Understanding Solid Angle and Proving its Equation

[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...

Any more info about the use of a solid angle??

Thx..

 

[Astronuc]

Start the area integral in spherical coordinates.

http://mathworld.wolfram.com/SolidAngle.html

A = r² d$\Omega$

for a sphere (i.e. in 3-D) the total solid angle is 4$\pi$

 

[Nomy-the wanderer]

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

Thx...

 

[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?

 

[Astronuc]

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.

 

[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...

 

[Astronuc]

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

A = r² d$\Omega$ should read

dA = r² d$\Omega$

and

A_sphere = 4$\pi$r²

 

[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..