Spherical sector

In geometry, a spherical sector,^[1] also known as a spherical cone,^[2] is a portion of a sphere or of a ball defined by a conical boundary with apex at the center of the sphere. It can be described as the union of a spherical cap and the cone formed by the center of the sphere and the base of the cap. It is the three-dimensional analogue of the sector of a circle.

Volume

If the radius of the sphere is denoted by $r$ and the height of the cap by $h$ , the volume of the spherical sector is $V={\frac {2\pi r^{2}h}{3}}\,.$

This may also be written as $V={\frac {2\pi r^{3}}{3}}(1-\cos \varphi )\,,$ where $φ$ is half the cone angle, i.e., $φ$ is the angle between the rim of the cap and the direction to the middle of the cap as seen from the sphere center. The limiting case is for $φ$ approaching 180 degrees, which then describes a complete sphere.

The height, $h$ is given by $h=r(1-\cos \varphi )\,.$

The volume $V$ of the sector is related to the area $A$ of the cap by: $V={\frac {rA}{3}}\,.$

Area

The curved surface area of the spherical sector (on the surface of the sphere, excluding the cone surface) is $A=2\pi rh\,.$

It is also $A=\Omega r^{2}$ where $Ω$ is the solid angle of the spherical sector in steradians, the SI unit of solid angle. One steradian is defined as the solid angle subtended by a cap area of $A = r 2$ .

Derivation

The volume can be calculated by integrating the differential volume element $dV=\rho ^{2}\sin \phi \,d\rho \,d\phi \,d\theta$ over the volume of the spherical sector, $V=\int _{0}^{2\pi }\int _{0}^{\varphi }\int _{0}^{r}\rho ^{2}\sin \phi \,d\rho \,d\phi \,d\theta =\int _{0}^{2\pi }d\theta \int _{0}^{\varphi }\sin \phi \,d\phi \int _{0}^{r}\rho ^{2}d\rho ={\frac {2\pi r^{3}}{3}}(1-\cos \varphi )\,,$ where the integrals have been separated, because the integrand can be separated into a product of functions each with one dummy variable.

The area can be similarly calculated by integrating the differential spherical area element $dA=r^{2}\sin \phi \,d\phi \,d\theta$ over the spherical sector, giving $A=\int _{0}^{2\pi }\int _{0}^{\varphi }r^{2}\sin \phi \,d\phi \,d\theta =r^{2}\int _{0}^{2\pi }d\theta \int _{0}^{\varphi }\sin \phi \,d\phi =2\pi r^{2}(1-\cos \varphi )\,,$ where $φ$ is inclination (or elevation) and $θ$ is azimuth (right). Notice $r$ is a constant. Again, the integrals can be separated.