# Volume of an n-ball

Volumes of balls in dimensions 0 through 25
Surface areas of hyperspheres in dimensions 0 through 25

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 via a two-dimension recurrence relation. Closed-form expressions involve the gamma, factorial, or double factorial function. The volume can also be expressed in terms of ${\displaystyle A_{n}}$, the area of the unit n-sphere.

## Formulas

The first volumes are as follows:

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)

### Two-dimension recurrence relation

As is proved below using a vector-calculus double integral in polar coordinates, 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 interleaved recurrence relation:

${\displaystyle V_{n}(R)={\begin{cases}1&{\text{if }}n=0,\\[0.5ex]2R&{\text{if }}n=1,\\[0.5ex]{\dfrac {2\pi }{n}}R^{2}\times V_{n-2}(R)&{\text{otherwise}}.\end{cases}}}$

This allows computation of Vn(R) in approximately n / 2 steps.

### Closed form

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 is offset from but otherwise 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) = (n1/2) · (n3/2) · … · 1/2 · π1/2 if n is a non-negative integer.

### Alternative forms

The volume can also be expressed in terms of an (n − 1)-ball using the one-dimension 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}}}

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}}}

### Approximation for high dimensions

Stirling's approximation 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 Vn(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 An − 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, An − 1(R) inherits formulas and recursion relationships from Vn(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 = dx1dxn 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 R2r2. 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

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

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, …, φn − 1, where the domain of each φ except φn − 1 is [0, π), and the domain of φn − 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:

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

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 Sn − 1(r) is an (n − 1)-sphere of radius r (being the surface of an n-ball 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 An − 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 ≤ rR 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 Lp norms

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

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

and an Lp ball is the set of all vectors whose Lp 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 Lp ball of radius R is

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

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

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

and

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

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

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

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 {2^{n}}{n!}}R^{n},\\V_{n}^{\infty }(R)&=2^{n}R^{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 Lp sphere of radius R (the boundary of an Lp n-ball of radius R) cannot be calculated by differentiating the volume of an Lp 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, …, pn, define the (p1, …, pn) 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]

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

#### Comparison to Lp norm

Using the harmonic mean ${\displaystyle p={\frac {n}{{\frac {1}{p_{1}}}+\cdots {\frac {1}{p_{n}}}}}}$ and defining ${\displaystyle R={\sqrt[{p}]{L}}}$, the similarity to the volume formula for the Lp ball becomes clear.

${\displaystyle V\left(\left\{x\in \mathbf {R} ^{n}:{\sqrt[{p}]{\vert x_{1}\vert ^{p_{1}}+\cdots +\vert x_{n}\vert ^{p_{n}}}}\leq R\right\}\right)={\frac {2^{n}\Gamma {\bigl (}{\tfrac {1}{p_{1}}}+1{\bigr )}\cdots \Gamma {\bigl (}{\tfrac {1}{p_{n}}}+1{\bigr )}}{\Gamma {\bigl (}{\tfrac {n}{p}}+1{\bigr )}}}R^{n}.}$

## References

1. ^ Equation 5.19.4, NIST Digital Library of Mathematical Functions. http://dlmf.nist.gov/5.19#E4, Release 1.0.6 of 2013-05-06.
2. ^ Smith, David J. and Vamanamurthy, Mavina K., "How Small Is a Unit Ball?", Mathematics Magazine, Volume 62, Issue 2, 1989, pp. 101–107, https://doi.org/10.1080/0025570X.1989.11977419.
3. ^ Dirichlet, P. G. Lejeune (1839). "Sur une nouvelle méthode pour la détermination des intégrales multiples" [On a novel method for determining multiple integrals]. Journal de Mathématiques Pures et Appliquées. 4: 164–168.