Spherical sector

File:Kugel-sektor.png

A spherical sector (blue)

File:Spherical sector.png

A spherical sector

In geometry, a spherical sector, also known as a spherical cone, is a portion of a ball that is bounded by a spherical cap and the cone that connects the centre of the sphere to the boundary 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 $$V=\frac{2\pi r^3}{3}(1-\cos \phi ),$$ where φ is half the cone aperture angle, i.e., φ is the angle between the rim of the cap and the axis 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 $$h=r(1-\cos \phi ).$$

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 cap (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 = r².

Derivation

The volume can be calculated by integrating the differential volume element $$dV={\rho }^2\sin \varphi d\rho d\varphi d\theta$$ over the volume of the spherical sector, $$V={\int }_0^{2\pi }{\int }_0^{\phi }{\int }_0^r{\rho }^2\sin \varphi d\rho d\varphi d\theta ={\int }_0^{2\pi }d\theta {\int }_0^{\phi }\sin \varphi d\varphi {\int }_0^r{\rho }^{2}d\rho =\frac{2\pi r^3}{3}(1-\cos \phi ),$$ 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 $$dA=r^2\sin \varphi d\varphi d\theta$$ over the spherical sector, giving $$A={\int }_0^{2\pi }{\int }_0^{\phi }{r}^2\sin \varphi d\varphi d\theta =r^2{\int }_0^{2\pi }d\theta {\int }_0^{\phi }\sin \varphi d\varphi =2\pi r^2(1-\cos \phi ),$$ where φ is inclination (or elevation) and θ is azimuth (right). Notice r is a constant. Again, the integrals can be separated.

See also

• Circular sector — the analogous 2D figure.
• Spherical cap
• Spherical segment
• Spherical wedge