Total Solid Angle of 6 Squares on Unit Sphere

[Niles]

Homework Statement

Hi

I am looking at a unit sphere. Two squares are projected onto the sphere on opposite ends, as shown in figure 1 (the figure only shows one square, the other one is at the opposite end).

There are two more sets of these squares, each set in its own dimension, so there are a total of six squares projected onto the sphere opposite to each other pairwise. The side of each square is 0.25 long. What I want to find is the total solid angle which is not covered by the squares.

Can I get a hint to what the easiest way to do this is? I guess I need to do the integral over θ and phi, but it isn't trivial to me what the boundaries are. Niles.

 

[tiny-tim]

Hi Niles!

I'd choose a quarter of a square (with a 90° angle at the centre), and multiply by 24.

Or maybe an eighth, and multiply by 48.

(Choosing half a square, cut diagonally, is a bad idea since the angle at the corner isn't 90° !)

 

[Niles]

Hi tiny-tim

Sorry for my late reply. I'm not sure I understand your suggestion, are you implying that I don't need to do any integrals? It just occurred to me that the projected squares can even be circles - does that simplify the matter?Niles.

 

[tiny-tim]

Hi Niles!

… are you implying that I don't need to do any integrals?

No, you do need to integrate …

your θ will go from 0 to 90° (or 45°), and your r will go from 0 to a function of θ that you'll need to find.

It just occurred to me that the projected squares can even be circles - does that simplify the matter?

You mean a circular cap?

Then yes, that's lot easier, since your r will be constant.

 

[Niles]

You mean a circular cap?

Yes, exactly! I was searching for that word. OK, I'll keep working on it, I'll let you know how it goes. Thanks.

Best.