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.