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

In geometry, a spherical cap, spherical dome, or spherical segment of one base is a portion of a sphere cut off by a 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 r{displaystyle r} of the sphere


• The radius a{displaystyle a} of the base of the cap


• The height h{displaystyle h} of the cap


• The polar angle θ{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 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=πh23(3r−h){displaystyle V={frac {pi h^{2}}{3}}(3r-h)} [1]	V=16πh(3a2+h2){displaystyle V={frac {1}{6}}pi h(3a^{2}+h^{2})}	V=π3r3(2+cos⁡θ)(1−cos⁡θ)2{displaystyle V={frac {pi }{3}}r^{3}(2+cos theta )(1-cos theta )^{2}}
Area	A=2πrh{displaystyle A=2pi rh}[1]	A=π(a2+h2){displaystyle A=pi (a^{2}+h^{2})}	A=2πr2(1−cos⁡θ){displaystyle A=2pi r^{2}(1-cos theta )}

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

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

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

r2=(r−h)2+a2=r2+h2−2rh+a2,{displaystyle r^{2}=(r-h)^{2}+a^{2}=r^{2}+h^{2}-2rh+a^{2},,}

so that

r=a2+h22h.{displaystyle r={frac {a^{2}+h^{2}}{2h}},.}

Substituting this into the formulas gives:

V=πh23(3a2+3h22h−h)=16πh(3a2+h2),{displaystyle V={frac {pi h^{2}}{3}}left({frac {3a^{2}+3h^{2}}{2h}}-hright)={frac {1}{6}}pi h(3a^{2}+h^{2}),,}

A=2π(a2+h2)2hh=π(a2+h2).{displaystyle A=2pi {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 Vsec{displaystyle V_{sec}} of the spherical sector, by an intuitive argument,^[2] as

A=3rVsec=3r2πr2h3=2πrh.{displaystyle A={frac {3}{r}}V_{sec}={frac {3}{r}}{frac {2pi r^{2}h}{3}}=2pi 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=13bh′{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:

Vsec=∑V=∑13bh′=∑13br=r3∑b=r3A{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 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)=r2−(x−r)2=2rx−x2{displaystyle f(x)={sqrt {r^{2}-(x-r)^{2}}}={sqrt {2rx-x^{2}}}}

for x∈[0,h]{displaystyle xin [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=2π∫0hf(x)1+f′(x)2dx{displaystyle A=2pi int _{0}^{h}f(x){sqrt {1+f'(x)^{2}}},dx}

The derivative of f{displaystyle f} is

f′(x)=r−x2rx−x2{displaystyle f'(x)={frac {r-x}{sqrt {2rx-x^{2}}}}}

and hence

1+f′(x)2=r22rx−x2{displaystyle 1+f'(x)^{2}={frac {r^{2}}{2rx-x^{2}}}}

The formula for the area is therefore

A=2π∫0h2rx−x2r22rx−x2dx=2π∫0hrdx=2πr[x]0h=2πrh{displaystyle A=2pi int _{0}^{h}{sqrt {2rx-x^{2}}}{sqrt {frac {r^{2}}{2rx-x^{2}}}},dx=2pi int _{0}^{h}r,dx=2pi rleft[xright]_{0}^{h}=2pi rh}

The volume is

V=π∫0hf(x)2dx=π∫0h(2rx−x2)dx=π[rx2−13x3]0h=πh23(3r−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)}

Applications

Volumes of union and intersection of two intersecting spheres

The volume of the union of two intersecting spheres
of radii r1{displaystyle r_{1}} and r2{displaystyle r_{2}} is
^[3]

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

where

V(1)=4π3r13+4π3r23{displaystyle V^{(1)}={frac {4pi }{3}}r_{1}^{3}+{frac {4pi }{3}}r_{2}^{3}}

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

V(2)=πh123(3r1−h1)+πh223(3r2−h2){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≤r1+r2{displaystyle dleq r_{1}+r_{2}} is the
distance between the two sphere centers, elimination of the variables h1{displaystyle h_{1}} and h2{displaystyle h_{2}} leads
to^[4][5]

V(2)=π12d(r1+r2−d)2(d2+2d(r1+r2)−3(r1−r2)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),.}

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 r{displaystyle r}, and caps with heights h1{displaystyle h_{1}} and h2{displaystyle h_{2}}, the area is

A=2πr|h1−h2|,{displaystyle A=2pi r|h_{1}-h_{2}|,,}

or, using geographic coordinates with latitudes ϕ1{displaystyle phi _{1}} and ϕ2{displaystyle phi _{2}},^[6]

A=2πr2|sin⁡ϕ1−sin⁡ϕ2|,{displaystyle A=2pi 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 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=πn−12rnΓ(n+12)∫0arccos⁡(r−hr)sinn⁡(t)dt{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 Γ{displaystyle Gamma } (the gamma function) is given by Γ(z)=∫0∞tz−1e−tdt{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 Cn=πn/2/Γ[1+n2]{displaystyle C_{n}={scriptstyle pi ^{n/2}/Gamma [1+{frac {n}{2}}]}} and the hypergeometric function 2F1{displaystyle {}_{2}F_{1}} or the regularized incomplete beta function Ix(a,b){displaystyle I_{x}(a,b)} as

V=Cnrn(12−r−hrΓ[1+n2]πΓ[n+12]2F1(12,1−n2;32;(r−hr)2))=12CnrnI(2rh−h2)/r2(n+12,12){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 An=2πn/2/Γ[n2]{displaystyle A_{n}={scriptstyle 2pi ^{n/2}/Gamma [{frac {n}{2}}]}} as

A=12Anrn−1I(2rh−h2)/r2(n−12,12){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 scriptstyle 0leq hleq r}.

Earlier in ^[9] (1986, USSR Academ. Press) the following formulas were derived:
A=Anpn−2(q),V=Cnpn(q){displaystyle A=A_{n}p_{n-2}(q),V=C_{n}p_{n}(q)}, where
q=1−h/r(0≤q≤1),pn(q)=(1−Gn(q)/Gn(1))/2{displaystyle q=1-h/r(0leq qleq 1),p_{n}(q)=(1-G_{n}(q)/G_{n}(1))/2},

Gn(q)=∫0q(1−t2)(n−1)/2dt{displaystyle G_{n}(q)=int limits _{0}^{q}(1-t^{2})^{(n-1)/2}dt}.

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

Gn(q)=∑i=0k(−1)i(ki)q2i+12i+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 ^[10] that, if n→∞{displaystyle nto infty } and qn=const.{displaystyle q{sqrt {n}}={text{const.}}}, then pn(q)→1−F(qn){displaystyle p_{n}(q)to 1-F({q{sqrt {n}}})} where F(){displaystyle F()} is the integral of the standard normal distribution.

A more quantitative way of writing this, is in ^[11] where the bound is
A/An=nΘ(1)⋅[(2−h/r)h/r]n/2{displaystyle A/A_{n}=n^{Theta (1)}cdot [(2-h/r)h/r]^{n/2}} is given.
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 nto infty }), the bound simplifies to nΘ(1)⋅e−(1−h/r)2n/2{displaystyle n^{Theta (1)}cdot e^{-(1-h/r)^{2}n/2}}.

See also

• 
  Circular segment — the analogous 2D object


• 
  Solid angle — contains formula for n-sphere caps


• Spherical segment


• Spherical sector


• Spherical wedge

References

Further reading

• 
  Richmond, Timothy J. (1984). "Solvent accessible surface area and excluded volume in proteins: Analytical equation for overlapping spheres and implications for the hydrophobic effect". J. Mol. Biol. 178 (1): 63–89. doi:10.1016/0022-2836(84)90231-6.
  


• 
  Lustig, Rolf (1986). "Geometry of four hard fused spheres in an arbitrary spatial configuration". Mol. Phys. 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". J. Phys. Chem. 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". Mol. Phys. 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". Int. J. Quantum Chem. 15 (5): 507–523. doi:10.1002/jcc.540150504.
  


• 
  Grant, J. A.; Pickup, B. T. (1995). "A Gaussian description of molecular shape". J. Phys. Chem. 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". Comput. Phys. Commun. 165: 59–96. Bibcode:2005CoPhC.165...59B. doi:10.1016/j.cpc.2004.08.002.
  

External links

Wikimedia Commons has media related to Spherical caps.

• 
  Weisstein, Eric W. "Spherical cap". MathWorld. Derivation and some additional formulas.


• 
  Online calculator for spherical cap volume and area.


• 
  Summary of spherical formulas.