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I have an angular distribution $\frac{s \sigma}{d\Omega} = \frac{d\sigma}{d \left(\cos\theta\right) d\phi}$. How can I calculate it over a circle which lies on the plane $X = dist$, has radius $r$ and its centre is located at $P(a,b)$ on this plane?

Edit: All I need to know is what are $\theta_1$, $\theta_2$, $\phi_1$ and $\phi_2$ in the integral $\int^{\cos\theta_2}_{\cos\theta_1}\int^{\phi_2}_{\phi_1}\frac{d\sigma}{d \left(\cos\theta\right) d\phi}d \left(\cos\theta\right) d\phi$.

user2738748
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  • What's $s$ and is $d\Omega = d(\theta\phi)$? If so, check that. – innisfree Nov 20 '15 at 13:21
  • I corrected the formula. I meant $d\Omega = d\theta d\phi$ – user2738748 Nov 20 '15 at 13:34
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    I don't think that $d\Omega = d\theta d\phi$. There should be a $\sin\theta$ term in there. If you integrate your form over all angles you won't get $4\pi$ steradians. – Bill N Nov 20 '15 at 13:44
  • One thing to note is that if you have a distribution density (whatever per steradian) you can't integrate it over a circle. You can integrate it over a band of some width (like a belt). The width of that band could approach zero, in which case you would represent it by d$A$. You'd have to express d$A$ in terms of your coordinate variables, taking into account the orientation of the circle. Watch out for factors of sin and cos! – garyp Nov 20 '15 at 14:23
  • Why can't I integrate it over a circle (at least theoretically)? I don't mean a circle which lays on the sphere (then it isn't a circle anymore). Sorry, I didn't make myself clear: all I need are the boundaries for calculating the integral. – user2738748 Nov 20 '15 at 14:41

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