# Volume element of the Sphere

If we consider the sphere on $E^3$ with Riemannian metric $G=dx + dy + dz$ then transforming to spherical coordinates we get $G=R^2 d{\theta} +R^2 sin(\theta) d{\phi}^2$. Hence the volume form is $R^2sin(\theta)d{\theta}∧d{\phi}$.

Now, I need to find new coordinates p and q, such that the volume form is just $dp ∧ dq$.

I have tried using;

$x={\alpha}Rsin({\beta}{\theta})cos({\gamma}{\phi})$, $y={\alpha}Rsin({\beta}{\theta})sin({\gamma}{\phi})$, $z={\alpha}Rcos({\beta}{\theta})$,

and then plugging this into $G=dx + dy + dz$ to get $G_s = {\alpha}^2 {\beta}^2 R^2 d{\theta}^2 + {\alpha}^2 {\gamma}^2 R^2 sin^2({\beta}{\theta})$ we then require this essentially to be equal to $d{\theta}^2 + d{\phi}^2$. However I can't get this to work, so I don't think this is the right approach. But i'm not sure how else to do it, i've considered stereo-graphical coordinates also..

It seems like a straightforward problem but i'm not having much luck, any guidance would be much appreciated.

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The volume form on $S^2$ cannot be written as the wedge product of two $1$-forms, because every $1$-form on $S^2$ vanishes somewhere by the Hairy Ball theorem.