# Reflectance of Round Moon vs Flat Moon

Treating the moon as an ideal, bright Lambertian reflector, I was trying to compare its efficiency as a light source to an equally-sized flat sheet of drywall. It's pretty straightforward to set the problem up, but I'm not so sure of my calculus. I got that the flat sheet of drywall would provide 50% more illumination in full moon conditions as compared to the spherical moon, treating both of them as smooth, diffuse lossless reflectors. I wonder if anyone wants to second-guess my math?

I talk about the problem on my blogsite here.

-
Is there a particular piece of your work that you're not sure of? General work-checking questions are kind of discouraged... – David Z Jan 27 '12 at 6:11
Would it be a better question if I didn't mention that I'd tried to solve it already? – Marty Green Jan 27 '12 at 10:26
Nah, I wouldn't say so. I guess I shouldn't say that work-checking questions are discouraged, really... but I figure if you're asking for someone to check your work, you must have some reason to suspect that it's incorrect, and it'll make a better question if you identify that reason. Think of it like this: "how do I do it?" < "what's wrong with my work?" < "did I use this particular concept correctly?" – David Z Jan 27 '12 at 18:40
Reason? Not really. I just normally assume I'm going to do something wrong when things get complicated. – Marty Green Jan 27 '12 at 21:02
I hope you understand the moon is not a Lambertian reflector. It has a strong retro-reflection component, giving it a "flat" look during full moon. – Mike Dunlavey Jan 28 '12 at 2:44

EDITED: Indeed, the flat Moon would provide 50% more illumination than the round Moon. I assumed that it is full moon now, so the Sun illuminates the Moon, and we are on axis $z$ connecting the Moon and the Sun (the origin is at the center of the Moon). If $\theta$ is the angle between axis $z$ and some direction, and illumination (total light energy per area - both incident and reflected, as there is no absorption) on the surface of the (round or flat) Moon equals $a$ for $\theta=0$. Then illumination on the surface of the round Moon for some $\theta$ equals $a\cos(\theta)$, as the same Sun light energy falls on a larger area. Therefore, the Lambertian light intensity in the direction $z$ will be proportional to $a\cos^2(\theta)$. Therefore, we need to compare an integral of $a$ over a unit circle with an integral of $a\cos^2(\theta)$ over the surface of the unit hemisphere. (Initially, I integrated $a\cos^2(\theta)$ over the unit circle, rather than the unit hemisphere, and offered a wrong result here).
I stand corrected. I should have integrated $\cos^2(\theta)$ over the surface of the round Moon, not over the unit circle. Indeed, the flat Moon would provide 50% more illumination than the round Moon. – akhmeteli Jan 27 '12 at 15:18