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How can I calculate the solid angle that a sphere of radius R subtends at a point P? I would expect the result to be a function of the radius and the distance (which I'll call d) between the center of the sphere and P. I would also expect this angle to be 4π when d < R, and 2π when d = R, and less than 2π when d > R.

I think what I really need is some pointers on how to solve the integral (taken from wikipedia) $\Omega = \iint_S \frac { \vec{r} \cdot \hat{n} \,dS }{r^3}$ given a parameterization of a sphere. I don't know how to start to set this up so any and all help is appreciated!

Ideally I would like to derive the answer from this surface integral, not geometrically, because there are other parametric surfaces I would like to know the solid angle for, which might be difficult if not impossible to solve without integration.

*I reposted this from mathoverflow because this isn't a research-level question.

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This is too simple a situation to apply calculus to. You know that the area (solid angle) of a circular cap of (angular) radius $\rho$ is $2\pi(1-\cos\rho)$, and you know that your ball of radius $R$, at a distance $d$ has an angular radius of $\rho=\arcsin(R/d)$ as seen from your point $P$. The proper formula falls out, valid only when $R\le d$.

You do realize, I hope, that what you need to parametrize is, in the words of the Wikipedia entry, “the projection of the surface $S$ to the unit sphere with center $P$”, and not anything on your ball of radius $R$. This projection is always a circular cap, as I say above; for a more complicated surface in space, it may be harder to describe the projection on the unit sphere.

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Welcome to the site Professor Lubin! –  Zev Chonoles Oct 17 '11 at 5:27
    
I don't see why I have to parameterize in terms of the projection... Doesn't the integral sum up little projected areas, of a surface with arbitrary shape in space? Thanks for your help. –  Ojotoxy Oct 17 '11 at 6:12
    
I think my question is actually very similar to this one but for a different shape. It sure looks like the surface, not it's projection, is parameterized. –  Ojotoxy Oct 17 '11 at 6:27
    
That paraboloid example involves a different shape, already given with a parametrization. In your case, you'd have to get a parametrization of the visible part of the viewed sphere. Much messier, don't you agree? –  Lubin Oct 17 '11 at 23:46
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