I particularly like this example since it's simple yet informative. The very obvious interpretation would be from the figure in the slide, where we take the cosine and sine product value and integrate over the spherical phi and theta; this is due to that we are looking at the hemisphere from its normal vector direction towards its 'sides' till alpha radian and the entire circle cutoff would be taken into account. Another interpretation would simply be retrieving the area from the intersecting circle with radius sine of alpha.
bbtong
^Thank you for the great explanation!
eddylan2894
If the source is not a disk but a hemisphere with the same r, would the irradiance integral be the same?
I particularly like this example since it's simple yet informative. The very obvious interpretation would be from the figure in the slide, where we take the cosine and sine product value and integrate over the spherical phi and theta; this is due to that we are looking at the hemisphere from its normal vector direction towards its 'sides' till alpha radian and the entire circle cutoff would be taken into account. Another interpretation would simply be retrieving the area from the intersecting circle with radius sine of alpha.
^Thank you for the great explanation!
If the source is not a disk but a hemisphere with the same r, would the irradiance integral be the same?