Light Source in Cylinder - Part 2

The general equation for calculating the power incident on the surface is (assuming an isotropic radiator):

$P = \int \int_S \,\, \frac{P_{total}}{4 \pi \, |\vec{D}|^2} \, (\hat{n} \cdot \hat{D}) \, dS \\ P_{total} = \text{Total radiator power} \\ \vec{D} = \text{Vector from radiator to point on surface} \\ dS = \text{Infinitesimal scalar surface area element} \\ \hat{D} = \text{Unit vector in the direction of } \vec{D} \\ \hat{n} = \text{Unit outward surface normal vector }$

For the bottom disk (let the radiator position be $\vec{Q}$ ):

$0 \leq r \leq 1 \\ 0 \leq \theta \leq 2 \pi \\ x = r \, cos \theta \\ y = r \, sin \theta \\ z = 0 \\ \vec{D} = (x - Q_x, y - Q_y, z - Q_z) \\ \hat{n} = (0,0,-1) \\ dS = r \, dr \, d\theta$

For the top disk:

$0 \leq r \leq 1 \\ 0 \leq \theta \leq 2 \pi \\ x = r \, cos \theta \\ y = r \, sin \theta \\ z = 1 \\ \vec{D} = (x - Q_x, y - Q_y, z - Q_z) \\ \hat{n} = (0,0,+1) \\ dS = r \, dr \, d\theta$

For the cylinder:

$0 \leq z \leq 1 \\ 0 \leq \theta \leq 2 \pi \\ x = cos \theta \\ y = sin \theta \\ z = z \\ \vec{D} = (x - Q_x, y - Q_y, z - Q_z) \\ \hat{n} = (x,y,0) \\ dS = d\theta \, dz$

Suppose the total radiated power is $1$ . The results (evaluated numerically) are:

$P_B = \text{Bottom disk power} \approx 0.153256 \\ P_T = \text{Top disk power} \approx 0.315708 \\ P_L = \text{Cylinder power for negative x part} \approx 0.109608 \\ P_R = \text{Cylinder power for positive x part} \approx 0.421557$