Volume of an n-ball

Volumes of balls in dimensions 0 through 25; unit ball in red.

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 Template:Math-ball is a ball in an Template:Math-dimensional Euclidean space. The volume of a Template:Math-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 Template:Mvar-ball of radius Template:Mvar is $R^n{V}_n,$ where $V_n$ is the volume of the [[unit ball|unit Template:Mvar-ball]], the Template:Mvar-ball of radius Template:Math.

The real number $V_n$ 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 $A_n$, the area of the [[unit n-sphere|unit Template:Mvar-sphere]].

Formulas

The first volumes are as follows:

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

Closed form

The Template:Math-dimensional volume of a Euclidean ball of radius Template:Math in Template:Math-dimensional Euclidean space is:

$V_n(R)=\frac{{\pi }^{n/2}}{\mathrm{\Gamma }(\frac{n}{2}+1)}{R}^n,$

where Template:Math is Euler's gamma function. The gamma function is offset from but otherwise extends the factorial function to non-integer arguments. It satisfies Template:Math if Template:Math is a positive integer and Template:Math if Template:Math is a non-negative integer.

Two-dimension recurrence relation

The volume can be computed without use of the Gamma function. As is proved below using a vector-calculus double integral in polar coordinates, the volume Template:Math of an Template:Math-ball of radius Template:Math can be expressed recursively in terms of the volume of an Template:Math-ball, via the interleaved recurrence relation:

$V_n(R)=\{\begin{array}{cc}1& \text{if }n=0,\\ 2R& \text{if }n=1,\\ \frac{2\pi }{n}{R}^2\times V_{n-2}(R)& \text{otherwise}.\end{array}$

This allows computation of Template:Math in approximately Template:Math steps.

Alternative forms

The volume can also be expressed in terms of an Template:Math-ball using the one-dimension recurrence relation:

$\begin{array}{cc}{V}_0(R)& =1,\\ V_n(R)& =\frac{\mathrm{\Gamma }(\frac{n}{2}+\frac{1}{2})\sqrt{\pi }}{\mathrm{\Gamma }(\frac{n}{2}+1)}R{V}_{n-1}(R).\end{array}$

Inverting the above, the radius of an Template:Math-ball of volume Template:Math can be expressed recursively in terms of the radius of an Template:Math- or Template:Math-ball:

$\begin{array}{cc}{R}_n(V)& =(\frac{1}{2}n{)}^{1/n}{\left(\mathrm{\Gamma }\right(\frac{n}{2}\left)V\right)}^{-2/(n(n-2))}{R}_{n-2}(V),\\ R_n(V)& =\frac{\mathrm{\Gamma }(\frac{n}{2}+1{)}^{1/n}}{\mathrm{\Gamma }(\frac{n}{2}+\frac{1}{2}{)}^{1/(n-1)}}{V}^{-1/(n(n-1))}{R}_{n-1}(V).\end{array}$

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. For non-negative integer Template:Mvar, these are:

$\begin{array}{cc}{V}_{2k}(R)& =\frac{{\pi }^k}{k!}{R}^{2k},\\ V_{2k+1}(R)& =\frac{2(k!)(4\pi {)}^k}{(2k+1)!}{R}^{2k+1}.\end{array}$

The volume can also be expressed in terms of double factorials. For a positive odd integer Template:Math, the double factorial is defined by

$(2k+1)!!=(2k+1)\cdot (2k-1)\cdots 5\cdot 3\cdot 1.$

The volume of an odd-dimensional ball is

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

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

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

the volume of an Template:Math-dimensional ball is, regardless of whether Template:Math is even or odd,

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

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

$\begin{array}{cc}{R}_n(V)& =\frac{\mathrm{\Gamma }(\frac{n}{2}+1{)}^{1/n}}{\sqrt{\pi }}{V}^{1/n}\\ & ={\left(\frac{n!!V}{2(2\pi {)}^{(n-1)/2}}\right)}^{1/n}\\ R_{2k}(V)& =\frac{(k!V{)}^{1/(2k)}}{\sqrt{\pi }},\\ R_{2k+1}(V)& ={\left(\frac{(2k+1)!V}{2(k!)(4\pi {)}^k}\right)}^{1/(2k+1)}.\end{array}$

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.

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

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

In particular, for any fixed value of Template:Math the volume tends to a limiting value of 0 as Template:Math goes to infinity. Which value of Template:Mvar maximizes Template:Math depends upon the value of Template:Mvar; for example, the volume Template:Math is increasing for Template:Math, achieves its maximum when Template:Math, and is decreasing for Template:Math.

Also, there is an asymptotic formula for the surface area$$\underset{n}{\lim }\frac{1}{n}\ln A_{n-1}(\sqrt{n})=\frac{1}{2}(\ln (2\pi )+1)$$

Relation with surface area

Surface areas of hyperspheres in dimensions 0 through 25

Let Template:Math denote the hypervolume of the [[n-sphere|Template:Math-sphere]] of radius Template:Math. The Template:Math-sphere is the Template:Math-dimensional boundary (surface) of the Template:Math-dimensional ball of radius Template:Math, and the sphere's hypervolume and the ball's hypervolume are related by:

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

Thus, Template:Math inherits formulas and recursion relationships from Template:Math, such as

$A_{n-1}(R)=\frac{2{\pi }^{n/2}}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}{R}^{n-1}.$

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 Template:Mathth power of the radius

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

$V_n(R)\propto R^n.$

The proportionality constant is the volume of the unit ball.

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

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

where Template:Math and the substitution Template:Math was made.

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

$V_n(R)=\underset{-R}{\overset{R}{\int }}{V}_{n-1}\left(\sqrt{R^2-x^2}\right)dx,$

it is possible by the inductive hypothesis to remove a factor of Template:Math from the radius of the Template:Math-ball to get:

$V_n(R)=R^{n-1}\underset{-R}{\overset{R}{\int }}{V}_{n-1}\left(\sqrt{1-{\left(\frac{x}{R}\right)}^2}\right)dx.$

Making the change of variables Template:Math leads to:

$V_n(R)=R^n\underset{-1}{\overset{1}{\int }}{V}_{n-1}\left(\sqrt{1-t^2}\right)dt=R^n{V}_n(1),$

which demonstrates the proportionality relation in dimension Template:Math. 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 Template:Math-ball and an Template:Math-ball can be given using the proportionality formula above and integration in cylindrical coordinates. Fix a plane through the center of the ball. Let Template:Math denote the distance between a point in the plane and the center of the sphere, and let Template:Math denote the azimuth. Intersecting the Template:Math-ball with the Template:Math-dimensional plane defined by fixing a radius and an azimuth gives an Template:Math-ball of radius Template:Math. The volume of the ball can therefore be written as an iterated integral of the volumes of the Template:Math-balls over the possible radii and azimuths:

$V_n(R)=\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{R}{\int }}{V}_{n-2}\left(\sqrt{R^2-r^2}\right)rdrd\theta ,$

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

$V_n(R)=2\pi V_{n-2}(R)\underset{0}{\overset{R}{\int }}{(1-{\left(\frac{r}{R}\right)}^2)}^{(n-2)/2}rdr.$

The integral can be evaluated by making the substitution Template:Math to get

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

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 Template:Math and Template:Math. The inductive step is similar to the above, but instead of applying proportionality to the volumes of the Template:Math-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 Template:Math-ball and an Template:Math-ball. As in the proof of the proportionality formula, the volume of an Template:Math-ball can be written as an integral over the volumes of Template:Math-balls. Instead of making a substitution, however, the proportionality relation can be applied to the volumes of the Template:Math-balls in the integrand:

$V_n(R)=V_{n-1}(R)\underset{-R}{\overset{R}{\int }}{(1-{\left(\frac{x}{R}\right)}^2)}^{(n-1)/2}dx.$

The integrand is an even function, so by symmetry the interval of integration can be restricted to Template:Math. On the interval Template:Math, it is possible to apply the substitution Template:Math. This transforms the expression into

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

The integral is a value of a well-known special function called the beta function Template:Math, and the volume in terms of the beta function is

$V_n(R)=V_{n-1}(R)\cdot R\cdot \mathrm{B}(\frac{n+1}{2},\frac{1}{2}).$

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

$V_n(R)=V_{n-1}(R)\cdot R\cdot \frac{\mathrm{\Gamma }(\frac{n}{2}+\frac{1}{2})\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(\frac{n}{2}+1)}.$

Using the value Template:Math gives the one-dimension recursion formula:

$V_n(R)=R\sqrt{\pi }\frac{\mathrm{\Gamma }(\frac{n}{2}+\frac{1}{2})}{\mathrm{\Gamma }(\frac{n}{2}+1)}{V}_{n-1}(R).$

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 $V_n(R)$ can be computed by integrating the volume element in spherical coordinates. The spherical coordinate system has a radial coordinate Template:Math and angular coordinates Template:Math, where the domain of each Template:Math except Template:Math is Template:Math, and the domain of Template:Math is Template:Math. The spherical volume element is:

$dV=r^{n-1}{\sin }^{n-2}({\phi }_1){\sin }^{n-3}({\phi }_2)\cdots \sin ({\phi }_{n-2})drd{\phi }_{1}d{\phi }_2\cdots d{\phi }_{n-1},$

and the volume is the integral of this quantity over Template:Math between 0 and Template:Math and all possible angles:

$V_n(R)=\underset{0}{\overset{R}{\int }}\underset{0}{\overset{\pi }{\int }}\cdots \underset{0}{\overset{2\pi }{\int }}{r}^{n-1}{\sin }^{n-2}({\phi }_1)\cdots \sin ({\phi }_{n-2})d{\phi }_{n-1}\cdots d{\phi }_{1}dr.$

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:

$V_n(R)=\left(\underset{0}{\overset{R}{\int }}{r}^{n-1}dr\right)(\underset{0}{\overset{\pi }{\int }}{\sin }^{n-2}({\phi }_1)d{\phi }_1)\cdots \left(\underset{0}{\overset{2\pi }{\int }}d{\phi }_{n-1}\right).$

The integral over the radius is Template:Math. The intervals of integration on the angular coordinates can, by the symmetry of the sine about Template:Sfrac, be changed to Template:Math:

$V_n(R)=\frac{R^n}{n}(2\underset{0}{\overset{\pi /2}{\int }}{\sin }^{n-2}({\phi }_1)d{\phi }_1)\cdots \left(4\underset{0}{\overset{\pi /2}{\int }}d{\phi }_{n-1}\right).$

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

$V_n(R)=\frac{R^n}{n}\mathrm{B}(\frac{n-1}{2},\frac{1}{2})\mathrm{B}(\frac{n-2}{2},\frac{1}{2})\cdots \mathrm{B}(1,\frac{1}{2})\cdot 2\mathrm{B}(\frac{1}{2},\frac{1}{2}).$

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

$V_n(R)=\frac{R^n}{n}\cdot \frac{\mathrm{\Gamma }(\frac{n}{2}-\frac{1}{2})\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}\cdot \frac{\mathrm{\Gamma }(\frac{n}{2}-1)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(\frac{n}{2}-\frac{1}{2})}\cdots \frac{\mathrm{\Gamma }(1)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }\left(\frac{3}{2}\right)}\cdot 2\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(1)}.$

This product telescopes. Combining this with the values Template:Math and Template:Math and the functional equation Template:Math leads to

$V_n(R)=\frac{2{\pi }^{n/2}{R}^n}{n\mathrm{\Gamma }\left(\frac{n}{2}\right)}=\frac{{\pi }^{n/2}{R}^n}{\mathrm{\Gamma }(\frac{n}{2}+1)}.$

Gaussian integrals

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

$f(x_1,\dots ,x_n)=\exp \left(-\frac{1}{2}\sum _{i=1}^n{x}_i^2\right).$

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:

$\underset{{𝐑}^n}{\int }fdV=\prod _{i=1}^n(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\exp (-\frac{1}{2}{x}_i^2)d{x}_i)=(2\pi {)}^{n/2},$

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

$\underset{{𝐑}^n}{\int }fdV=\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{S^{n-1}(r)}{\int }\exp (-\frac{1}{2}{r}^2)dAdr,$

where Template:Math is an Template:Math-sphere of radius Template:Math (being the surface of an Template:Math-ball of radius Template:Math) and Template:Math is the area element (equivalently, the Template:Math-dimensional volume element). The surface area of the sphere satisfies a proportionality equation similar to the one for the volume of a ball: If Template:Math is the surface area of an Template:Math-sphere of radius Template:Math, then:

$A_{n-1}(r)=r^{n-1}{A}_{n-1}(1).$

Applying this to the above integral gives the expression

$(2\pi {)}^{n/2}=\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{S^{n-1}(r)}{\int }\exp (-\frac{1}{2}{r}^2)dAdr=A_{n-1}(1)\underset{0}{\overset{\mathrm{\infty }}{\int }}\exp (-\frac{1}{2}{r}^2)r^{n-1}dr.$

Substituting Template:Math:

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\exp (-\frac{1}{2}{r}^2)r^{n-1}dr=2^{(n-2)/2}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-t}{t}^{(n-2)/2}dt.$

The integral on the right is the gamma function evaluated at Template:Math.

Combining the two results shows that

$A_{n-1}(1)=\frac{2{\pi }^{n/2}}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}.$

To derive the volume of an Template:Math-ball of radius Template:Math from this formula, integrate the surface area of a sphere of radius Template:Math for Template:Math and apply the functional equation Template:Math:

$V_n(R)=\underset{0}{\overset{R}{\int }}\frac{2{\pi }^{n/2}}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}{r}^{n-1}dr=\frac{2{\pi }^{n/2}}{n\mathrm{\Gamma }\left(\frac{n}{2}\right)}{R}^n=\frac{{\pi }^{n/2}}{\mathrm{\Gamma }(\frac{n}{2}+1)}{R}^n.$

Geometric proof

The relations $V_{n+1}(R)=\frac{R}{n+1}{A}_n(R)$ and $A_{n+1}(R)=(2\pi R)V_n(R)$ 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 $R$ is obtained from a unit ball $B_n$ by rescaling all directions in $R$ times, $V_n(R)$ is proportional to $R^n$, which implies $\frac{d{V}_n(R)}{dR}=\frac{n}{R}{V}_n(R)$. Also, $A_{n-1}(R)=\frac{d{V}_n(R)}{dR}$ because a ball is a union of concentric spheres and increasing radius by ε corresponds to a shell of thickness ε. Thus, $V_n(R)=\frac{R}{n}{A}_{n-1}(R)$; equivalently, $V_{n+1}(R)=\frac{R}{n+1}{A}_n(R)$.

$A_{n+1}(R)=(2\pi R)V_n(R)$ follows from existence of a volume-preserving bijection between the unit sphere $S_{n+1}$ and $S_1\times B_n$:

$(x,y,\overrightarrow{z})\mapsto (\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}},\overrightarrow{z})$

($\overrightarrow{z}$ is an n-tuple; $|(x,y,\overrightarrow{z})|=1$; 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 $1/\sqrt{x^2+y^2}$ times in the direction of constant $x^2+y^2$) that exactly matches the compression in the direction of the gradient of $|\overrightarrow{z}|$ on $S_n$ (the relevant angles being equal). For $S_2$, a similar argument was originally made by Archimedes in On the Sphere and Cylinder.

Balls in Template:Math norms

There are also explicit expressions for the volumes of balls in [[Lp space|Template:Math norm]]s. The Template:Math norm of the vector Template:Math in Template:Math is

$\Vert x{\Vert }_p=(\sum _{i=1}^n|x_i{|}^p{)}^{1/p},$

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

The volume of an Template:Math ball of radius Template:Math is

$V_n^p(R)=\frac{\left(2\mathrm{\Gamma }\right(\frac{1}{p}+1){)}^n}{\mathrm{\Gamma }(\frac{n}{p}+1)}{R}^n.$

These volumes satisfy recurrence relations similar to those for Template:Math:

$V_n^p(R)=\frac{\left(2\mathrm{\Gamma }\right(\frac{1}{p}+1){)}^{p}p}{n}{R}^p{V}_{n-p}^p(R)$

and

$V_n^p(R)=2\frac{\mathrm{\Gamma }(\frac{1}{p}+1)\mathrm{\Gamma }(\frac{n-1}{p}+1)}{\mathrm{\Gamma }(\frac{n}{p}+1)}R{V}_{n-1}^p(R),$

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

$V_n^p(R)=\frac{2}{(\genfrac{}{}{0ex}{}{n/p}{1/p})}R{V}_{n-1}^p(R).$

For Template:Math, one recovers the recurrence for the volume of a Euclidean ball because Template:Math.

For example, in the cases Template:Math (taxicab norm) and Template:Math (max norm), the volumes are:

$\begin{array}{cc}{V}_n^1(R)& =\frac{2^n}{n!}{R}^n,\\ V_n^{\mathrm{\infty }}(R)& =2^n{R}^n.\end{array}$

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

Relation with surface area

For most values of Template:Math, the surface area $A_{n-1}^p(R)$ of an Template:Math sphere of radius Template:Math (the boundary of an Template:Math Template:Math-ball of radius Template:Math) cannot be calculated by differentiating the volume of an Template:Math 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 Template:Math-norm varies from point to point. For Template:Math and Template:Math, this factor is one. However, if Template:Math then the correction factor is Template:Math: the surface area of an Template:Math sphere of radius Template:Math in Template:Math is Template:Math times the derivative of the volume of an Template:Math ball. This can be seen most simply by applying the divergence theorem to the vector field Template:Math to get

$n{V}_n^1(R)=$Template:Oiint $=$ Template:Oiint $=\frac{R}{\sqrt{n}}$Template:Oiint $=\frac{R}{\sqrt{n}}{A}_{n-1}^1(R).$

For other values of Template:Math, the constant is a complicated integral.

Generalizations

The volume formula can be generalized even further. For positive real numbers Template:Math, define the Template:Math ball with limit Template:Math to be

$B_{p_1,\dots ,p_n}(L)=\{x=(x_1,\dots ,x_n)\in {𝐑}^n:|x_1{|}^{p_1}+\cdots +|x_n{|}^{p_n}\le L\}.$

The volume of this ball has been known since the time of Dirichlet:

$V(B_{p_1,\dots ,p_n}(L))=\frac{2^n\mathrm{\Gamma }(\frac{1}{p_1}+1)\cdots \mathrm{\Gamma }(\frac{1}{p_n}+1)}{\mathrm{\Gamma }(\frac{1}{p_1}+\cdots +\frac{1}{p_n}+1)}{L}^{\frac{1}{p_1}+\cdots +\frac{1}{p_n}}.$

Comparison to Template:Math norm

Using the harmonic mean $p=\frac{n}{\frac{1}{p_1}+\cdots \frac{1}{p_n}}$ and defining $R=\sqrt[p]{L}$, the similarity to the volume formula for the Template:Math ball becomes clear.

$V\left(\{x\in {𝐑}^n:\sqrt[p]{|x_1{|}^{p_1}+\cdots +|x_n{|}^{p_n}}\le R\}\right)=\frac{2^n\mathrm{\Gamma }(\frac{1}{p_1}+1)\cdots \mathrm{\Gamma }(\frac{1}{p_n}+1)}{\mathrm{\Gamma }(\frac{n}{p}+1)}{R}^n.$

See also

• [[n-sphere|Template:Math-sphere]]
• Sphere packing
• Hamming bound

References

Further reading

• Template:Cite journal, bibliography, accessible to layman.
  • Updated version: Template:Cite book, bibliography, accessible to layman.

External links

• Derivation in hyperspherical coordinates Template:In lang
• Hypersphere on Wolfram MathWorld
• Volume of the Hypersphere at Math Reference