Given the unit sphere in $\mathbb{R}^3$, we want to pick points uniformly across its surface. My first thought was to parameterize by spherical coordinates $$ (\theta,\phi) \mapsto (\cos\theta\sin \phi, \sin \theta \sin\phi, \cos\phi) $$
and choose $\theta \sim U[0,2\pi), \phi\sim U[0,\pi]$, from which a random point on the sphere can be selected. But according to this article, the area element of the spherical parameterization is given by $\sin\phi \ d\theta d\phi$. Since this quantity depends on $\phi$ itself, the resulting distribution over the sphere will not be uniform. This I can follow.
The site then suggests a change of coordinates given by $$ (\theta, \phi) = \big(2\pi u, \cos^{-1}(2v-1)\big), \quad \text{where }u,v\sim U(0,1) $$
As far as I understand, the area element using the new $u,v$ coordinate system should be constant (i.e. should not depend on $u,v$). Now the Jacobian of the above transformation is given by $$ J = \begin{bmatrix} 2\pi & 0\\ 0 & \frac{2}{\sqrt{1-(2v-1)^2}} \end{bmatrix} = \begin{bmatrix} 2\pi & 0\\ 0 & \frac{1}{\sqrt{(1-v)v}} \end{bmatrix} $$ so $d\theta d\phi = \det(J) \ dudv$. Finally, we can calculate the new area element $$ \sin\phi \ d\theta d\phi = \sin\cos^{-1}(2v-1)\det(J)\ dudv = 2\pi \sin\cos^{-1}(2v-1)\cdot \frac{1}{\sqrt{(1-v)v}} \ dudv $$
This is clearly dependent on $v$, hence is not constant. Where have I gone wrong? I've spent much time on this but with no progress.
I know that there are very similar questions here and here, but as far as I can tell none of the answers illuminate my question.
Well it turns out that nothing I did above is wrong, it's just that I can simplify the above expression even more. Recall the fact that (drawing a picture of the positive half of unit circle really helps here)
$$ \sin \arccos x = \sqrt{1-x^2} $$
From this it follows that
$$ \sin \phi \ d\theta d\phi = 2\pi \sqrt{1-(2v-1)^2}\cdot \frac{1}{\sqrt(1-v)v} \ dudv= 2\pi \cdot 2 \ dudv= 4\pi \ dudv $$
This expression is constant, so the suggested change of variables does in fact work.