# Surface integrals in spherical coordinates

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?

$$S=\int_{S\subset\mathcal{R}^3}{\rm d}S=\int\int \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2}\;{\rm d}x\,{\rm d}y$$
• 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.. – leosenko Sep 5 '19 at 16:27
• I would think so. You should obtain $\sqrt{2}\pi$ as a result. – Quanto Sep 5 '19 at 16:43