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In this question, one of the answers (starting with "Note that the parametrization of sphere is given by...") suggested to compute the surface area by doing the following:

With the parameterization: $$r(u,v)=(r\sin u\cos v,r\sin u\sin v,r\cos u)$$ where $0\leq u\leq \pi$ and $0\leq v\leq2\pi$ the surface area is given by the formula $$\int_0^{2\pi}\int_0^{\pi}||r_u\times r_v|| du\hspace{1mm} dv$$

What theorem makes this formally valid? I'm familiar with one where the mapping $r$ would have to be a homeomorphism from $\mathbb{R}^2$ to the (punctured) unit sphere (in which case this surface area was demonstrated with a completely different function $r$).

I'm thinking that perhaps there's a way to claim this yields a homeomorphism from $\mathbb{R}^2$ (via a set homeomorphic to $\mathbb{R}^2$) to a different portion of the unit sphere (i.e. the unit sphere with some set of Lebesgue measure $0$ removed; Although it seems that the continuity of the inverse would be problematic for $x=0$, where $x$ is the first coordinate)?

Is that the case? Or is there some other theorem that leads to this formula?

Note -- I asked the original poster, only to realize that the answer had been posted over a decade ago...

Anon
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  • you'll need a good definition of surface area to actually prove this; Hausdorff measure? – FShrike Jun 22 '25 at 11:18
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    Restricting to the open rectangle $0 < u < \pi$ and $0 < v < 2\pi$ yields a diffeomorphic parametrisation of the sphere minus the Greenwich meridian. Then Gram it. – Dermot Craddock Jun 22 '25 at 12:27
  • @DermotCraddock Thanks -- that sounds about what I imagined would need to be removed. Do you have a reference or sketch of the proof that it is a diffeomorphism? – Anon Jun 22 '25 at 12:45
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    If we also let $r$ range over $(0,+\infty)$, we get the spherical coordinates in $\mathbb{R}^3$ (minus a closed half-plane). If you already know that is a diffeomorphism, you're pretty much done. Otherwise check that that's a bijection (see wiki page linked), and that the Jacobian determinant is nonzero. – Dermot Craddock Jun 22 '25 at 13:01
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    Before you start asking about theorems, the first thing you need to fix are the definitions. How are you defining the surface area rigorously? Somewhat intuitive explanation are given here and here in old answers of mine. For a precise measure-theoretic definition, see here. In fact, see all the links in the comments here for various relevant theorems concerning integration on manifolds. – peek-a-boo Jun 23 '25 at 23:39

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