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If I am given a surface in spherical coordinates $(r,\theta,\varphi)$, such that it is parametrised as:

$$ \begin{align} r&=r(\theta,\varphi)\\ \theta&=\theta\\ \varphi&=\varphi \end{align} $$

What is the area $S$ of such surface?

If I use the standard surface integral: $$S=\int_{S\subset\mathcal{R}^3}{\rm d}S=\int_{0}^{2\pi}\int_{0}^{\pi}\sqrt{1+\left(\frac{\partial r}{\partial \theta}\right)^2 + \left(\frac{\partial r}{\partial \varphi}\right)^2}\;{\rm d}\theta\,{\rm d}\varphi$$

But this seems to be incorrect, instead, I found it is supposed to be: $$S=\int_{0}^{2\pi}\int_{0}^{\pi}\sqrt{r^2+\left(\frac{\partial r}{\partial \theta}\right)^2 + \frac{1}{\sin^2\theta}\left(\frac{\partial r}{\partial \varphi}\right)^2}\;r\sin\theta\;{\rm d}\theta\,{\rm d}\varphi$$

which, if one takes out $r^2$ from the square root, gives the jacobian of spherical coordinate transformation and under the square root one seems to be left with the square of gradient of $r$ in spherical coordinates.

It does not make really sense to me because the standard approach to surface integral that i know (the previous equation) already seems to contain all the necessary transformations in the determinant of the Gram matrix. I have never come accross something like this, whenever i parametrised the surface using curvilinear coordindates, the Gram matrix approach was the one

Can someone explain why the first approach does not work in this case and why one needs to use the second way?

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The standard surface integral, or the first approach, is for the cartesian coordinates, i.e.,

$$S=\int_{S\subset\mathcal{R}^3}{\rm d}S=\int\int \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2}\;{\rm d}x\,{\rm d}y$$

It can not be simply recast into one with spherical coordinates.

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  • $\begingroup$ If I have a surface $z=\sqrt{x^2+y^2}$ bounded by the interior of a cylinder $x^2+y^2=2x$ I can parametrise the surface $x = 1+r\cos\varphi,\,y=r\sin\varphi,\,z=\sqrt{r^2+2r\cos\varphi+1}$, I get ${\rm d} S=\sqrt{2}r{\rm d}r\,{\rm d}\varphi$ using the first approach quite easily, no additional jacobian or Lame's parameters.. $\endgroup$ – leosenko Sep 5 '19 at 16:27
  • $\begingroup$ I would think so. You should obtain $\sqrt{2}\pi$ as a result. $\endgroup$ – Quanto Sep 5 '19 at 16:43
  • $\begingroup$ yes, but why can't i use this approach for the case in my question? In both problems, you go into curvilinear coordinates. $\endgroup$ – leosenko Sep 5 '19 at 16:44
  • $\begingroup$ There may be confusion. My point is the standard formula works only for Cartesian, not spherical coordinates. $\endgroup$ – Quanto Sep 5 '19 at 16:49
  • $\begingroup$ Mmm, I am well confused. With the determinant of Gram matrix you end up integrating in curvilinear coordinates anyway, at least I think, which happens exactly in the example i gave in the first comment, that is not Cartesian coordinate system... $\endgroup$ – leosenko Sep 5 '19 at 16:53

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