Solid angle subtended by an infinite cylinder

The formula for calculating the solid angle is

$\Omega = \displaystyle \iint_{A} \dfrac{(n \cdot (p_0 - p) )}{ | p_0 - p |^3 } dA$

where $p$ is the position vector of a point on the surface of the object (the infinite cylinder in this case), $p_0$ is the given point where we want to calculate the solid angle, $n$ is the unit normal vector pointing outward of the object, and $A$ is the region on the object that satisfies $n \cdot ( p_0 - p ) \ge 0$ .

Using cylindrical coordinates, a point on the surface of the given cylinder is given by $p ( t, z ) = ( \cos t, \sin t , z )$ , and the unit outward normal unit vector is $n = ( \cos t, \sin t, 0 )$ . The differential area $dA = dt \hspace{4pt} dz$ . The dot product in the numerator is $n \cdot ( p_0 - p) = (\cos t, \sin t, 0) \cdot (5 - \cos t, -\sin t, -z) = 5 \cos t - 1$ . Therefore, the range for the variable $t$ is $[ - \alpha , \alpha ]$ where $\alpha = \cos^{-1} \dfrac{1}{5}$ . The integral becomes,

$\Omega = \displaystyle \int_{z=-\infty}^{\infty} \int_{ - \alpha }^{ \alpha } \dfrac{ 5 \cos t - 1}{ (26 - 10 \cos t + z^2)^{\frac{3}{2}} } dt \hspace{4pt} dz$

Performing the integral with respect to $z$ first, the above integral becomes,

$\Omega = \displaystyle \int_{ - \alpha }^{ \alpha } \dfrac{ (5 \cos t - 1)}{ 13 - 5 \cos t } dt$

Subtracting and adding ${12}$ in the numerator and dividing out, this becomes,

$\Omega =\displaystyle \int_{ - \alpha }^{ \alpha } -1 + \dfrac{ 12 }{ 13 - 5 \cos t } dt$

For the last part, we use the substitution $u = \tan(\dfrac{t}{2})$ , then $du = \dfrac{1}{2} (u^2 + 1) dt$ , and $\cos t = \dfrac{1 - u^2}{1 + u^2}$ .

So that, $\displaystyle \int \dfrac{12}{13 - 5 \cos t } dt = \displaystyle \int \dfrac{24}{ 8 + 18 u^2 } = \dfrac{4}{3} \displaystyle \int \dfrac{du }{ (\dfrac{2}{3})^2 + u^2} = 2 \tan^{-1} \dfrac{3}{2} u$

Substituting the limits in the last expression yields our final result

$\Omega = 4 \tan^{-1} ( \frac{3}{2} \tan( \frac{1}{2} \alpha ) ) - 2 \alpha \approx \boxed{0.805431683}$

Conceptualize a unit sphere around the given point $\vec{P}$ . Calculate the solid angle as follows:

$\large{\Omega = \int_0^{\pi} \int_{\pi/2}^{3 \pi/2} M(\theta,\phi) \, sin \phi \, d\theta \, d\phi}$

To calculate $M(\theta,\phi)$ , project the vector $\vec{u}$ (defined below) from $\vec{P}$ through space and see if it intersects the infinite cylinder:

$\large{ u_x = cos \theta \, sin \phi \\ u_y = sin \theta \, sin \phi \\ (P_x + \alpha \, u_x)^2 + (P_y + \alpha \, u_y)^2 = 1 \\ A = u_x^2 + u_y^2 \\ B = 2 P_x \, u_x + 2 P_y \, u_y \\ C = P_x^2 + P_y^2 - 1 \\ M(\theta,\phi) = 1 \,\,\,\, \text{for} \,\, B^2 \geq 4 A C \\ M(\theta,\phi) = 0 \,\,\,\, \text{otherwise} }$