BREAKING NEWS
Spherical cap

Summary

An example of a spherical cap in blue (and another in red.)
3D model of a spherical cap.

In geometry, a spherical cap or spherical dome is a portion of a sphere or of a ball cut off by a plane. It is also a spherical segment of one base, i.e., bounded by a single plane. If the plane passes through the center of the sphere, so that the height of the cap is equal to the radius of the sphere, the spherical cap is called a hemisphere.

Volume and surface area

The volume of the spherical cap and the area of the curved surface may be calculated using combinations of

• The radius ${\displaystyle r}$ of the sphere
• The radius ${\displaystyle a}$ of the base of the cap
• The height ${\displaystyle h}$ of the cap
• The polar angle ${\displaystyle \theta }$ between the rays from the center of the sphere to the apex of the cap (the pole) and the edge of the disk forming the base of the cap
Using ${\displaystyle r}$ and ${\displaystyle h}$ Using ${\displaystyle a}$ and ${\displaystyle h}$ Using ${\displaystyle r}$ and ${\displaystyle \theta }$
Volume ${\displaystyle V={\frac {\pi h^{2}}{3}}(3r-h)}$ [1] ${\displaystyle V={\frac {1}{6}}\pi h(3a^{2}+h^{2})}$ ${\displaystyle V={\frac {\pi }{3}}r^{3}(2+\cos \theta )(1-\cos \theta )^{2}}$
Area ${\displaystyle A=2\pi rh}$[1] ${\displaystyle A=\pi (a^{2}+h^{2})}$ ${\displaystyle A=2\pi r^{2}(1-\cos \theta )}$

If ${\displaystyle \phi }$ denotes the latitude in geographic coordinates, then ${\displaystyle \theta +\phi =\pi /2=90^{\circ }\,}$, and ${\displaystyle \cos \theta =\sin \phi }$.

The relationship between ${\displaystyle h}$ and ${\displaystyle r}$ is relevant as long as ${\displaystyle 0\leq h\leq 2r}$. For example, the red section of the illustration is also a spherical cap for which ${\displaystyle h>r}$.

The formulas using ${\displaystyle r}$ and ${\displaystyle h}$ can be rewritten to use the radius ${\displaystyle a}$ of the base of the cap instead of ${\displaystyle r}$, using the Pythagorean theorem:

${\displaystyle r^{2}=(r-h)^{2}+a^{2}=r^{2}+h^{2}-2rh+a^{2}\,,}$

so that

${\displaystyle r={\frac {a^{2}+h^{2}}{2h}}\,.}$

Substituting this into the formulas gives:

${\displaystyle V={\frac {\pi h^{2}}{3}}\left({\frac {3a^{2}+3h^{2}}{2h}}-h\right)={\frac {1}{6}}\pi h(3a^{2}+h^{2})\,,}$
${\displaystyle A=2\pi {\frac {(a^{2}+h^{2})}{2h}}h=\pi (a^{2}+h^{2})\,.}$

Deriving the surface area intuitively from the spherical sector volume

Note that aside from the calculus based argument below, the area of the spherical cap may be derived from the volume ${\displaystyle V_{sec}}$ of the spherical sector, by an intuitive argument,[2] as

${\displaystyle A={\frac {3}{r}}V_{sec}={\frac {3}{r}}{\frac {2\pi r^{2}h}{3}}=2\pi rh\,.}$

The intuitive argument is based upon summing the total sector volume from that of infinitesimal triangular pyramids. Utilizing the pyramid (or cone) volume formula of ${\displaystyle V={\frac {1}{3}}bh'}$, where ${\displaystyle b}$ is the infinitesimal area of each pyramidal base (located on the surface of the sphere) and ${\displaystyle h'}$ is the height of each pyramid from its base to its apex (at the center of the sphere). Since each ${\displaystyle h'}$, in the limit, is constant and equivalent to the radius ${\displaystyle r}$ of the sphere, the sum of the infinitesimal pyramidal bases would equal the area of the spherical sector, and:

${\displaystyle V_{sec}=\sum {V}=\sum {\frac {1}{3}}bh'=\sum {\frac {1}{3}}br={\frac {r}{3}}\sum b={\frac {r}{3}}A}$

Deriving the volume and surface area using calculus

Rotating the green area creates a spherical cap with height ${\displaystyle h}$ and sphere radius ${\displaystyle r}$.

The volume and area formulas may be derived by examining the rotation of the function

${\displaystyle f(x)={\sqrt {r^{2}-(x-r)^{2}}}={\sqrt {2rx-x^{2}}}}$

for ${\displaystyle x\in [0,h]}$, using the formulas the surface of the rotation for the area and the solid of the revolution for the volume. The area is

${\displaystyle A=2\pi \int _{0}^{h}f(x){\sqrt {1+f'(x)^{2}}}\,dx}$

The derivative of ${\displaystyle f}$ is

${\displaystyle f'(x)={\frac {r-x}{\sqrt {2rx-x^{2}}}}}$

and hence

${\displaystyle 1+f'(x)^{2}={\frac {r^{2}}{2rx-x^{2}}}}$

The formula for the area is therefore

${\displaystyle A=2\pi \int _{0}^{h}{\sqrt {2rx-x^{2}}}{\sqrt {\frac {r^{2}}{2rx-x^{2}}}}\,dx=2\pi \int _{0}^{h}r\,dx=2\pi r\left[x\right]_{0}^{h}=2\pi rh}$

The volume is

${\displaystyle V=\pi \int _{0}^{h}f(x)^{2}\,dx=\pi \int _{0}^{h}(2rx-x^{2})\,dx=\pi \left[rx^{2}-{\frac {1}{3}}x^{3}\right]_{0}^{h}={\frac {\pi h^{2}}{3}}(3r-h)}$

Applications

Volumes of union and intersection of two intersecting spheres

The volume of the union of two intersecting spheres of radii ${\displaystyle r_{1}}$ and ${\displaystyle r_{2}}$ is [3]

${\displaystyle V=V^{(1)}-V^{(2)}\,,}$

where

${\displaystyle V^{(1)}={\frac {4\pi }{3}}r_{1}^{3}+{\frac {4\pi }{3}}r_{2}^{3}}$

is the sum of the volumes of the two isolated spheres, and

${\displaystyle V^{(2)}={\frac {\pi h_{1}^{2}}{3}}(3r_{1}-h_{1})+{\frac {\pi h_{2}^{2}}{3}}(3r_{2}-h_{2})}$

the sum of the volumes of the two spherical caps forming their intersection. If ${\displaystyle d\leq r_{1}+r_{2}}$ is the distance between the two sphere centers, elimination of the variables ${\displaystyle h_{1}}$ and ${\displaystyle h_{2}}$ leads to[4][5]

${\displaystyle V^{(2)}={\frac {\pi }{12d}}(r_{1}+r_{2}-d)^{2}\left(d^{2}+2d(r_{1}+r_{2})-3(r_{1}-r_{2})^{2}\right)\,.}$

Volume of a spherical cap with a curved base

The volume of a spherical cap with a curved base can be calculated by considering two spheres with radii ${\displaystyle r_{1}}$ and ${\displaystyle r_{2}}$, separated by some distance ${\displaystyle d}$, and for which their surfaces intersect at ${\displaystyle x=h}$. That is, the curvature of the base comes from sphere 2. The volume is thus the difference between sphere 2's cap (with height ${\displaystyle (r_{2}-r_{1})-(d-h)}$) and sphere 1's cap (with height ${\displaystyle h}$),

{\displaystyle {\begin{aligned}V&={\frac {\pi h^{2}}{3}}(3r_{1}-h)-{\frac {\pi [(r_{2}-r_{1})-(d-h)]^{2}}{3}}[3r_{2}-((r_{2}-r_{1})-(d-h))]\,,\\V&={\frac {\pi h^{2}}{3}}(3r_{1}-h)-{\frac {\pi }{3}}(d-h)^{3}\left({\frac {r_{2}-r_{1}}{d-h}}-1\right)^{2}\left[{\frac {2r_{2}+r_{1}}{d-h}}+1\right]\,.\end{aligned}}}

This formula is valid only for configurations that satisfy ${\displaystyle 0 and ${\displaystyle d-(r_{2}-r_{1}). If sphere 2 is very large such that ${\displaystyle r_{2}\gg r_{1}}$, hence ${\displaystyle d\gg h}$ and ${\displaystyle r_{2}\approx d}$, which is the case for a spherical cap with a base that has a negligible curvature, the above equation is equal to the volume of a spherical cap with a flat base, as expected.

Areas of intersecting spheres

Consider two intersecting spheres of radii ${\displaystyle r_{1}}$ and ${\displaystyle r_{2}}$, with their centers separated by distance ${\displaystyle d}$. They intersect if

${\displaystyle |r_{1}-r_{2}|\leq d\leq r_{1}+r_{2}}$

From the law of cosines, the polar angle of the spherical cap on the sphere of radius ${\displaystyle r_{1}}$ is

${\displaystyle \cos \theta ={\frac {r_{1}^{2}-r_{2}^{2}+d^{2}}{2r_{1}d}}}$

Using this, the surface area of the spherical cap on the sphere of radius ${\displaystyle r_{1}}$ is

${\displaystyle A_{1}=2\pi r_{1}^{2}\left(1+{\frac {r_{2}^{2}-r_{1}^{2}-d^{2}}{2r_{1}d}}\right)}$

Surface area bounded by parallel disks

The curved surface area of the spherical segment bounded by two parallel disks is the difference of surface areas of their respective spherical caps. For a sphere of radius ${\displaystyle r}$, and caps with heights ${\displaystyle h_{1}}$ and ${\displaystyle h_{2}}$, the area is

${\displaystyle A=2\pi r|h_{1}-h_{2}|\,,}$

or, using geographic coordinates with latitudes ${\displaystyle \phi _{1}}$ and ${\displaystyle \phi _{2}}$,[6]

${\displaystyle A=2\pi r^{2}|\sin \phi _{1}-\sin \phi _{2}|\,,}$

For example, assuming the Earth is a sphere of radius 6371 km, the surface area of the arctic (north of the Arctic Circle, at latitude 66.56° as of August 2016[7]) is 2π·63712|sin 90° − sin 66.56°| = 21.04 million km2, or 0.5·|sin 90° − sin 66.56°| = 4.125% of the total surface area of the Earth.

This formula can also be used to demonstrate that half the surface area of the Earth lies between latitudes 30° South and 30° North in a spherical zone which encompasses all of the Tropics.

Generalizations

Sections of other solids

The spheroidal dome is obtained by sectioning off a portion of a spheroid so that the resulting dome is circularly symmetric (having an axis of rotation), and likewise the ellipsoidal dome is derived from the ellipsoid.

Hyperspherical cap

Generally, the ${\displaystyle n}$-dimensional volume of a hyperspherical cap of height ${\displaystyle h}$ and radius ${\displaystyle r}$ in ${\displaystyle n}$-dimensional Euclidean space is given by:[citation needed] ${\displaystyle V={\frac {\pi ^{\frac {n-1}{2}}\,r^{n}}{\,\Gamma \left({\frac {n+1}{2}}\right)}}\int \limits _{0}^{\arccos \left({\frac {r-h}{r}}\right)}\sin ^{n}(t)\,\mathrm {d} t}$ where ${\displaystyle \Gamma }$ (the gamma function) is given by ${\displaystyle \Gamma (z)=\int _{0}^{\infty }t^{z-1}\mathrm {e} ^{-t}\,\mathrm {d} t}$.

The formula for ${\displaystyle V}$ can be expressed in terms of the volume of the unit n-ball ${\displaystyle C_{n}={\scriptstyle \pi ^{n/2}/\Gamma [1+{\frac {n}{2}}]}}$ and the hypergeometric function ${\displaystyle {}_{2}F_{1}}$ or the regularized incomplete beta function ${\displaystyle I_{x}(a,b)}$ as

${\displaystyle V=C_{n}\,r^{n}\left({\frac {1}{2}}\,-\,{\frac {r-h}{r}}\,{\frac {\Gamma [1+{\frac {n}{2}}]}{{\sqrt {\pi }}\,\Gamma [{\frac {n+1}{2}}]}}{\,\,}_{2}F_{1}\left({\tfrac {1}{2}},{\tfrac {1-n}{2}};{\tfrac {3}{2}};\left({\tfrac {r-h}{r}}\right)^{2}\right)\right)={\frac {1}{2}}C_{n}\,r^{n}I_{(2rh-h^{2})/r^{2}}\left({\frac {n+1}{2}},{\frac {1}{2}}\right)}$,

and the area formula ${\displaystyle A}$ can be expressed in terms of the area of the unit n-ball ${\displaystyle A_{n}={\scriptstyle 2\pi ^{n/2}/\Gamma [{\frac {n}{2}}]}}$ as

${\displaystyle A={\frac {1}{2}}A_{n}\,r^{n-1}I_{(2rh-h^{2})/r^{2}}\left({\frac {n-1}{2}},{\frac {1}{2}}\right)}$ ,

where ${\displaystyle 0\leq h\leq r}$.

Earlier in [8] (1986, USSR Academ. Press) the following formulas were derived: ${\displaystyle A=A_{n}p_{n-2}(q),V=C_{n}p_{n}(q)}$, where ${\displaystyle q=1-h/r(0\leq q\leq 1),p_{n}(q)=(1-G_{n}(q)/G_{n}(1))/2}$,

${\displaystyle G_{n}(q)=\int \limits _{0}^{q}(1-t^{2})^{(n-1)/2}dt}$.

For odd ${\displaystyle n=2k+1:}$

${\displaystyle G_{n}(q)=\sum _{i=0}^{k}(-1)^{i}{\binom {k}{i}}{\frac {q^{2i+1}}{2i+1}}}$.

Asymptotics

It is shown in [9] that, if ${\displaystyle n\to \infty }$ and ${\displaystyle q{\sqrt {n}}={\text{const.}}}$, then ${\displaystyle p_{n}(q)\to 1-F({q{\sqrt {n}}})}$ where ${\displaystyle F()}$ is the integral of the standard normal distribution.

A more quantitative bound is ${\displaystyle A/A_{n}=n^{\Theta (1)}\cdot [(2-h/r)h/r]^{n/2}}$. For large caps (that is when ${\displaystyle (1-h/r)^{4}\cdot n=O(1)}$ as ${\displaystyle n\to \infty }$), the bound simplifies to ${\displaystyle n^{\Theta (1)}\cdot e^{-(1-h/r)^{2}n/2}}$. [10]

References

1. ^ a b Polyanin, Andrei D; Manzhirov, Alexander V. (2006), Handbook of Mathematics for Engineers and Scientists, CRC Press, p. 69, ISBN 9781584885023.
2. ^ Shekhtman, Zor. "Unizor - Geometry3D - Spherical Sectors". YouTube. Zor Shekhtman. Retrieved 31 Dec 2018.
3. ^ Connolly, Michael L. (1985). "Computation of molecular volume". Journal of the American Chemical Society. 107 (5): 1118–1124. doi:10.1021/ja00291a006.
4. ^ Pavani, R.; Ranghino, G. (1982). "A method to compute the volume of a molecule". Computers & Chemistry. 6 (3): 133–135. doi:10.1016/0097-8485(82)80006-5.
5. ^ Bondi, A. (1964). "Van der Waals volumes and radii". The Journal of Physical Chemistry. 68 (3): 441–451. doi:10.1021/j100785a001.
6. ^ Scott E. Donaldson, Stanley G. Siegel (2001). Successful Software Development. ISBN 9780130868268. Retrieved 29 August 2016.
7. ^ "Obliquity of the Ecliptic (Eps Mean)". Neoprogrammics.com. Retrieved 2014-05-13.
8. ^ Chudnov, Alexander M. (1986). "On minimax signal generation and reception algorithms (rus.)". Problems of Information Transmission. 22 (4): 49–54.
9. ^ Chudnov, Alexander M (1991). "Game-theoretical problems of synthesis of signal generation and reception algorithms (rus.)". Problems of Information Transmission. 27 (3): 57–65.
10. ^ Anja Becker, Léo Ducas, Nicolas Gama, and Thijs Laarhoven. 2016. New directions in nearest neighbor searching with applications to lattice sieving. In Proceedings of the twenty-seventh annual ACM-SIAM symposium on Discrete algorithms (SODA '16), Robert Krauthgamer (Ed.). Society for Industrial and Applied Mathematics, Philadelphia, PA, USA, 10-24.