# Solid angle

1. Oct 8, 2005

### Nomy-the wanderer

So what i know that a solid angle is to sphere as the curve is to a circle...

curve= rΘ, and (differential solid angle) dΩ=2ΠsinΘdΘ

I need to prove it, and i'm a bit rusty and i donno where to start, i wonder if there's any usefull links or tips...

Thx..

2. Oct 8, 2005

### Astronuc

Staff Emeritus
3. Oct 9, 2005

### Nomy-the wanderer

Astronuc what would my life be without u?? :D

Thx...

4. Oct 11, 2005

### Nomy-the wanderer

Well, i was thinking that doesn't the equation i wrote in the 1st post seem close to the parameter of a circle???And if that circle small, that its parameter would be almost equal to its area???

5. Oct 11, 2005

### Astronuc

Staff Emeritus
Area of a circle in polar coordinates is just the integral of "r d$\theta$ dr", with r limits of 0,r, and $\theta$ from 0, 2$\pi$ so one should end up with $\pi$r2.

Similarly in spherical coordinates the integrand is r2 sin $\phi$ d$\phi$ d$\theta$, and to find the area, one simply integrates over the two angle with r fixed, and the area should be 4$\pi$r2

$\phi$ limits -$\pi$, $\pi$ and $\theta$ limits 0, 2$\pi$

So in some sense, finding the circumference of a circle, is analogous to finding the area of a sphere.

Last edited: Oct 11, 2005
6. Oct 11, 2005

### Nomy-the wanderer

I do understand that, maybe i wasn't clear enough...I just thought it was quite similar to the relation i wanted to get to...

7. Oct 11, 2005

### Astronuc

Staff Emeritus
I just notice an error in one of the expression I posted.

A = r2 d$\Omega$ should read

dA = r2 d$\Omega$

and

Asphere = 4$\pi$r2

8. Oct 12, 2005

### Nomy-the wanderer

Alright that's what i proved, but after i proved it, i just thought something naive, so don't bother

Thx Astronuc..

File size:
56.7 KB
Views:
583