Question on solid angle of sphere.

[yungman]

I understand the surface of the sphere is 4\pi sr. where the area of one sr is r^2.

My question is why d\Omega = sin \theta \;d \theta \;d\phi? Can anyone show me how to derive this.

Is it because surface area dS = (Rd\theta)(R\; sin\;\theta\;d\phi)\;\hbox { so if }\; \Omega = \frac S {R^2} \;\Rightarrow \; d\Omega = \frac {d\;S}{R^2}= sin \theta d\theta d \phi

 

[BruceW]

\vec{r} = r( cos(\phi)sin(\theta) \hat{i} + sin(\phi)sin(\theta) \hat{j} + cos(\theta) \hat{k} )
and the surface element perpendicular to \hat{r} is:
d\Omega = \frac{\partial \vec{r}}{\partial \theta} \ d\theta \wedge \frac{\partial \vec{r}}{\partial \phi} \ d\phi
which gives:
d\Omega = sin(\theta) d\theta d\phi \hat{r}

EDIT: I can't remember which order the cross product should go, but it should be such that d\Omega is positive.