Spherical sector
# Spherical sector

In geometry, a spherical sector, also known as a spherical cone, 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 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.

