integral of clamped cosine over hemisphere

[shuipi]

Say I have a point with a normal direction N, and a hemisphere of outgoing directions Lo around N leaving this point. Both N and Lo are just normalized directions. From the books it says the integral of N dot Lo for all possible Lo over the hemisphere is Pi, but I always get 2Pi, can you help with the math?

Assume the angle from N to any Lo is theta, so the integral should be

2Pi * Integrate(cosine(theta) * dtheta)

where theta goes from 0 to Pi / 2, and the 2Pi represents all possible azimuths of Lo.

Since the antiderivative of cosine is sine, sin(Pi / 2) - sin(0) = 1

So I get a result of 2Pi

 

[arjan de lumens]

Let's see: You say "the 2Pi represents all possible azimuths of Lo." Well, in that case, you are integrating over a spherical-coordinate parameterization of the hemisphere, which is probably NOT what you want. From the looks of your problem, you should integrate over the actual surface area of the hemisphere. What this means in this case is: each angle Theta defines a circle on your hemisphere; you must use the circumference of this circle as a term in your integral (this replaces the 2Pi "all-possible-azimuths" term in the caclulation you have attempted so far). This circle has a circumference of 2*Pi*sin(Theta). For every point on the circle, the dot-product has a value of cos(Theta).

This gives the integral:
Integrate[ 2*Pi*sin(Theta)*cos(Theta) * d Theta]
with theta going from 0 to Pi/2, which should give you the Pi result you are seeking.

 

[shuipi]

Thanks for the explanation. I think my mistake was using dPhi * dTheta as the differential solid angle, which in fact should be sin(Theta) * dPhi * dTheta. Phi being the azimuth.