Question: What is the formula to calculate the volume of a sphere?

Spherical cap


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

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
≤
2
r


{\displaystyle 0\leq h\leq 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:





r

2


=
(
r
−
h

)

2


+

a

2


=

r

2


+

h

2


−
2
r
h
+

a

2



,


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


so that




r
=




a

2


+

h

2




2
h




.


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


Substituting this into the formulas gives:




V
=



π

h

2



3



(




3

a

2


+
3

h

2




2
h



−
h

)

=


1
6


π
h
(
3

a

2


+

h

2


)

,


{\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})\,,}






A
=
2
π



(

a

2


+

h

2


)


2
h



h
=
π
(

a

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 




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

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)}


Applications
Volumes of union and intersection of two intersecting spheres
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)\,.}


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 




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 6371km, 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
=




π



n
−
1

2





r

n





Γ

(



n
+
1

2


)





∫

0


arccos
⁡

(



r
−
h

r


)




sin

n


⁡
(
t
)


d

t


{\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


∞



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


]




{\displaystyle C_{n}={\scriptstyle \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


]




{\displaystyle A_{n}={\scriptstyle 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 \scriptstyle 0\leq h\leq r}

.
Earlier in [9] (1986, USSR Academ. Press) the following formulas were derived: 




A
=

A

n



p

n
−
2


(
q
)
,
V
=

C

n



p

n


(
q
)


{\displaystyle A=A_{n}p_{n-2}(q),V=C_{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 \limits _{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}}}

.
Asymptotics
It is shown in [10] that, 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.
A more quantitative way of writing this, is in [11] where the bound is




A

/


A

n


=

n

Θ
(
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 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}}

.
See also

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


Further reading



 Derivation and some additional formulas.
.
.