Prove that solid angle of any closed surface is 4pi

[Adjax]

I googled a lot on proof of Gauss theorem and nearly every other proof (on web and so on books) state that solid angle of closed surface is 4pi but I can't find the proof of this nowhere !

I tried setting up the integral but don't know how to proceed furthur :
Ω=∫(cosθ/r^2)*dA

Also The one more thing which confuses me is http://mathworld.wolfram.com/SolidAngle.html proves solid angle of some other 3d shapes , those shapes also have close surfaces then why their solid angle is not equal to 4pi sr?

 

[Nidum]

Also The one more thing which confuses me is http://mathworld.wolfram.com/SolidAngle.html proves solid angle of some other 3d shapes , those shapes also have close surfaces then why their solid angle is not equal to 4pi sr?

I think that these calculations are for the solid angle subtended by just one facet on the 3D shapes .

Add the solid angles subtended by all the facets on one of the shapes together and it comes to 4pi .

 

[Adjax]

@Nidum : Uh my bad, but how shall one prove that for any closed surface solid angle is always 4pi sr ?

 

[Nidum]

The mathematicians can help you with formal proof but intuitively you can see why it is true by considering the arbitrary shape to be inside a sphere and the central point of the sphere to be inside the arbitrary shape . I'll leave you to think about that .

 

[olivermsun]

I googled a lot on proof of Gauss theorem and nearly every other proof (on web and so on books) state that solid angle of closed surface is 4pi but I can't find the proof of this nowhere !

By solid angle of closed surface, do you mean the solid angle subtended/covered by a surface, similar to the arc length subtended by a (plane) angle?

 

[Paulo Buchsbaum]

The proof cannot be made with integrals, because a random surface haven't a defined functional form.

The below proof is geometric and more than enough to convince the reader.

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Imagine any convex surface C and any point P inside it. Now imagine a sphere s of radius r totally contained in the closed surface that has as its center the reference point P. Now get a microcone from a piece of the surface closed in C (Area A) with the vertex in the center (P). It crosses the inner sphere in area a. The microcone is small enough to approximate the piece in C by a small piece of a sphere of radius R with P center.

Thus the microcones are similar and it is obvious that the solid angle is the same. (A / R² = a /r²). If we extended the sphere of radius R and get all the cones, we would reach the same 4π. When we take the next microcone from the outer surface, the larger radius would change, but in the same way we could associate with the angle of the inner sphere. So when we go through the entire outer surface exhaustively, we will reach the same 4π.

How to ensure that the smaller sphere will not have holes or overlaps? Holes are impossible because the extension of the smaller sphere must cross the closed surface. overlaps are impossible because the segment starting from P crosses the closed surface only once before leaving the closed convex surface.

Rigorously, there is a problem in the proof, because in fact the solid angle is the scalar product between the vector unit radius and the a vector normal to the surface with module area, divided by the square of the radius R, which turns to A cos θ / R². However, there is no problem because everything happens as if considering the microarea in C perpendicular to the radius R, multiplying by the cos θ. This does not change the opening of the angle.

For any external point P the same solid angle would give 0 in relation to a convex surface C:

A small sphere around P is also constructed. Every microcone crossing the sphere s toward P would cross surface C twice, one entering and the other leaving. When it comes in, the solid angle is negative (opposite to normal array outside surface) and when it leaves the surface, it is negative. By similar reasoning, it is seen that the angle module is the same on the small sphere, so the values are always canceled.

All that has been said above is valid for non-convex surfaces. When P is inside, the difference is that the number of crosses with s is odd, half entering and half leaving and one more leaving. Thus all are canceled, less the last one, reducing to the proof that already has been made. When P is outside the surface, the number of crosses is always even, half entering and half leaving, all canceling out.