# Why/how does integral of solid angle = pi?

1. Jun 14, 2013

### ck99

Hi folks, can someone help explain this in words of one syllable or less? I am looking at a text that compares flux and intensity of a distant source, and it states that

∫∫dΩ = ∏

I know that

dΩ = sinθ dθ d∅

but I don't understand where the given result comes from. What are the limits of integration here, and how does it all work? I assume that we treat the distant source as a circle projected onto our sphere of view, but in that case surely the size of that circle must come into the answer somehow? (EG a large source like the sun would subtend a larger area than Jupiter, they can't both have an area of ∏!)

2. Jun 14, 2013

### HallsofIvy

Since you know that $d\Omega= sin(\theta)d\theta d\phi$ you know that $\int\int d\Omega= \int\int sin(\theta) d\theta d\phi$. And that, assuming for the moment that the limits in one variable do not involve the other, is $-cos(\theta)\phi$. Since we are free to orient a coordinate system any way we want, we can assume that the lower limits of integration are $\theta= 0$, $\phi= 0$. And we then want upper limits so that $-cos(\theta)\phi= \pi$. There are several different combinations of $\theta$ and $\phi$ that will do that. Taking $\phi= \pi$, the largest possible, we would need $-cos(\theta)+ cos(0)= -cos(\theta)+ 1= 1$, $cos(\theta)= 0$ which gives $\theta= \pi/2$. But taking $\phi= \pi/2$ we would need $-cos(\theta)+ cos(0)= 1- cos(\theta)= 2$, $cos(\theta)= -1$ so $\theta= \pi$.

$\theta$ from 0 to $\pi/2$ and $\phi$ from 0 to $\pi$ or $\theta$ from 0 to $\pi$ and $\phi$ from 0 to $\pi/2$. Which do you prefer? Since you are free to orient you coordinate system however you like are they really different?

Last edited by a moderator: Jun 14, 2013
3. Jun 14, 2013

### ck99

Thank you for your response; I understand now how that result is reached but now I don't see how it relates to my problem. In this text, it says that flux F = L/(4∏D2) where D is the distance from the source and L is its luminosity. Then it says that intensity I = F/Ω, which implies that I = L/4D2 if Ω = ∏, but in fact it states that I = L/4R2 where R is the radius of the source!

This last result would be true if Ω = R2 / ∏D2 so I did some research, and the closest I can find is that the solid angle subtended by a distant source is given by area/D2 or ∏R2/D2. (that last result comes from page 5 of this link http://www.drdrbill.com/downloads/optics/photometry/Solid_Angle.pdf )

So I have three different answers, and I can't see how they link together!