Given is the unit sphere and a function $f(\alpha, \beta)$ on the sphere. $\alpha$ is the angle from the north pole (i.e. $\alpha=\pi/2 -$ latitude) and $\beta$ is the longitude. How do I calculate the surface integral of $f$ over the whole surface of the sphere?
I know that this is wrong:
$$\int_0^{2 \pi} \int_0^\pi f(\alpha, \beta) \ \ \ d\alpha \ d\beta $$
Is this integral:
$$\int_0^{2 \pi} \int_0^\pi f(\alpha, \beta) \ \ \ \sin \alpha \ \ \ d\alpha \ d\beta $$
the correct one? At least for $f=1$ the result seems correct.
You can figure out the right size of the differential area element with the sketch below
The result is
$$ {\rm d}A = ({\rm d}\alpha)(\sin\alpha \ {\rm d}\beta) = \sin\alpha \ {\rm d}\alpha {\rm d}\beta $$
So that the integral on a unit sphere of the function $f$ is
$$ \int_0^\pi{\rm d\alpha}\int_0^{2\pi}{\rm d}\beta ~\sin\alpha \ f(\alpha, \beta) $$
Formally, you can get to the same result through the determinant of the Jacobian matrix