What is the metric tensor on a spherical surface?

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What is the metric tensor on a spherical surface?
 
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I learned this like 2 minutes ago but I believe the following is correct:

A parametrisation of the sphere of radius [itex]\rho[/itex] centered on the origin is

[tex]f(\theta, \phi)=(\rho \sin(\theta) \cos(\phi) , \rho \sin(\theta)\sin(\phi),\rho \cos(\theta))[/tex]

where I am using this convention for the spherical angles : http://en.wikipedia.org/wiki/Spherical_coordinates#Spherical_coordinates

The components of the metric tensor are then

[tex]g_{11}(\theta, \phi) = \langle \frac{\partial f}{\partial \theta}, \frac{\partial f}{\partial \theta} \rangle[/tex]
[tex]g_{12}(\theta, \phi) = \langle \frac{\partial f}{\partial \theta}, \frac{\partial f}{\partial \phi} \rangle[/tex]
[tex]g_{21}(\theta, \phi) = \langle \frac{\partial f}{\partial \phi}, \frac{\partial f}{\partial \theta} \rangle[/tex]
[tex]g_{22}(\theta, \phi) = \langle \frac{\partial f}{\partial \phi}, \frac{\partial f}{\partial \phi} \rangle[/tex]The matrix form is then

[tex]G(\theta, \phi)=\left( \begin {array} {cc} g_{11}(\theta, \phi) & g_{12}(\theta, \phi) \\ g_{21}(\theta, \phi) & g_{22}(\theta, \phi) \end {array} \right)[/tex]

All you got to do is calculate the derivatives. Have fun. :p
 
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