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I'm looking for an average value of sine of an angle between two rays, lying within a cone with a certain angle.

Given a cone with an aperture of ${2\chi}$ and two rays lying within the cone. The rays can be represented as vectors in a spherical coordinate system:

$$ {\vec{e_1}=\lbrace1,\phi_1,\theta_1 \rbrace},{\vec{e_2}=\lbrace1,\phi_2,\theta_2 \rbrace} $$

where ${\phi_1,\phi_2\in[0,2\pi], \theta_1,\theta_2\in[0,\chi]}$ (assuming the axis of the cone is aligned with the z axis). The distribution of every angle is uniform. It is needed to find the average value of the sine of the angle between the rays (vectors).

We can get an answer by simply integrating the sine within a needed area: $$ {{\frac{1}{4\pi^2\chi^2}}\int\limits_{0}^{\chi}\int\limits_{0}^{\chi}\int\limits_{0}^{2\pi}\int\limits_{0}^{2\pi}\sin(\vec{e_1},\vec{e_2})d\phi_1\phi_2\theta_1d\theta_2} $$

We can get cosine of an angle between the vectors using the dot product:

$${\cos(\vec{e_1},\vec{e_2})=\frac{(\vec{e_1},\vec{e_2})}{|\vec{e_1}||\vec{e_2}|}}=\sin\theta_{1}\sin\theta_{2}\cos\left(\phi_{1}-\phi_{2}\right)+\cos\theta_{1}\cos\theta_{2}$$

Using this an average value of cosine can be easily got:

$$ {{\frac{1}{4\pi^2\chi^2}}\int\limits_{0}^{\chi}\int\limits_{0}^{\chi}\int\limits_{0}^{2\pi}\int\limits_{0}^{2\pi}\cos(\vec{e_1},\vec{e_2})d\phi_1\phi_2\theta_1d\theta_2=\frac{\sin^2\chi}{\chi^2}} $$

But I didn't have much luck trying to get the average value of sine. I can't solve this:

$$ {{\frac{1}{4\pi^2\chi^2}}\int\limits_{0}^{\chi}\int\limits_{0}^{\chi}\int\limits_{0}^{2\pi}\int\limits_{0}^{2\pi}\sqrt{1-\left(\sin\theta_{1}\sin\theta_{2}\cos\left(\phi_{1}-\phi_{2}\right)+\cos\theta_{1}\cos\theta_{2}\right)^{2}}d\phi_1\phi_2\theta_1d\theta_2} $$

Using the Monte-Carlo simulation I got this curve:

Monte Carlo simulations of an average sine of an angle between two rays in a cone

Any help would be appreciated!

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    Why is there a limit on both $\theta$ and $\phi$? Assuming the axis of the cone is aligned with the zenith direction, the polar angle should range from $0$ to $\chi$ while the azimuthal angle should span its whole range of $2\pi$. (I'm using terminology from [Wikipedia's page on spherical coordinates](https://en.wikipedia.org/wiki/Spherical_coordinate_system) because conventions differ on which angle is $\theta$ and which is $\phi$.)2012-05-11
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    Oops, it seems as if you are right and my limits for angles should be reviewed. Thank you.2012-05-11
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    Are the ray vectors uniform random variates in the interior of the spherical cone, or on the surface of it? These give rise to two different measures, $$ dV=r^2\sin\phi\,dr\,d\phi\,d\theta=dr\,dS \qquad\text{and}\qquad dS=r^2\sin\phi\,d\phi\,d\theta=r^2\,d\Omega $$ The former is $dr$ times the latter which, in turn, is $r^2$ times the solid angle element $d\Omega$. Which of these you are using not only will affect the numerical answer but possibly also its closed form integrability. [This](http://en.wikipedia.org/wiki/Solid_angle#Latitude-longitude_rectangle) might also help you solve it.2012-05-11
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    If you use $\theta\in[0,2\pi)$ for the longitudinal rotation between the two points (which is uniformly distributed) and $\phi_1,\phi_2\in[0,\pi]$ for complement of the latitude of each point, i.e. the angle from the north pole, then a point would have cartesian coordinates $(x,y,z)=(\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi)$ and the angle $\alpha$ between our points has cosine $\cos\alpha=\cos\phi_1\cos\phi_2+\sin\phi_1\sin\phi_2\cos\theta$. This symmetry eliminates one longitudinal $\theta$ from the integral. I think you are missing $\sin\phi_1\sin\phi_2$ in your area elements.2012-05-11
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    Where did this problem arise?2012-05-12
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    In a percolation problem. I'm calculating the average number of bonds in a 3d set of randomly disturbed 1d sticks. I can send you one of my papers.2012-05-12

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