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 -ball is a ball in an -dimensional Euclidean space. The volume of a -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 -ball of radius is $R^nV_n,$ where $V_n$ is the volume of the unit -ball, the -ball of radius .

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 -sphere.

Formulas

The first volumes are as follows:

Two-dimension recurrence relation

As is proved below using a vector-calculus double integral in polar coordinates, the volume of an -ball of radius can be expressed recursively in terms of the volume of an -ball, via the interleaved recurrence relation:
:
V_n(R) = \begin
1 &\text n=0,\\ .5ex2R &\text n=1,\\ .5ex\dfracR^2 \times V_(R) &\text.
\end

This allows computation of in approximately steps.

Closed form

The -dimensional volume of a Euclidean ball of radius in -dimensional Euclidean space is:
:$V_n(R) = \fracR^n,$
where is Euler's gamma function. The gamma function is offset from but otherwise extends the factorial function to non-integer arguments. It satisfies if is a positive integer and if is a non-negative integer.

Alternative forms

The volume can also be expressed in terms of an -ball using the one-dimension recurrence relation:
:$\begin V_0(R) &= 1, \\ V_n(R) &= \fracR\, V_(R). \end$

Inverting the above, the radius of an -ball of volume can be expressed recursively in terms of the radius of an - or -ball:
:$\begin R_n(V) &= \bigl(\tfrac12n\bigr)^\left(\Gamma\bigl(\tfrac n2\bigr) V\right)^R_(V), \\ R_n(V) &= \fracV^R_(V). \end$

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:
:$\begin V_(R) &= \fracR^, \\ V_(R) &= \fracR^. \end$

The volume can also be expressed in terms of double factorials. For an odd integer , the double factorial is defined by
:$(2k + 1)!! = (2k + 1) \cdot (2k - 1) \dotsm 5 \cdot 3 \cdot 1.$
The volume of an odd-dimensional ball is
:$V_(R) = \fracR^.$
There are multiple conventions for double factorials of even integers. Under the convention in which the double factorial satisfies
:$(2k)!! = (2k) \cdot (2k - 2) \dotsm 4 \cdot 2 \cdot \sqrt = 2^k \cdot k! \cdot \sqrt,$
the volume of an -dimensional ball is, regardless of whether is even or odd,
:$V_n(R) = \fracR^n.$

Instead of expressing the volume of the ball in terms of its radius , the formulas can be inverted to express the radius as a function of the volume:
:$\begin R_n(V) &= \fracV^ \\ &= \left(\frac\right)^ \\ R_(V) &= \frac, \\ R_(V) &= \left(\frac\right)^.\end$

Approximation for high dimensions

Stirling's formula for the gamma function can be used to approximate the volume when the number of dimensions is high.
:$V_n(R) \sim \frac\left(\frac\right)^R^n.$
:$R_n(V) \sim (\pi n)^\sqrt V^.$
In particular, for any fixed value of the volume tends to a limiting value of 0 as goes to infinity. For example, the volume is increasing for , achieves its maximum when , and is decreasing for .

Relation with surface area

Let denote the hypervolume of the -sphere of radius . The -sphere is the -dimensional boundary (surface) of the -dimensional ball of radius , and the sphere's hypervolume and the ball's hypervolume are related by:
:$A_(R) = \frac V_(R) = \fracV_(R).$

Thus, inherits formulas and recursion relationships from , such as
:$A_(R) = \fracR^.$
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 th power of the radius

An important step in several proofs about volumes of -balls, and a generally useful fact besides, is that the volume of the -ball of radius is proportional to :
:$V_n(R) \propto R^n.$
The proportionality constant is the volume of the unit ball.

This is a special case of a general fact about volumes in -dimensional space: If
is a body (measurable set) in that space and is the body obtained by stretching in all directions by the factor then the volume of equals times the volume of . This is a direct consequence of the change of variables formula:
:$V(RK) = \int_ dx = \int_K R^n\, dy = R^n V(K)$
where and the substitution was made.

Another proof of the above relation, which avoids multi-dimensional integration, uses induction: The base case is , where the proportionality is obvious. For the inductive step, assume that proportionality is true in dimension . Note that the intersection of an ''n''-ball with a hyperplane is an -ball. When the volume of the -ball is written as an integral of volumes of -balls:
:$V_n(R) = \int_^R V_\!\left(\sqrt\right) dx,$
it is possible by the inductive hypothesis to remove a factor of from the radius of the -ball to get:
:$V_n(R) = R^\! \int_^R V_\!\left(\sqrt\right) dx.$
Making the change of variables leads to:
:$V_n(R) = R^n\! \int_^1 V_\!\left(\sqrt\right) dt = R^n V_n(1),$
which demonstrates the proportionality relation in dimension . 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 -ball and an -ball can be given using the proportionality formula above and integration in cylindrical coordinates. Fix a plane through the center of the ball. Let denote the distance between a point in the plane and the center of the sphere, and let denote the azimuth. Intersecting the -ball with the -dimensional plane defined by fixing a radius and an azimuth gives an -ball of radius . The volume of the ball can therefore be written as an iterated integral of the volumes of the -balls over the possible radii and azimuths:
:$V_n(R) = \int_0^ \int_0^R V_\!\left(\sqrt\right) r\,dr\,d\theta,$
The azimuthal coordinate can be immediately integrated out. Applying the proportionality relation shows that the volume equals
:$V_n(R) = 2\pi V_(R) \int_0^R \left(1 - \left(\frac\right)^2\right)^\,r\,dr.$
The integral can be evaluated by making the substitution to get
:\begin
V_n(R) &= 2\pi V_(R) \cdot \left \frac\left(1 - \left(\frac\right)^2\right)^\right^ \\
&= \frac V_(R),
\end
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 and . The inductive step is similar to the above, but instead of applying proportionality to the volumes of the -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 -ball and an -ball. As in the proof of the proportionality formula, the volume of an -ball can be written as an integral over the volumes of -balls. Instead of making a substitution, however, the proportionality relation can be applied to the volumes of the -balls in the integrand:
:$V_n(R) = V_(R) \int_^R \left(1 - \left(\frac\right)^2\right)^ \,dx.$
The integrand is an even function, so by symmetry the interval of integration can be restricted to . On the interval , it is possible to apply the substitution . This transforms the expression into
:$V_(R) \cdot R \cdot \int_0^1 (1-u)^u^\,du$
The integral is a value of a well-known special function called the beta function , and the volume in terms of the beta function is
:$V_n(R) = V_(R) \cdot R \cdot \Beta\left(\tfrac2, \tfrac12\right).$
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
:$V_n(R) = V_(R) \cdot R \cdot \frac.$
Using the value gives the one-dimension recursion formula:
:$V_n(R) = R\sqrt\frac V_(R).$

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 and angular coordinates , where the domain of each except is , and the domain of is . The spherical volume element is:
:$dV = r^\sin^(\varphi_1)\sin^(\varphi_2) \cdots \sin(\varphi_)\,dr\,d\varphi_1\,d\varphi_2 \cdots d\varphi_,$
and the volume is the integral of this quantity over between 0 and and all possible angles:
:$V_n(R) = \int_0^R \int_0^\pi \cdots \int_0^ r^\sin^(\varphi_1) \cdots \sin(\varphi_)\,d\varphi_ \cdots d\varphi_1\,dr.$
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:
:$V_n(R) = \left(\int_0^R r^\,dr\right)\!\left(\int_0^\pi \sin^(\varphi_1)\,d\varphi_1\right)\cdots\left(\int_0^ d\varphi_\right).$
The integral over the radius is . The intervals of integration on the angular coordinates can, by the symmetry of the sine about , be changed to :
:$V_n(R) = \frac \left(2\int_0^ \sin^(\varphi_1)\,d\varphi_1\right) \cdots \left(4\int_0^ d\varphi_\right).$
Each of the remaining integrals is now a particular value of the beta function:
:$V_n(R) = \frac \Beta\bigl(\tfrac2, \tfrac12\bigr) \Beta\bigl(\tfrac2, \tfrac12\bigr) \cdots \Beta\bigl(1, \tfrac12\bigr) \cdot 2\,\Beta\bigl(\tfrac12, \tfrac12\bigr).$

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

:$V_n(R) = \frac \cdot \frac \cdot \frac \cdots \frac \cdot 2 \frac.$

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

:$V_n(R) = \frac = \frac.$

Gaussian integrals

The volume formula can be proven directly using Gaussian integrals. Consider the function:
:$f(x_1, \ldots, x_n) = \exp\biggl(\biggr).$
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:
:$\int_ f \,dV = \prod_^n \left(\int_^\infty \exp\left(-\tfrac12 x_i^2\right)\,dx_i\right) = (2\pi)^,$
where is the -dimensional volume element. Using rotational invariance, the same integral can be computed in spherical coordinates:
:$\int_ f \,dV = \int_0^\infty \int_ \exp\left(-\tfrac12 r^2\right) \,dA\,dr,$
where is an -sphere of radius (being the surface of an -ball of radius ) and is the area element (equivalently, the -dimensional volume element). The surface area of the sphere satisfies a proportionality equation similar to the one for the volume of a ball: If is the surface area of an -sphere of radius , then:
:$A_(r) = r^ A_(1).$
Applying this to the above integral gives the expression
:$(2\pi)^ = \int_0^\infty \int_ \exp\left(-\tfrac12 r^2\right) \,dA\,dr = A_(1) \int_0^\infty \exp\left(-\tfrac12 r^2\right)\,r^\,dr.$
Substituting :
:$\int_0^\infty \exp\left(-\tfrac12 r^2\right)\,r^\,dr = 2^ \int_0^\infty e^ t^\,dt.$
The integral on the right is the gamma function evaluated at .

Combining the two results shows that
:$A_(1) = \frac.$
To derive the volume of an -ball of radius from this formula, integrate the surface area of a sphere of radius for and apply the functional equation :
:$V_n(R) = \int_0^R \frac \,r^\,dr = \fracR^n = \fracR^n.$

Geometric proof

The relations $V_(R) = \fracA_n(R)$ and $A_(R) = (2\pi R)V_n(R)$ 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 $\frac = \frac V_n(R)$. Also, $A_(R) = \frac$ because a ball is a union of concentric spheres and increasing radius by ''ε'' corresponds to a shell of thickness ''ε''. Thus, $V_(R) = \fracA_(R)$; equivalently, $V_(R) = \fracA_n(R)$.

$A_(R) = (2\pi R)V_n(R)$ follows from existence of a volume-preserving bijection between the unit sphere $S_$ and $S_1 \times B_n$:
: $(x,y,\vec) \mapsto \left(\frac,\frac,\vec\right)$
($\vec$ is an ''n''-tuple; $, (x,y,\vec), =1$; 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 $1/\!\sqrt$ times in the direction of constant $x^2+y^2$) that exactly matches the compression in the direction of the gradient of $, \vec,$ 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 norms

There are also explicit expressions for the volumes of balls in norms. The norm of the vector in is
:$\, x\, _p = \biggl(\sum_^n , x_i , ^p \biggr)^,$
and an ball is the set of all vectors whose norm is less than or equal to a fixed number called the radius of the ball. The case is the standard Euclidean distance function, but other values of occur in diverse contexts such as information theory, coding theory, and dimensional regularization.

The volume of an ball of radius is

:$V^p_n(R) = \fracR^n.$

These volumes satisfy recurrence relations similar to those for :

:$V^p_n(R) = \frac R^p \, V^p_(R)$

and

:$V^p_n(R) = 2 \frac R \, V^p_(R),$

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

:$V^p_n(R) = \frac R \, V^p_(R).$

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

For example, in the cases (taxicab norm) and ( max norm), the volumes are:
:$\begin V^1_n(R) &= \fracR^n, \\ V^\infty_n(R) &= 2^n R^n. \end$
These agree with elementary calculations of the volumes of cross-polytopes and hypercubes.

Relation with surface area

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

:$nV^1_n(R) =$ $=$ $= \frac$ $= \frac A^1_(R).$
For other values of , the constant is a complicated integral.

Generalizations

The volume formula can be generalized even further. For positive real numbers , define the ball with limit to be
:$B_(L) = \left\.$
The volume of this ball has been known since the time of Dirichlet:
:$V\bigl(B_(L)\bigr) = \frac L^.$

Comparison to norm

Using the harmonic mean $p = \frac$ and defining R = \sqrt /math>, the similarity to the volume formula for the ball becomes clear.
:$V\left(\left\\right) = \frac R^n.$

See also

* -sphere
* Sphere packing
* Hamming bound

References

External links

Derivation in hyperspherical coordinates
{{in lang, fr


on Wolfram MathWorld


at Math Reference

Multi-dimensional geometry