\right)d\theta\wedge d\phi \end{align} $$
I know the correct result should be $d^2S=\sin\theta \, d\theta\wedge d\phi$, and that it certainly should not depend on $\phi$. But I can't quite see where I went wrong. I suppose I am looking at using a push-forward from $\theta\phi$ space to the surface of a 3D sphere, and then I am trying to pull-back the area element from the 3d space, but this statement will still lead to the same calculations.