Surface integral in spherical coordinates

Let us consider a surface integral


where $ \Sigma $ is a surface which have a parameterization described in terms of angles $ \theta $ and $ \phi $ in spherical coordinates.
$ x=r \cos \theta sin \phi $
$ y=r \sin \theta \sin \phi  $
$ z = r \cos \phi $
and let
$ \chi(r,\phi,\theta)=(r \cos \theta sin \phi, r \sin \theta \sin \phi,r \cos \phi) $
We are interested in a formula for evaluating a surface integral where r is a function of angular variables
$ r=\rho(\theta,\phi) $
We have $ \frac{\partial \chi}{\partial r}(r,\phi,\theta)=( \cos \theta sin \phi,  \sin \theta \sin \phi, \cos \phi)=e_r $
Let consider $ \psi(\phi,\theta)}=(\rho(\theta,\phi)\cos \theta sin \phi, \rho(\theta,\phi) \sin \theta \sin \phi,\rho(\theta,\phi) \cos \phi)=\rho(\theta,\phi)e_r $
We want to find the expression of

$$d\sigma=||\frac{\partial \psi}{\partial \theta}\times\frac{\partial \psi}{\partial \phi}||d\theta d\phi=\rho\sqrt{(\rho^2+\frac{\partial \rho}{\partial \theta}^2)\sin^2\theta+\frac{\partial \rho}{\partial \phi}^2}d\theta d\phi$$

We have

$$\frac{\partial \psi}{\partial \theta}=\frac{\partial \rho}{\partial\theta}e_r+\rho\frac{\partial e_r}{\partial\theta}=\frac{\partial \rho}{\partial\theta}e_r+\rho e_{\theta}=4\pi ^2a^2$$
$$\frac{\partial \psi}{\partial \phi}=\frac{\partial \rho}{\partial \phi}e_r+\rho\frac{\partial e_r}{\partial\phi}=\frac{\partial \rho}{\partial \phi}e_r+\rho \sin\theta e_{\phi}$$
$$\frac{\partial \psi}{\partial \theta}\times\frac{\partial \psi}{\partial \phi}=<br />
(\frac{\partial \rho}{\partial \theta}e_r+\rho e_{\theta})\times(\frac{\partial \rho}{\partial \phi}e_r+\rho\sin\theta e_{\phi})<br />
=\rho\frac{\partial \rho}{\partial \phi}e_{\phi}\times e_r+\rho^2 \sin\theta e_{\theta}\times e_{\phi}+\rho\sin\theta \frac{\partial \rho}{\partial \theta}e_r\times e_{\phi}$$

We use orthogonal coordinatesAs $ e_r,e_{\theta},e_{\phi} $ are orthonormal vectors, so are $ e_r\times e_{\theta}, e_{\theta}\times e_{\phi},e_{\phi}\times e_r $ so the norm of $ \frac{\partial \psi}{\partial \theta}\times\frac{\partial \psi}{\partial \phi} $ is the square root of sum of squares of its components.
Finally we have

$$=\int_DF(\rho(\theta,\phi)\cos \theta sin \phi, \rho(\theta,\phi) \sin \theta \sin \phi,\rho(\theta,\phi) \cos \phi)\rho\sqrt{(\rho^2+\frac{\partial \rho}{\partial \theta}^2)\sin^2\theta+\frac{\partial \rho}{\partial \phi}^2}d\theta d\phi$$

We now evaluate the surface of sphere $ x^2+y^2+z^2=a^2 $
We have in this situation $ r^2=a^2 $ so $ r=\rho(\theta,\phi)=a $ so $ d\sigma=a^2\sin\theta d\theta d\phi $
We have

$$\int_{\Sigma}d\sigma=\int\int_Da^2\sin\theta d\theta d\phi=\int_0^{2\pi}\int_0^{\pi}a^2\sin\theta d\theta d\phi=4\pi a^2$$

Now let us see another example.
Consider the torus of equation


We have


and we get



$$d\sigma=\rho\sqrt{(\rho^2+\frac{\partial \rho}{\partial \theta}^2)\sin^2\theta+\frac{\partial \rho}{\partial \phi}^2}d\theta d\phi=4a^2\sin^2\theta d\theta d\phi$$


$$Area_{torus}=\int_0^{2\pi}\int_{0}^{\pi}4a^2\sin^2\theta d \theta d\phi=4a^2*2\pi*2<br />

This result is consistent with area of surface of rotation equals to the product of length of the curve $ 2\pi a $ by the length $ 2\pi a  $ of the mass center curve

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