Spherical cap (original) (raw)

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Section of a sphere

An example of a spherical cap in blue (and another in red)

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 (forming a great circle), 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

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

These variables are inter-related through the formulas a = r sin ⁡ θ {\displaystyle a=r\sin \theta } , h = r ( 1 − cos ⁡ θ ) {\displaystyle h=r(1-\cos \theta )} , 2 h r = a 2 + h 2 {\displaystyle 2hr=a^{2}+h^{2}} , and 2 h a = ( a 2 + h 2 ) sin ⁡ θ {\displaystyle 2ha=(a^{2}+h^{2})\sin \theta } .

| | Using r {\displaystyle r} and h {\displaystyle h} | Using a {\displaystyle a} and h {\displaystyle h} | Using r {\displaystyle r} and θ {\displaystyle \theta } | | | ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | | Volume | V = π h 2 3 ( 3 r − h ) {\displaystyle V={\frac {\pi h^{2}}{3}}(3r-h)} [1] | V = 1 6 π h ( 3 a 2 + h 2 ) {\displaystyle V={\frac {1}{6}}\pi h(3a^{2}+h^{2})} | V = π 3 r 3 ( 2 + cos ⁡ θ ) ( 1 − cos ⁡ θ ) 2 {\displaystyle V={\frac {\pi }{3}}r^{3}(2+\cos \theta )(1-\cos \theta )^{2}} | | Area | A = 2 π r h {\displaystyle A=2\pi rh} [1] | A = π ( a 2 + h 2 ) {\displaystyle A=\pi (a^{2}+h^{2})} | A = 2 π r 2 ( 1 − cos ⁡ θ ) {\displaystyle A=2\pi r^{2}(1-\cos \theta )} | | Constraints | 0 ≤ h ≤ 2 r {\displaystyle 0\leq h\leq 2r} | 0 ≤ a , 0 ≤ h {\displaystyle 0\leq a,\;0\leq h} | 0 ≤ θ ≤ π , 0 ≤ r {\displaystyle 0\leq \theta \leq \pi ,\;0\leq r} |

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

Deriving the surface area intuitively from the spherical sector volume

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

A = 3 r V s e c = 3 r 2 π r 2 h 3 = 2 π r h . {\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 V = 1 3 b h ′ {\displaystyle V={\frac {1}{3}}bh'} , where b {\displaystyle b} is the infinitesimal area of each pyramidal base (located on the surface of the sphere) and h ′ {\displaystyle h'} is the height of each pyramid from its base to its apex (at the center of the sphere). Since each h ′ {\displaystyle h'} , in the limit, is constant and equivalent to the radius r {\displaystyle r} of the sphere, the sum of the infinitesimal pyramidal bases would equal the area of the spherical sector, and:

V s e c = ∑ V = ∑ 1 3 b h ′ = ∑ 1 3 b r = r 3 ∑ b = r 3 A {\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

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

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

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

for x ∈ [ 0 , h ] {\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

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

The derivative of f {\displaystyle f} is

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

and hence

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

The formula for the area is therefore

A = 2 π ∫ 0 h 2 r x − x 2 r 2 2 r x − x 2 d x = 2 π ∫ 0 h r d x = 2 π r [ x ] 0 h = 2 π r h {\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

V = π ∫ 0 h f ( x ) 2 d x = π ∫ 0 h ( 2 r x − x 2 ) d x = π [ r x 2 − 1 3 x 3 ] 0 h = π h 2 3 ( 3 r − h ) {\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)}

The moments of inertia of a spherical cap (where the z-axis is the symmetrical axis) about the principal axes (center) of the sphere are:

J z z , cap = m h ( 3 h 2 − 15 h R + 20 R 2 ) 10 ( 3 R − h ) {\displaystyle J_{zz,{\text{cap}}}={\frac {mh\left(3h^{2}-15hR+20R^{2}\right)}{10\left(3R-h\right)}}}

J x x , cap = J y y , cap = m ( − 9 h 3 + 45 h 2 R − 80 h R 2 + 60 R 3 ) 20 ( 3 R − h ) {\displaystyle J_{xx,{\text{cap}}}=J_{yy,{\text{cap}}}={\frac {m\left(-9h^{3}+45h^{2}R-80hR^{2}+60R^{3}\right)}{20\left(3R-h\right)}}}

where m and h are, respectively, the mass and height of the spherical cap and R is the radius of the entire sphere.

Volumes of union and intersection of two intersecting spheres

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

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

where

V ( 1 ) = 4 π 3 r 1 3 + 4 π 3 r 2 3 {\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

V ( 2 ) = π h 1 2 3 ( 3 r 1 − h 1 ) + π h 2 2 3 ( 3 r 2 − h 2 ) {\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 d ≤ r 1 + r 2 {\displaystyle d\leq r_{1}+r_{2}} is the distance between the two sphere centers, elimination of the variables h 1 {\displaystyle h_{1}} and h 2 {\displaystyle h_{2}} leads to[4][5]

V ( 2 ) = π 12 d ( r 1 + r 2 − d ) 2 ( d 2 + 2 d ( r 1 + r 2 ) − 3 ( r 1 − r 2 ) 2 ) . {\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

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The volume of a spherical cap with a curved base can be calculated by considering two spheres with radii r 1 {\displaystyle r_{1}} and r 2 {\displaystyle r_{2}} , separated by some distance d {\displaystyle d} , and for which their surfaces intersect at x = h {\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 ( r 2 − r 1 ) − ( d − h ) {\displaystyle (r_{2}-r_{1})-(d-h)} ) and sphere 1's cap (with height h {\displaystyle h} ),

V = π h 2 3 ( 3 r 1 − h ) − π [ ( r 2 − r 1 ) − ( d − h ) ] 2 3 [ 3 r 2 − ( ( r 2 − r 1 ) − ( d − h ) ) ] , V = π h 2 3 ( 3 r 1 − h ) − π 3 ( d − h ) 3 ( r 2 − r 1 d − h − 1 ) 2 [ 2 r 2 + r 1 d − h + 1 ] . {\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 0 < d < r 2 {\displaystyle 0<d<r_{2}} and d − ( r 2 − r 1 ) < h ≤ r 1 {\displaystyle d-(r_{2}-r_{1})<h\leq r_{1}} . If sphere 2 is very large such that r 2 ≫ r 1 {\displaystyle r_{2}\gg r_{1}} , hence d ≫ h {\displaystyle d\gg h} and r 2 ≈ d {\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

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

| r 1 − r 2 | ≤ d ≤ r 1 + r 2 {\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 r 1 {\displaystyle r_{1}} is

cos ⁡ θ = r 1 2 − r 2 2 + d 2 2 r 1 d {\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 r 1 {\displaystyle r_{1}} is

A 1 = 2 π r 1 2 ( 1 + r 2 2 − r 1 2 − d 2 2 r 1 d ) {\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

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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 r {\displaystyle r} , and caps with heights h 1 {\displaystyle h_{1}} and h 2 {\displaystyle h_{2}} , the area is

A = 2 π r | h 1 − h 2 | , {\displaystyle A=2\pi r|h_{1}-h_{2}|\,,}

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

A = 2 π r 2 | sin ⁡ ϕ 1 − sin ⁡ ϕ 2 | , {\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 2_π_ ⋅ 63712 |sin 90° − sin 66.56°| = 21.04 million km2 (8.12 million mi2), 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.

Sections of other solids

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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.

Generally, the n {\displaystyle n} -dimensional volume of a hyperspherical cap of height h {\displaystyle h} and radius r {\displaystyle r} in n {\displaystyle n} -dimensional Euclidean space is given by:[8] V = π n − 1 2 r n Γ ( n + 1 2 ) ∫ 0 arccos ⁡ ( r − h r ) sin n ⁡ ( θ ) d θ {\displaystyle V={\frac {\pi ^{\frac {n-1}{2}}\,r^{n}}{\,\Gamma \left({\frac {n+1}{2}}\right)}}\int _{0}^{\arccos \left({\frac {r-h}{r}}\right)}\sin ^{n}(\theta )\,\mathrm {d} \theta } where Γ {\displaystyle \Gamma } (the gamma function) is given by Γ ( z ) = ∫ 0 ∞ t z − 1 e − t d t {\displaystyle \Gamma (z)=\int _{0}^{\infty }t^{z-1}\mathrm {e} ^{-t}\,\mathrm {d} t} .

The formula for V {\displaystyle V} can be expressed in terms of the volume of the unit n-ball C n = π n / 2 / Γ [ 1 + n 2 ] {\textstyle C_{n}=\pi ^{n/2}/\Gamma [1+{\frac {n}{2}}]} and the hypergeometric function 2 F 1 {\displaystyle {}_{2}F_{1}} or the regularized incomplete beta function I x ( a , b ) {\displaystyle I_{x}(a,b)} as V = C n r n ( 1 2 − r − h r Γ [ 1 + n 2 ] π Γ [ n + 1 2 ] 2 F 1 ( 1 2 , 1 − n 2 ; 3 2 ; ( r − h r ) 2 ) ) = 1 2 C n r n I ( 2 r h − h 2 ) / r 2 ( n + 1 2 , 1 2 ) , {\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 A {\displaystyle A} can be expressed in terms of the area of the unit n-ball A n = 2 π n / 2 / Γ [ n 2 ] {\textstyle A_{n}={2\pi ^{n/2}/\Gamma [{\frac {n}{2}}]}} as A = 1 2 A n r n − 1 I ( 2 r h − h 2 ) / r 2 ( n − 1 2 , 1 2 ) , {\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 0 ≤ h ≤ r {\displaystyle 0\leq h\leq r} .

A. Chudnov[9] derived the following formulas: A = A n r n − 1 p n − 2 ( q ) , V = C n r n p n ( q ) , {\displaystyle A=A_{n}r^{n-1}p_{n-2}(q),\,V=C_{n}r^{n}p_{n}(q),} where q = 1 − h / r ( 0 ≤ q ≤ 1 ) , p n ( q ) = ( 1 − G n ( q ) / G n ( 1 ) ) / 2 , {\displaystyle q=1-h/r(0\leq q\leq 1),p_{n}(q)=(1-G_{n}(q)/G_{n}(1))/2,} G n ( q ) = ∫ 0 q ( 1 − t 2 ) ( n − 1 ) / 2 d t . {\displaystyle G_{n}(q)=\int _{0}^{q}(1-t^{2})^{(n-1)/2}dt.}

For odd n = 2 k + 1 {\displaystyle n=2k+1} : G n ( q ) = ∑ i = 0 k ( − 1 ) i ( k i ) q 2 i + 1 2 i + 1 . {\displaystyle G_{n}(q)=\sum _{i=0}^{k}(-1)^{i}{\binom {k}{i}}{\frac {q^{2i+1}}{2i+1}}.}

If n → ∞ {\displaystyle n\to \infty } and q n = const. {\displaystyle q{\sqrt {n}}={\text{const.}}} , then p n ( q ) → 1 − F ( q n ) {\displaystyle p_{n}(q)\to 1-F({q{\sqrt {n}}})} where F {\displaystyle F} is the integral of the standard normal distribution.[10]

A more quantitative bound is A / ( A n r n − 1 ) = n Θ ( 1 ) ⋅ [ ( 2 − h / r ) h / r ] n / 2 {\displaystyle A/(A_{n}r^{n-1})=n^{\Theta (1)}\cdot [(2-h/r)h/r]^{n/2}} . For large caps (that is when ( 1 − h / r ) 4 ⋅ n = O ( 1 ) {\displaystyle (1-h/r)^{4}\cdot n=O(1)} as n → ∞ {\displaystyle n\to \infty } ), the bound simplifies to n Θ ( 1 ) ⋅ e − ( 1 − h / r ) 2 n / 2 {\displaystyle n^{\Theta (1)}\cdot e^{-(1-h/r)^{2}n/2}} .[11]

• Circular segment — the analogous 2D object
• Solid angle — contains formula for n-sphere caps
• Spherical segment
• Spherical sector
• Spherical wedge

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. Archived from the original on 2021-12-22. 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. ^ Li, S. (2011). "Concise Formulas for the Area and Volume of a Hyperspherical Cap". Asian Journal of Mathematics and Statistics: 66–70. doi:10.3923/ajms.2011.66.70.
9. ^ Chudnov, Alexander M. (1986). "On minimax signal generation and reception algorithms (engl. transl.)". Problems of Information Transmission. 22 (4): 49–54.
10. ^ Chudnov, Alexander M (1991). "Game-theoretical problems of synthesis of signal generation and reception algorithms (engl. transl.)". Problems of Information Transmission. 27 (3): 57–65.
11. ^ Becker, Anja; Ducas, Léo; Gama, Nicolas; Laarhoven, Thijs (10 January 2016). Krauthgamer, Robert (ed.). New directions in nearest neighbor searching with applications to lattice sieving. Twenty-seventh Annual ACM-SIAM Symposium on Discrete Algorithms (SODA '16), Arlington, Virginia. Philadelphia: Society for Industrial and Applied Mathematics. pp. 10–24. ISBN 978-1-61197-433-1.

• Richmond, Timothy J. (1984). "Solvent accessible surface area and excluded volume in proteins: Analytical equation for overlapping spheres and implications for the hydrophobic effect". Journal of Molecular Biology. 178 (1): 63–89. doi:10.1016/0022-2836(84)90231-6. PMID 6548264.
• Lustig, Rolf (1986). "Geometry of four hard fused spheres in an arbitrary spatial configuration". Molecular Physics. 59 (2): 195–207. Bibcode:1986MolPh..59..195L. doi:10.1080/00268978600102011.
• Gibson, K. D.; Scheraga, Harold A. (1987). "Volume of the intersection of three spheres of unequal size: a simplified formula". The Journal of Physical Chemistry. 91 (15): 4121–4122. doi:10.1021/j100299a035.
• Gibson, K. D.; Scheraga, Harold A. (1987). "Exact calculation of the volume and surface area of fused hard-sphere molecules with unequal atomic radii". Molecular Physics. 62 (5): 1247–1265. Bibcode:1987MolPh..62.1247G. doi:10.1080/00268978700102951.
• Petitjean, Michel (1994). "On the analytical calculation of van der Waals surfaces and volumes: some numerical aspects". Journal of Computational Chemistry. 15 (5): 507–523. doi:10.1002/jcc.540150504.
• Grant, J. A.; Pickup, B. T. (1995). "A Gaussian description of molecular shape". The Journal of Physical Chemistry. 99 (11): 3503–3510. doi:10.1021/j100011a016.
• Busa, Jan; Dzurina, Jozef; Hayryan, Edik; Hayryan, Shura (2005). "ARVO: A fortran package for computing the solvent accessible surface area and the excluded volume of overlapping spheres via analytic equations". Computer Physics Communications. 165 (1): 59–96. Bibcode:2005CoPhC.165...59B. doi:10.1016/j.cpc.2004.08.002.
• Strobl, Severin; Formella, Arno; Pöschel, Thorsten (2016). "Exact calculation of the overlap volume of spheres and mesh elements". J. Comput. Phys. 311: 158–172. doi:10.1016/j.jcp.2016.02.003. Provides formulas for intersection sphere caps.
• Weisstein, Eric W. "Spherical cap". MathWorld. Derivation and some additional formulas.
• Online calculator for spherical cap volume and area.
• Summary of spherical formulas.