what is the solid angle subtended by a cylinder of length H and radius R at the centre of the cylinder? Assume the cylinder is aligned along z axis.
$\Omega =\int_{\phi}\int_{\theta}\sin(\phi)d\phi d\theta$
So far I got to assuming that the limits for the polar angle are :
$(\pi/2)-\arctan(L/2R)$ to $(\pi/2)+\arctan(L/2R)$, and the limits for the azimuthal angle are 0 to $2\pi$.
Is that right?
thank
mike
Best Answer
Why not to use simple geometry instead of using calculus
Notice,
solid angle subtended by any circular plane of radius $r$ at any point lying at a normal distance $h$ from its center is given as $$\color{red}{\Omega=2\pi\left(1-\frac{h}{\sqrt{h^2+r^2}}\right)}$$
Now, both the circular ends, each having radius $R$, of the cylinder are open & are at a normal distance $\frac{H}{2}$ from center of cylinder, hence the solid angle $(\Omega_{end})$ subtended by each circular end of the cylinder $$\Omega_{end}=2\pi\left(1-\frac{\frac{H}{2}}{\sqrt{\left(\frac{H}{2}\right)^2+R^2}}\right)$$ $$=2\pi\left(1-\frac{H}{\sqrt{H^2+4R^2}}\right)$$ Now, assume that the cylinder is closed at both the ends then the total solid angle subtended by the closed cylinder at the center $$=4\pi\ sr$$
Hence, the solid angle subtended by the cylinder (open at both ends) at its center $$=\text{(total solid angle subtended by a closed cylinder)}-\text{(solid angle subtended by both the circular ends)}$$ $$=4\pi-2\times 2\pi\left(1-\frac{H}{\sqrt{H^2+4R^2}}\right)$$ $$=\color{blue}{\frac{4\pi H}{\sqrt{H^2+4R^2}}}$$