Volume of an n-ball

In geometry, a ball is a region in a space comprising all points within a fixed distance, called the radius, from a given point; that is, it is the region enclosed by a sphere or hypersphere. An n-ball is a ball in an n-dimensional Euclidean space. The volume of a n-ball is the Lebesgue measure of this ball, which generalizes to any dimension the usual volume of a ball in 3-dimensional space. The volume of a n-ball of radius R is ${\displaystyle R^{n}V_{n},}$ where ${\displaystyle V_{n}}$ is the volume of the unit n-ball, the n-ball of radius 1.

The real number ${\displaystyle V_{n}}$ can be expressed by a expression involving the gamma function. It can be computed with several recurrence relations. It can be expressed in terms of ${\displaystyle A_{n},}$ the area of the unit n-sphere.

Formulas[]

The volume[]

The n-dimensional volume of a Euclidean ball of radius R in n-dimensional Euclidean space is:^[1]

${\displaystyle V_{n}(R)={\frac {\pi ^{n/2}}{\Gamma {\bigl (}{\tfrac {n}{2}}+1{\bigr )}}}R^{n},}$

where Γ is Euler's gamma function. The gamma function extends the factorial function to non-integer arguments. It satisfies Γ(n) = (n − 1)! if n is a positive integer and Γ(n + 1/2) = (n − 1/2) · (n − 3/2) · … · 1/2 · π^1/2 if n is a non-negative integer.

Alternative forms[]

Using explicit formulas for particular values of the gamma function at the integers and half-integers gives formulas for the volume of a Euclidean ball in terms of factorials. These are:

${\displaystyle {\begin{aligned}V_{2k}(R)&={\frac {\pi ^{k}}{k!}}R^{2k},\\V_{2k+1}(R)&={\frac {2(k!)(4\pi )^{k}}{(2k+1)!}}R^{2k+1}.\end{aligned}}}$

The volume can also be expressed in terms of double factorials. For an odd integer 2k + 1, the double factorial is defined by

${\displaystyle (2k+1)!!=(2k+1)\cdot (2k-1)\dotsm 5\cdot 3\cdot 1.}$

The volume of an odd-dimensional ball is

${\displaystyle V_{2k+1}(R)={\frac {2(2\pi )^{k}}{(2k+1)!!}}R^{2k+1}.}$

There are multiple conventions for double factorials of even integers. Under the convention in which the double factorial satisfies

${\displaystyle (2k)!!=(2k)\cdot (2k-2)\dotsm 4\cdot 2\cdot {\sqrt {2/\pi }}=2^{k}\cdot k!\cdot {\sqrt {2/\pi }},}$

the volume of an n-dimensional ball is, regardless of whether n is even or odd,

${\displaystyle V_{n}(R)={\frac {2(2\pi )^{(n-1)/2}}{n!!}}R^{n}.}$

Instead of expressing the volume V of the ball in terms of its radius R, the formulas can be inverted to express the radius as a function of the volume:

${\displaystyle {\begin{aligned}R_{n}(V)&={\frac {\Gamma {\bigl (}{\tfrac {n}{2}}+1{\bigr )}^{1/n}}{\sqrt {\pi }}}V^{1/n}\\&=\left({\frac {n!!V}{2(2\pi )^{(n-1)/2}}}\right)^{1/n}\\R_{2k}(V)&={\frac {(k!V)^{1/(2k)}}{\sqrt {\pi }}},\\R_{2k+1}(V)&=\left({\frac {(2k+1)!V}{2(k!)(4\pi )^{k}}}\right)^{1/(2k+1)}.\end{aligned}}}$

Recurrence relations[]

The volume V of an n-ball of radius R can be expressed recursively in terms of the volume of an (n − 2)-ball, via the two-term recurrence relation:

${\displaystyle {\begin{aligned}V_{0}(R)&=1,\\[2mu]V_{1}(R)&=2R,\\V_{n}(R)&={\frac {2\pi }{n}}R^{2}V_{n-2}(R).\end{aligned}}}$

The volume can also be expressed in terms of an (n − 1)-ball using the one-term recurrence relation:

${\displaystyle {\begin{aligned}V_{0}(R)&=1,\\V_{n}(R)&={\frac {\Gamma {\bigl (}{\tfrac {n}{2}}+{\tfrac {1}{2}}{\bigr )}{\sqrt {\pi }}}{\Gamma {\bigl (}{\tfrac {n}{2}}+1{\bigr )}}}R\,V_{n-1}(R).\end{aligned}}}$

Inverting the above, the radius of an n-ball of volume V can be expressed recursively in terms of the radius of an (n − 2)- or (n − 1)-ball:

${\displaystyle {\begin{aligned}R_{n}(V)&={\bigl (}{\tfrac {1}{2}}n{\bigr )}^{1/n}\left(\Gamma {\bigl (}{\tfrac {n}{2}}{\bigr )}V\right)^{-2/(n(n-2))}R_{n-2}(V),\\R_{n}(V)&={\frac {\Gamma {\bigl (}{\tfrac {n}{2}}+1{\bigr )}^{1/n}}{\Gamma {\bigl (}{\tfrac {n}{2}}+{\tfrac {1}{2}}{\bigr )}^{1/(n-1)}}}V^{-1/(n(n-1))}R_{n-1}(V).\end{aligned}}}$

Low dimensions[]

In low dimensions, these volume and radius formulas simplify to the following.

Dimension	Volume of a ball of radius R	Radius of a ball of volume V
0	${\displaystyle 1}$	(all 0-balls have volume 1)
1	${\displaystyle 2R}$	${\displaystyle {\frac {V}{2}}=0.5\times V}$
2	${\displaystyle \pi R^{2}\approx 3.142\times R^{2}}$	${\displaystyle {\frac {V^{1/2}}{\sqrt {\pi }}}\approx 0.564\times V^{\frac {1}{2}}}$
3	${\displaystyle {\frac {4\pi }{3}}R^{3}\approx 4.189\times R^{3}}$	${\displaystyle \left({\frac {3V}{4\pi }}\right)^{1/3}\approx 0.620\times V^{1/3}}$
4	${\displaystyle {\frac {\pi ^{2}}{2}}R^{4}\approx 4.935\times R^{4}}$	${\displaystyle {\frac {(2V)^{1/4}}{\sqrt {\pi }}}\approx 0.671\times V^{1/4}}$
5	${\displaystyle {\frac {8\pi ^{2}}{15}}R^{5}\approx 5.264\times R^{5}}$	${\displaystyle \left({\frac {15V}{8\pi ^{2}}}\right)^{1/5}\approx 0.717\times V^{1/5}}$
6	${\displaystyle {\frac {\pi ^{3}}{6}}R^{6}\approx 5.168\times R^{6}}$	${\displaystyle {\frac {(6V)^{1/6}}{\sqrt {\pi }}}\approx 0.761\times V^{1/6}}$
7	${\displaystyle {\frac {16\pi ^{3}}{105}}R^{7}\approx 4.725\times R^{7}}$	${\displaystyle \left({\frac {105V}{16\pi ^{3}}}\right)^{1/7}\approx 0.801\times V^{1/7}}$
8	${\displaystyle {\frac {\pi ^{4}}{24}}R^{8}\approx 4.059\times R^{8}}$	${\displaystyle {\frac {(24V)^{1/8}}{\sqrt {\pi }}}\approx 0.839\times V^{1/8}}$
9	${\displaystyle {\frac {32\pi ^{4}}{945}}R^{9}\approx 3.299\times R^{9}}$	${\displaystyle \left({\frac {945V}{32\pi ^{4}}}\right)^{1/9}\approx 0.876\times V^{1/9}}$
10	${\displaystyle {\frac {\pi ^{5}}{120}}R^{10}\approx 2.550\times R^{10}}$	${\displaystyle {\frac {(120V)^{1/10}}{\sqrt {\pi }}}\approx 0.911\times V^{1/10}}$
11	${\displaystyle {\frac {64\pi ^{5}}{10395}}R^{11}\approx 1.884\times R^{11}}$	${\displaystyle \left({\frac {10395V}{64\pi ^{5}}}\right)^{1/11}\approx 0.944\times V^{1/11}}$
12	${\displaystyle {\frac {\pi ^{6}}{720}}R^{12}\approx 1.335\times R^{12}}$	${\displaystyle {\frac {(720V)^{1/12}}{\sqrt {\pi }}}\approx 0.976\times V^{1/12}}$
13	${\displaystyle {\frac {128\pi ^{6}}{135135}}R^{13}\approx 0.911\times R^{13}}$	${\displaystyle \left({\frac {135135V}{128\pi ^{6}}}\right)^{1/13}\approx 1.007\times V^{1/13}}$
14	${\displaystyle {\frac {\pi ^{7}}{5040}}R^{14}\approx 0.599\times R^{14}}$	${\displaystyle {\frac {(5040V)^{1/14}}{\sqrt {\pi }}}\approx 1.037\times V^{1/14}}$
15	${\displaystyle {\frac {256\pi ^{7}}{2027025}}R^{15}\approx 0.381\times R^{15}}$	${\displaystyle \left({\frac {2027025V}{256\pi ^{7}}}\right)^{1/15}\approx 1.066\times V^{1/15}}$
n	Vn(R)	Rn(V)

High dimensions[]

Stirling's formula for the gamma function can be used to approximate the volume when the number of dimensions is high.

${\displaystyle V_{n}(R)\sim {\frac {1}{\sqrt {n\pi }}}\left({\frac {2\pi e}{n}}\right)^{n/2}R^{n}.}$

${\displaystyle R_{n}(V)\sim (\pi n)^{1/(2n)}{\sqrt {\frac {n}{2\pi e}}}V^{1/n}.}$

In particular, for any fixed value of R the volume tends to a limiting value of 0 as n goes to infinity. For example, the volume V_n(1) is increasing for 0 ≤ n ≤ 5, achieves its maximum when n = 5, and is decreasing for n ≥ 5.^[2]

Relation with surface area[]

Let A_{n − 1}(R) denote the hypervolume of the (n − 1)-sphere of radius R. The (n − 1)-sphere is the (n − 1)-dimensional boundary (surface) of the n-dimensional ball of radius R, and the sphere's hypervolume and the ball's hypervolume are related by:

${\displaystyle A_{n-1}(R)={\frac {d}{dR}}V_{n}(R)={\frac {n}{R}}V_{n}(R).}$

Thus, A_{n − 1}(R) inherits formulas and recursion relationships from V_n(R), such as

${\displaystyle A_{n-1}(R)={\frac {2\pi ^{n/2}}{\Gamma {\bigl (}{\tfrac {n}{2}}{\bigr )}}}R^{n-1}.}$

There are also formulas in terms of factorials and double factorials.

Proofs[]

There are many proofs of the above formulas.

The volume is proportional to the nth power of the radius[]

An important step in several proofs about volumes of n-balls, and a generally useful fact besides, is that the volume of the n-ball of radius R is proportional to Rⁿ:

${\displaystyle V_{n}(R)\propto R^{n}.}$

The proportionality constant is the volume of the unit ball.

This is a special case of a general fact about volumes in n-dimensional space: If K is a body (measurable set) in that space and RK is the body obtained by stretching in all directions by the factor R then the volume of RK equals Rⁿ times the volume of K. This is a direct consequence of the change of variables formula:

${\displaystyle V(RK)=\int _{RK}dx=\int _{K}R^{n}\,dy=R^{n}V(K)}$

where dx = dx₁…dx_n and the substitution x = Ry was made.

Another proof of the above relation, which avoids multi-dimensional integration, uses induction: The base case is n = 0, where the proportionality is obvious. For the inductive step, assume that proportionality is true in dimension n − 1. Note that the intersection of an n-ball with a hyperplane is an (n − 1)-ball. When the volume of the n-ball is written as an integral of volumes of (n − 1)-balls:

${\displaystyle V_{n}(R)=\int _{-R}^{R}V_{n-1}\!\left({\sqrt {R^{2}-x^{2}}}\right)dx,}$

it is possible by the inductive hypothesis to remove a factor of R from the radius of the (n − 1)-ball to get:

${\displaystyle V_{n}(R)=R^{n-1}\!\int _{-R}^{R}V_{n-1}\!\left({\sqrt {1-\left({\frac {x}{R}}\right)^{2}}}\right)dx.}$

Making the change of variables t = x/R leads to:

${\displaystyle V_{n}(R)=R^{n}\!\int _{-1}^{1}V_{n-1}\!\left({\sqrt {1-t^{2}}}\right)dt=R^{n}V_{n}(1),}$

which demonstrates the proportionality relation in dimension n. By induction, the proportionality relation is true in all dimensions.

The two-dimension recursion formula[]

A proof of the recursion formula relating the volume of the n-ball and an (n − 2)-ball can be given using the proportionality formula above and integration in cylindrical coordinates. Fix a plane through the center of the ball. Let r denote the distance between a point in the plane and the center of the sphere, and let θ denote the azimuth. Intersecting the n-ball with the (n − 2)-dimensional plane defined by fixing a radius and an azimuth gives an (n − 2)-ball of radius √R² − r². The volume of the ball can therefore be written as an iterated integral of the volumes of the (n − 2)-balls over the possible radii and azimuths:

${\displaystyle V_{n}(R)=\int _{0}^{2\pi }\int _{0}^{R}V_{n-2}\!\left({\sqrt {R^{2}-r^{2}}}\right)r\,dr\,d\theta ,}$

The azimuthal coordinate can be immediately integrated out. Applying the proportionality relation shows that the volume equals

${\displaystyle V_{n}(R)=2\pi V_{n-2}(R)\int _{0}^{R}\left(1-\left({\frac {r}{R}}\right)^{2}\right)^{(n-2)/2}\,r\,dr.}$

The integral can be evaluated by making the substitution u = 1 − (r/R)²
to get

${\displaystyle {\begin{aligned}V_{n}(R)&=2\pi V_{n-2}(R)\cdot \left[-{\frac {R^{2}}{n}}\left(1-\left({\frac {r}{R}}\right)^{2}\right)^{n/2}\right]_{r=0}^{r=R}\\&={\frac {2\pi R^{2}}{n}}V_{n-2}(R),\end{aligned}}}$

which is the two-dimension recursion formula.

The same technique can be used to give an inductive proof of the volume formula. The base cases of the induction are the 0-ball and the 1-ball, which can be checked directly using the facts Γ(1) = 1 and Γ(3/2) = 1/2 · Γ(1/2) = √π/2. The inductive step is similar to the above, but instead of applying proportionality to the volumes of the (n − 2)-balls, the inductive hypothesis is applied instead.

The one-dimension recursion formula[]

The proportionality relation can also be used to prove the recursion formula relating the volumes of an n-ball and an (n − 1)-ball. As in the proof of the proportionality formula, the volume of an n-ball can be written as an integral over the volumes of (n − 1)-balls. Instead of making a substitution, however, the proportionality relation can be applied to the volumes of the (n − 1)-balls in the integrand:

${\displaystyle V_{n}(R)=V_{n-1}(R)\int _{-R}^{R}\left(1-\left({\frac {x}{R}}\right)^{2}\right)^{(n-1)/2}\,dx.}$

The integrand is an even function, so by symmetry the interval of integration can be restricted to [0, R]. On the interval [0, R], it is possible to apply the substitution u = (x/R)²
. This transforms the expression into

${\displaystyle V_{n-1}(R)\cdot R\cdot \int _{0}^{1}(1-u)^{(n-1)/2}u^{-{\frac {1}{2}}}\,du}$

The integral is a value of a well-known special function called the beta function Β(x, y), and the volume in terms of the beta function is

${\displaystyle V_{n}(R)=V_{n-1}(R)\cdot R\cdot \mathrm {B} {\bigl (}{\tfrac {n+1}{2}},{\tfrac {1}{2}}{\bigr )}.}$

The beta function can be expressed in terms of the gamma function in much the same way that factorials are related to binomial coefficients. Applying this relationship gives

${\displaystyle V_{n}(R)=V_{n-1}(R)\cdot R\cdot {\frac {\Gamma {\bigl (}{\tfrac {n}{2}}+{\tfrac {1}{2}}{\bigr )}\Gamma {\bigl (}{\tfrac {1}{2}}{\bigr )}}{\Gamma {\bigl (}{\tfrac {n}{2}}+1{\bigr )}}}.}$

Using the value Γ(1/2) = √π gives the one-dimension recursion formula:

${\displaystyle V_{n}(R)=R{\sqrt {\pi }}{\frac {\Gamma {\bigl (}{\tfrac {n}{2}}+{\tfrac {1}{2}}{\bigr )}}{\Gamma {\bigl (}{\tfrac {n}{2}}+1{\bigr )}}}V_{n-1}(R).}$

As with the two-dimension recursive formula, the same technique can be used to give an inductive proof of the volume formula.

Direct integration in spherical coordinates[]

The volume of the n-ball ${\displaystyle V_{n}(R)}$ can be computed by integrating the volume element in spherical coordinates. The spherical coordinate system has a radial coordinate r and angular coordinates φ₁, …, φ_n − 1, where the domain of each φ except φ_n − 1 is [0, π), and the domain of φ_n − 1 is [0, 2π). The spherical volume element is:

${\displaystyle dV=r^{n-1}\sin ^{n-2}(\varphi _{1})\sin ^{n-3}(\varphi _{2})\cdots \sin(\varphi _{n-2})\,dr\,d\varphi _{1}\,d\varphi _{2}\cdots d\varphi _{n-1},}$

and the volume is the integral of this quantity over r between 0 and R and all possible angles:

${\displaystyle V_{n}(R)=\int _{0}^{R}\int _{0}^{\pi }\cdots \int _{0}^{2\pi }r^{n-1}\sin ^{n-2}(\varphi _{1})\cdots \sin(\varphi _{n-2})\,d\varphi _{n-1}\cdots d\varphi _{1}\,dr.}$

Each of the factors in the integrand depends on only a single variable, and therefore the iterated integral can be written as a product of integrals:

${\displaystyle V_{n}(R)=\left(\int _{0}^{R}r^{n-1}\,dr\right)\!\left(\int _{0}^{\pi }\sin ^{n-2}(\varphi _{1})\,d\varphi _{1}\right)\cdots \left(\int _{0}^{2\pi }d\varphi _{n-1}\right).}$

The integral over the radius is Rⁿ/n. The intervals of integration on the angular coordinates can, by the symmetry of the sine about π/2, be changed to [0, π/2]:

${\displaystyle V_{n}(R)={\frac {R^{n}}{n}}\left(2\int _{0}^{\pi /2}\sin ^{n-2}(\varphi _{1})\,d\varphi _{1}\right)\cdots \left(4\int _{0}^{\pi /2}d\varphi _{n-1}\right).}$

Each of the remaining integrals is now a particular value of the beta function:

${\displaystyle V_{n}(R)={\frac {R^{n}}{n}}\mathrm {B} {\bigl (}{\tfrac {n-1}{2}},{\tfrac {1}{2}}{\bigr )}\mathrm {B} {\bigl (}{\tfrac {n-2}{2}},{\tfrac {1}{2}}{\bigr )}\cdots \mathrm {B} {\bigl (}1,{\tfrac {1}{2}}{\bigr )}\cdot 2\,\mathrm {B} {\bigl (}{\tfrac {1}{2}},{\tfrac {1}{2}}{\bigr )}.}$

The beta functions can be rewritten in terms of gamma functions:

${\displaystyle V_{n}(R)={R^{n}}{n}\cdot {\frac {\Gamma {\bigl (}{\tfrac {n}{2}}-{\tfrac {1}{2}}{\bigr )}\Gamma {\bigl (}{\tfrac {1}{2}}{\bigr )}}{\Gamma {\bigl (}{\tfrac {n}{2}}{\bigr )}}}\cdot {\frac {\Gamma {\bigl (}{\tfrac {n}{2}}-1{\bigr )}\Gamma {\bigl (}{\tfrac {1}{2}}{\bigr )}}{\Gamma {\bigl (}{\tfrac {n}{2}}-{\tfrac {1}{2}}{\bigr )}}}\cdots {\frac {\Gamma (1)\Gamma {\bigl (}{\tfrac {1}{2}}{\bigr )}}{\Gamma {\bigl (}{\tfrac {3}{2}}{\bigr )}}}\cdot 2{\frac {\Gamma {\bigl (}{\tfrac {1}{2}}{\bigr )}\Gamma {\bigl (}{\tfrac {1}{2}}{\bigr )}}{\Gamma (1)}}.}$

This product telescopes. Combining this with the values Γ(1/2) = √π and Γ(1) = 1 and the functional equation zΓ(z) = Γ(z + 1) leads to

${\displaystyle V_{n}(R)={\frac {2\pi ^{n/2}R^{n}}{n\,\Gamma {\bigl (}{\tfrac {n}{2}}{\bigr )}}}={\frac {\pi ^{n/2}R^{n}}{\Gamma {\bigl (}{\tfrac {n}{2}}+1{\bigr )}}}.}$

Gaussian integrals[]

The volume formula can be proven directly using Gaussian integrals. Consider the function:

${\displaystyle f(x_{1},\ldots ,x_{n})=\exp {\biggl (}{-{\tfrac {1}{2}}\sum _{i=1}^{n}x_{i}^{2}}{\biggr )}.}$

This function is both rotationally invariant and a product of functions of one variable each. Using the fact that it is a product and the formula for the Gaussian integral gives:

${\displaystyle \int _{\mathbf {R} ^{n}}f\,dV=\prod _{i=1}^{n}\left(\int _{-\infty }^{\infty }\exp \left(-{\tfrac {1}{2}}x_{i}^{2}\right)\,dx_{i}\right)=(2\pi )^{n/2},}$

where dV is the n-dimensional volume element. Using rotational invariance, the same integral can be computed in spherical coordinates:

${\displaystyle \int _{\mathbf {R} ^{n}}f\,dV=\int _{0}^{\infty }\int _{S^{n-1}(r)}\exp \left(-{\tfrac {1}{2}}r^{2}\right)\,dA\,dr,}$

where S^{n − 1}(r) is an (n − 1)-sphere of radius r and dA is the area element (equivalently, the (n − 1)-dimensional volume element). The surface area of the sphere satisfies a proportionality equation similar to the one for the volume of a ball: If A_{n − 1}(r) is the surface area of an (n − 1)-sphere of radius r, then:

${\displaystyle A_{n-1}(r)=r^{n-1}A_{n-1}(1).}$

Applying this to the above integral gives the expression

${\displaystyle (2\pi )^{n/2}=\int _{0}^{\infty }\int _{S^{n-1}(r)}\exp \left(-{\tfrac {1}{2}}r^{2}\right)\,dA\,dr=A_{n-1}(1)\int _{0}^{\infty }\exp \left(-{\tfrac {1}{2}}r^{2}\right)\,r^{n-1}\,dr.}$

Substituting t = r²/2:

${\displaystyle \int _{0}^{\infty }\exp \left(-{\tfrac {1}{2}}r^{2}\right)\,r^{n-1}\,dr=2^{(n-2)/2}\int _{0}^{\infty }e^{-t}t^{(n-2)/2}\,dt.}$

The integral on the right is the gamma function evaluated at n/2.

Combining the two results shows that

${\displaystyle A_{n-1}(1)={\frac {2\pi ^{n/2}}{\Gamma {\bigl (}{\tfrac {n}{2}}{\bigr )}}}.}$

To derive the volume of an n-ball of radius R from this formula, integrate the surface area of a sphere of radius r for 0 ≤ r ≤ R and apply the functional equation zΓ(z) = Γ(z + 1):

${\displaystyle V_{n}(R)=\int _{0}^{R}{\frac {2\pi ^{n/2}}{\Gamma {\bigl (}{\tfrac {n}{2}}{\bigr )}}}\,r^{n-1}\,dr={\frac {2\pi ^{n/2}}{n\,\Gamma {\bigl (}{\tfrac {n}{2}}{\bigr )}}}R^{n}={\frac {\pi ^{n/2}}{\Gamma {\bigl (}{\tfrac {n}{2}}+1{\bigr )}}}R^{n}.}$

Geometric proof[]

The relations ${\displaystyle V_{n+1}(R)={\frac {R}{n+1}}A_{n}(R)}$ and ${\displaystyle A_{n+1}(R)=(2\pi R)V_{n}(R)}$ and thus the volumes of n-balls and areas of n-spheres can also be derived geometrically. As noted above, because a ball of radius ${\displaystyle R}$ is obtained from a unit ball ${\displaystyle B_{n}}$ by rescaling all directions in ${\displaystyle R}$ times, ${\displaystyle V_{n}(R)}$ is proportional to ${\displaystyle R^{n}}$, which implies ${\displaystyle {\frac {dV_{n}(R)}{dR}}={\frac {n}{R}}V_{n}(R)}$. Also, ${\displaystyle A_{n-1}(R)={\frac {dV_{n}(R)}{dR}}}$ because a ball is a union of concentric spheres and increasing radius by ε corresponds to a shell of thickness ε. Thus, ${\displaystyle V_{n}(R)={\frac {R}{n}}A_{n-1}(R)}$; equivalently, ${\displaystyle V_{n+1}(R)={\frac {R}{n+1}}A_{n}(R)}$.

${\displaystyle A_{n+1}(R)=(2\pi R)V_{n}(R)}$ follows from existence of a volume-preserving bijection between the unit sphere ${\displaystyle S_{n+1}}$ and ${\displaystyle S_{1}\times B_{n}}$:

${\displaystyle (x,y,{\vec {z}})\mapsto \left({\frac {x}{\sqrt {x^{2}+y^{2}}}},{\frac {y}{\sqrt {x^{2}+y^{2}}}},{\vec {z}}\right)}$

(${\displaystyle {\vec {z}}}$ is an n-tuple; ${\displaystyle |(x,y,{\vec {z}})|=1}$; we are ignoring sets of measure 0). Volume is preserved because at each point, the difference from isometry is a stretching in the xy plane (in ${\textstyle 1/\!{\sqrt {x^{2}+y^{2}}}}$ times in the direction of constant ${\displaystyle x^{2}+y^{2}}$) that exactly matches the compression in the direction of the gradient of ${\displaystyle |{\vec {z}}|}$ on ${\displaystyle S_{n}}$ (the relevant angles being equal). For ${\displaystyle S_{2}}$, a similar argument was originally made by Archimedes in On the Sphere and Cylinder.

Balls in L^p norms[]

There are also explicit expressions for the volumes of balls in L^p norms. The L^p norm of the vector x = (x₁, …, x_n) in Rⁿ is

${\displaystyle \|x\|_{p}={\biggl (}\sum _{i=1}^{n}|x_{i}|^{p}{\biggr )}^{\!1/p},}$

and an L^p ball is the set of all vectors whose L^p norm is less than or equal to a fixed number called the radius of the ball. The case p = 2 is the standard Euclidean distance function, but other values of p occur in diverse contexts such as information theory, coding theory, and dimensional regularization.

The volume of an L^p ball of radius R is

${\displaystyle V_{n}^{p}(R)={\frac {{\Bigl (}2\,\Gamma {\bigl (}{\tfrac {1}{p}}+1{\bigr )}R{\Bigr )}^{n}}{\Gamma {\bigl (}{\tfrac {n}{p}}+1{\bigr )}}}.}$

These volumes satisfy recurrence relations similar to those for p = 2:

${\displaystyle V_{n}^{p}(R)={\frac {{\Bigl (}2\,\Gamma {\bigl (}{\tfrac {1}{p}}+1{\bigr )}R{\Bigr )}^{p}p}{n}}V_{n-p}^{p}(R)}$

and

${\displaystyle V_{n}^{p}(R)=2R{\frac {\Gamma {\bigl (}{\tfrac {1}{p}}+1{\bigr )}\Gamma {\bigl (}{\tfrac {n-1}{p}}+1{\bigr )}}{\Gamma {\bigl (}{\tfrac {n}{p}}+1{\bigr )}}}V_{n-1}^{p}(R),}$

which can be written more concisely using a generalized binomial coefficient,

${\displaystyle V_{n}^{p}(R)={\frac {2R}{\dbinom {n/p}{1/p}}}V_{n-1}^{p}(R).}$

For p = 2, one recovers the recurrence for the volume of a Euclidean ball because 2Γ(3/2) = √π.

For example, in the cases p = 1 (taxicab norm) and p = ∞ (max norm), the volumes are:

${\displaystyle {\begin{aligned}V_{n}^{1}(R)&={\frac {(2R)^{n}}{n!}},\\V_{n}^{\infty }(R)&=(2R)^{n}.\end{aligned}}}$

These agree with elementary calculations of the volumes of cross-polytopes and hypercubes.

Relation with surface area[]

For most values of p, the surface area ${\displaystyle A_{n-1}^{p}(R)}$ of an L^p sphere of radius R (the boundary of an L^p n-ball of radius R) cannot be calculated by differentiating the volume of an L^p ball with respect to its radius. While the volume can be expressed as an integral over the surface areas using the coarea formula, the coarea formula contains a correction factor that accounts for how the p-norm varies from point to point. For p = 2 and p = ∞, this factor is one. However, if p = 1 then the correction factor is √n: the surface area of an L¹ sphere of radius R in Rⁿ is √n times the derivative of the volume of an L¹ ball. This can be seen most simply by applying the divergence theorem to the vector field F(x) = x to get

${\displaystyle nV_{n}^{1}(R)=}$${\displaystyle \iiint _{V}\left(\mathbf {\nabla } \cdot \mathbf {F} \right)\,dV=}$ ${\displaystyle \scriptstyle S}$ ${\displaystyle (\mathbf {F} \cdot \mathbf {n} )\,dS}$ ${\displaystyle =}$ ${\displaystyle \scriptstyle S}$ ${\displaystyle {\frac {1}{\sqrt {n}}}(|x_{1}|+\cdots +|x_{n}|)\,dS}$ ${\displaystyle ={\frac {R}{\sqrt {n}}}}$ ${\displaystyle \scriptstyle S}$ ${\displaystyle \,dS}$ ${\displaystyle ={\frac {R}{\sqrt {n}}}A_{n-1}^{1}(R).}$

For other values of p, the constant is a complicated integral.

Generalizations[]

The volume formula can be generalized even further. For positive real numbers p₁, …, p_n, define the (p₁, …, p_n) ball with limit L ≥ 0 to be

${\displaystyle B_{p_{1},\ldots ,p_{n}}(L)=\left\{x=(x_{1},\ldots ,x_{n})\in \mathbf {R} ^{n}:\vert x_{1}\vert ^{p_{1}}+\cdots +\vert x_{n}\vert ^{p_{n}}\leq L\right\}.}$

The volume of this ball has been known since the time of Dirichlet:^[3]

${\displaystyle V{\bigl (}B_{p_{1},\ldots ,p_{n}}(L){\bigr )}=2^{n}{\frac {\Gamma {\bigl (}1+{\tfrac {1}{p_{1}}}{\bigr )}\cdots \Gamma {\bigl (}1+{\tfrac {1}{p_{n}}}{\bigr )}}{\Gamma {\bigl (}1+{\tfrac {1}{p_{1}}}+\cdots +{\tfrac {1}{p_{n}}}{\bigr )}}}L^{{\tfrac {1}{p_{1}}}+\cdots +{\tfrac {1}{p_{n}}}}.}$

See also[]

• n-sphere
• Sphere packing
• Hamming bound

References[]

External links[]

• Derivation in hyperspherical coordinates (in French)
• Hypersphere on Wolfram MathWorld
• Volume of the Hypersphere at Math Reference