Say I have 2 astronomical angle pairs defining a confined region on the visible hemisphere:
(minAzimuth, minElevation) & (maxAzimuth, maxElevation)
How can we calculate the solid angle of the projected area of the spherical rectangle on the unit sphere?
I assume that azimuth makes little difference and results in a linear multiple: (maxAzimuth - minAzimuth)/(2*Pi).
My intuition tells me that the elevation factor can be computed as: Sin(maxElevation) - Sin(minElevation)
Thus the answer as a proportion of 1 would be: (Sin(maxElevation) - Sin(minElevation)) * (maxAzimuth - minAzimuth)/(2*Pi).
and for the answer in steradians multiply by 2*Pi.
Is this correct?
Many thanks for your efforts.
"Yes", the formula in your original question appears to be correct.
Let $S$ be a sphere, $\ell$ a line through the center of $S$, and $C$ the circumscribed cylinder with axis $\ell$. A remarkable theorem of Archimedes asserts that axial projection away from $\ell$ is area preserving.
Assume $S$ is a unit sphere and the region in question is defined by $$ \theta_{1} \leq \text{Azimuth} \leq \theta_{2},\qquad \phi_{1} \leq \text{Elevation} \leq \phi_{2}. $$ The area of the "rectangular" patch on the cylinder, and hence of the "rectangular" patch on the sphere, is $$ (\theta_{2} - \theta_{1})(\sin\phi_{2} - \sin\phi_{1}). $$ The fraction of the upper hemisphere covered is, as you initially suspected, $$ (\theta_{2} - \theta_{1})(\sin\phi_{2} - \sin\phi_{1})/(2\pi). $$
When you obtained the extra factor of three, is it possible you were computing a volume?