How to calculate the solid angle subtended by the shaded portion of the circular plane at a point $P(0,0,h)$. This composite plane figure is obtained by removing the right $\Delta ABC$ $(AB=BC)$ from a circle with a radius $R$ & centered at the origin O? (as shown in the diagram above). Given $\frac{R}{h}=3$
Note: The point $P(0, 0, h)$ is lying at a height $h$ from the origin (centre O) on the z-axis normal to the plane of paper.
Best Answer
The solid angles $\Omega_1$ of the circle and the solid angle $\Omega_2$ of the plane triangle are both given by the formulas in Wikipedia (cone and Oosterom-Strackee, respectively). The result is obtained by subtracting both values. $$ \Omega = \Omega_1-\Omega_2. $$ The solid angle of the circle is $$ \Omega_1=2\pi (1-\cos \varphi) $$ where $\tan\varphi = R/h$, so $$ \Omega_1=2\pi (1-\frac{1}{\sqrt{1+(R/h)^2}}) . $$ The Cartesian coordinates of the 3 corners of the triangle are $\vec A=\vec R_3=(-R,0,h)$, $\vec C=\vec R_1=(R,0,h)$, $\vec B=\vec R_2=(0,R,h)$. The lengths are $$ R_1=R_2=R_3=\sqrt{R^2+h^2}. $$ The triple vector product of the parallelepiped is $$ [\vec R_1\vec R_2\vec R_3] = 2R^2h. $$ The v-Oosterom-Strackee formula yields $$ \tan \frac{\Omega_2}{2} = \frac{[\vec R_1\vec R_2\vec R_3]}{R_1R_2R_3 + (\vec R_1\cdot \vec R_2)R_3 +(\vec R_2\cdot \vec R_3)R_1 +(\vec R_3\cdot \vec R_1)R_2 } $$ $$ = \frac{2R^2h}{(R^2+h^2)^{3/2}+h^2\sqrt{R^2+h^2}+h^2\sqrt{R^2+h^2}+(h^2-R^2)\sqrt{R^2+h^2}} $$ $$ =\frac{R^2}{2h\sqrt{R^2+h^2}} $$ $$ =\frac{\tan^2 \varphi\cos\varphi}{2}. $$ $$ \Omega_2=2\arctan\frac{\sin^2\varphi }{2\cos\varphi}. $$