Computing irradiance at a point

[TheWheel]

I'm having trobule trying to work out the solution to a quite simple radiometry problem I've found in the book Physically Based Rendering. The problem is this one

a) Compute the irradiance at a point due to a unit-radius disk h units directly above its normal with constant outgoing radiance 10 J/m2 sr. Do the computation twice, once as an integral over solid angle and once as an integral over area.

b) Similarly, compute the irradiance at a point due to a square quadrilateral with the same outgoing radiance that has sides of length 1 and is 1 unit directly above the point in the direction of its surface normal.

Can someone explain the solution to these problems?

Thank you in advance.

 

[Valerie Park]

The solution to a) can be computed as follows: First, compute the irradiance as an integral over solid angle. The equation for the irradiance at a point due to a surface element with outgoing radiance L and area dA is:E = L * cos(θ) * dω Where θ is the angle between the normal of the surface element and the direction from the point to the surface element, and dω is the solid angle subtended by the surface element. For a unit-radius disk h units directly above its normal, the area of the disk is πr2 and the solid angle subtended by the disk is 2π. Therefore, the irradiance at the point due to the disk is:E = 10 J/m2 sr * cos(θ) * 2π Where θ is the angle between the disk normal and the direction from the point to the disk. Since the disk is h units directly above its normal, the angle θ is equal to arccos(h). Therefore, the irradiance at the point due to the disk is:E = 10 J/m2 sr * cos(arccos(h)) * 2π = 10 J/m2 sr * h * 2π = 20πh J/m2Now, compute the irradiance as an integral over area. The equation for the irradiance at a point due to a surface element with outgoing radiance L and area dA is:E = L * dA For a unit-radius disk h units directly above its normal, the area of the disk is πr2 and the area of the disk that is visible from the point is the area of a circle segment given by:dA = 2πr2 * (1 - cos(arccos(h))) = 2πr2 * (1 - h) Therefore, the irradiance at the point due to the disk is:E = 10 J/m2 sr * 2πr2 * (1 - h) = 20πr2 (1