> The way to do it is to integrate over hemisphere at that point acording to 
> light
> transport equation. In 2D scene it would be I(1.0 * cos x  dx) and we have 
> to do
> it from -pi/2 to pi/2. Right?
>
> The value of that is sin (pi/2) - sin(-pi/2) = 2. Hmm. Not right. Or is 
> it? Most
> likely I'm integrating incorrectly or the wrong thing.

You calculated the total light power collected at that point from all 
direction (and hence the amount emitted assuming 100% reflectance).  The 
correct answer is 2.  You'll get the same answer if you try to work out how 
much light power is emitted from a point on the sky, in fact the integral is 
the same.

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