 Solid angle of a circular cone

A right circular cone with vertex $O$ selects all directions within a half-opening angle $\theta$ of its axis. Here $\theta$ is measured from the axis to a boundary direction, so the full opening in a meridian plane is $2\theta$. For $0\le \theta \le \pi$, interpret the selected directions as a spherical cap; for $\theta >\pi /2$, this is a region larger than a hemisphere.

The spherical coordinate system gives the area element $dA=R^2\sin \alpha d\alpha d\varphi$ on the sphere. Dividing by $R^2$ and integrating over the cap yieldsThe factor $\sin \alpha$ matters: a thin angular strip near a pole contains fewer directions than an equally wide strip near the equator. Treating equal intervals of both spherical coordinates as equal solid angles would overcount the polar directions.

At $\theta =45^{\circ }$, the solid angle is $\pi (2-\sqrt{2})\approx 1.8403$ steradians, or about $14.64\%$ of all directions. At $\theta =90^{\circ }$ it is $2\pi$ steradians, and at $180^{\circ }$ it is $4\pi$ steradians. For a small $\theta$ measured in radians, the Taylor series $\cos \theta =1-{\theta }^2/2+O({\theta }^4)$ gives the small-angle approximation $\Omega \approx \pi {\theta }^2$. Doubling a narrow half-opening angle therefore approximately quadruples its coverage.