Total emitted power of diffuse area light

Question

I am reading the book Physically Based Rendering (Pharr, Humphreys). In the chapter on lights, they talk about approximating the total emitted power of different kinds of lights. For example, a point light's total power is intensity * 4 * pi. Here 4pi represents a solid angle over the entire sphere. This makes sense to me because intensity * solid angle = power (or radiant flux if you will). You can see this by the units as well. Intensity is W/sr and solid angle is sr, so W/sr * sr = W and power is measured in watts. It checks out.

However, I don't understand the corresponding calculation for the DiffuseAreaLight. From my understanding of the book they calculate the total power emitted from a diffuse area light as emitted radiance * area * pi. Since the unit of radiance is W/(sr*m^2) multiplying area gives W/sr. This makes me think the pi factor represents solid angle - but why only 1pi? I would have guessed 2pi since each point on the area light would radiate in a complete hemisphere (corresponding to 2pi steradians).

You can find the actual code mentioned in the book here.

What am I misunderstanding? Why does total emitted power = emitted radiance * area * pi for diffuse area lights make sense?

Answer 1

Radiance (in terms of flux) has the following definition:

$L_o = \frac{\mathrm{d}\Phi}{\mathrm{d}\omega^\perp\mathrm{d}A} = \frac{\mathrm{d}\Phi}{\cos{\theta} \mathrm{d}\omega \mathrm{d}A}$.

Thus in order to get the total emitted power we need to integrate over the area of the light and we need to integrate over the projected solid angle of the hemisphere (in the direction of the surface normal). The radiance for the area lights in pbrt is constant. This gives us:

$\Phi = \int_A \int_{H^2(\mathbf{n})} L_o\mathrm{d}\omega^\perp \mathrm{d}A = \int_A \int_{H^2(\mathbf{n})} L_o \cos\theta \mathrm{d}\omega \mathrm{d}A = \int_A L_o \int_0^{2\pi} \int_0^{\pi/2} \sin\theta \cos\theta \mathrm{d}\theta \mathrm{d}\phi \mathrm{d}A = \int_A L_o \pi \mathrm{d}A = A \cdot L_o \cdot \pi$

Answer 2

I think the assumption (perhaps stated perhaps not, I dont' have the text handy) is that the radiance is emitted in a cosine-lobe distribution. This means that there's falloff in proportion to the cosine of the angle between the emitter's normal, and the direction of emission.

If you look in the global illumination compendium, under Hemispherical Geometry, equation set (30), you'll see that the integral over the hemisphere, modulated by a cosine lobe, is exactly pi.