Surface integral in spherical coordinates

Submitted by Structure on Thu, 01/29/2009 - 20:17.

Let us consider a surface integral

$\int_{\Sigma}F(x,y,z)d\sigma$

where $\Sigma$ is a surface which have a parameterization described in terms of angles $\theta$ and $\phi$ in spherical coordinates.
Let
$x=r \cos \theta sin \phi$
$y=r \sin \theta \sin \phi$
$z = r \cos \phi$
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

$\int_{\Sigma}F(x,y,z)d\sigma=$

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

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

$(x^2+y^2+z^2)^2=4a^2(x^2+y^2)$

We have

$\rho^4=4a^2\rho^2\sin^2\theta$

and we get

$\rho(\theta,\phi)=2a\sin\theta$

Then

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

$Area_{torus}=\int_0^{2\pi}\int_{0}^{\pi}4a^2\sin^2\theta d \theta d\phi=4a^2*2\pi*2<br /> \frac{\pi}{2}\frac{1}{2}=4\pi^2a^2$

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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Surface integral in spherical coordinates