# Volume of an n-ball

In geometry, a ball is a region in space comprising all points within a fixed distance from a given point; that is, it is the region enclosed by a sphere or hypersphere. Volume of an n-ball_sentence_0

An n-ball is a ball in n-dimensional Euclidean space. Volume of an n-ball_sentence_1

The volume of a unit n-ball is an important expression that occurs in formulas throughout mathematics; it generalizes the notion of the volume enclosed by a sphere in 3-dimensional space. Volume of an n-ball_sentence_2

## Formulas Volume of an n-ball_section_0

### The volume Volume of an n-ball_section_1

The n-dimensional volume of a Euclidean ball of radius R in n-dimensional Euclidean space is: Volume of an n-ball_sentence_3

where Γ is Leonhard Euler's gamma function. Volume of an n-ball_sentence_4

The gamma function extends the factorial function to non-integer arguments. Volume of an n-ball_sentence_5

It satisfies Γ(n) = (n − 1)! Volume of an n-ball_sentence_6

if n is a positive integer and Γ(n + 1/2) = (n − 1/2) · (n − 3/2) · … · 1/2 · π if n is a non-negative integer. Volume of an n-ball_sentence_7

### Alternative forms Volume of an n-ball_section_2

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 that do not require an evaluation of the gamma function. Volume of an n-ball_sentence_8

They can instead be expressed in terms of the double factorial, which is defined as 0!! Volume of an n-ball_sentence_9

= 1 and for n > 0, Volume of an n-ball_sentence_10

Volume of an n-ball_description_list_0

• (2k + 1)!! = 1 · 3 · 5 ·  ⋅⋅⋅  · (2k − 1) · (2k + 1).Volume of an n-ball_item_0_0

The formula for volume may be expressed as: Volume of an n-ball_sentence_11

which can be combined into a single formula: Volume of an n-ball_sentence_12

Instead of expressing the volume V of the ball in terms of its radius R, the formula can be inverted to express the radius as a function of the volume: Volume of an n-ball_sentence_13

This formula, too, can be separated into even- and odd-dimensional cases using factorials and double factorials in place of the gamma function: Volume of an n-ball_sentence_14

### Recursions Volume of an n-ball_section_3

The volume satisfies several recursive formulas. Volume of an n-ball_sentence_15

These formulas can either be proved directly or proved as consequences of the general volume formula above. Volume of an n-ball_sentence_16

The simplest to state is a formula for the volume of an n-ball in terms of the volume of an (n − 2)-ball of the same radius: Volume of an n-ball_sentence_17

There is also a formula for the volume of an n-ball in terms of the volume of an (n − 1)-ball of the same radius: Volume of an n-ball_sentence_18

Using explicit formulas for the gamma function again shows that the one-dimension recursion formula can also be written as: Volume of an n-ball_sentence_19

The radius of an n-ball of volume V may be expressed recursively in terms of the radius of an (n − 1)-ball or an (n − 2)-ball. Volume of an n-ball_sentence_20

These formulas may be derived from the explicit formula for Rn(V) above. Volume of an n-ball_sentence_21

Using explicit formulas for the gamma function shows that the one-dimension recursion formula is equivalent to Volume of an n-ball_sentence_22

and that the two-dimension recursion formula is equivalent to Volume of an n-ball_sentence_23

Defining a recurrence relation Volume of an n-ball_sentence_24

### Low dimensions Volume of an n-ball_section_4

In low dimensions, these volume and radius formulas simplify to the following. Volume of an n-ball_sentence_25

### High dimensions Volume of an n-ball_section_5

Suppose that R is fixed. Volume of an n-ball_sentence_26

Then the volume of an n-ball of radius R approaches zero as n tends to infinity. Volume of an n-ball_sentence_27

This can be shown using the two-dimension recursion formula. Volume of an n-ball_sentence_28

At each step, the new factor being multiplied into the volume is proportional to 1 / n, where the constant of proportionality 2πR is independent of n. Eventually, n is so large that the new factor is less than 1. Volume of an n-ball_sentence_29

From then on, the volume of an n-ball must decrease at least geometrically, and therefore it tends to zero. Volume of an n-ball_sentence_30

A variant on this proof uses the one-dimension recursion formula. Volume of an n-ball_sentence_31

Here, the new factor is proportional to a quotient of gamma functions. Volume of an n-ball_sentence_32

Gautschi's inequality bounds this quotient above by n. The argument concludes as before by showing that the volumes decrease at least geometrically. Volume of an n-ball_sentence_33

A more precise description of the high dimensional behavior of the volume can be obtained using Stirling's approximation. Volume of an n-ball_sentence_34

It implies the asymptotic formula: Volume of an n-ball_sentence_35

The error in this approximation is a factor of 1 + O(n). Volume of an n-ball_sentence_36

Stirling's approximation is in fact an underestimate of the gamma function, so the above formula is an upper bound. Volume of an n-ball_sentence_37

This provides another proof that the volume of the ball decreases exponentially: When n is sufficiently large, the factor R√2πe/n is less than one, and then the same argument as before applies. Volume of an n-ball_sentence_38

If instead V is fixed while n is large, then by Stirling's approximation again, the radius of an n-ball of volume V is approximately Volume of an n-ball_sentence_39

### Relation with surface area Volume of an n-ball_section_6

Let An(R) denote the surface area of the n-sphere of radius R in (n+1)-dimensional Euclidean space. Volume of an n-ball_sentence_40

The n-sphere is the boundary of the (n + 1)-ball of radius R. The (n + 1)-ball is a union of concentric spheres, and consequently the surface area and the volume are related by: Volume of an n-ball_sentence_41

Combining this with the explicit formula for the volume of an (n + 1)-ball gives Volume of an n-ball_sentence_42

Since the volume is proportional to a power of the radius, the above relation leads to a simple equation relating the surface area of an n-ball and the volume of an (n + 1)-ball. Volume of an n-ball_sentence_43

By applying the two-dimension recursion formula, it also gives an equation relating the surface area of an n-ball and the volume of an (n − 1)-ball. Volume of an n-ball_sentence_44

These formulas, together with the volume and surface area of zero-dimensional balls, can be used as a system of recurrence relations for the volumes and surface areas of balls: Volume of an n-ball_sentence_45

### Dimension maximizing the volume of a fixed-radius ball Volume of an n-ball_section_7

Suppose that R is a fixed positive real number, and consider the volume Vn(R) as a function of the positive integer dimension n. Since the volume of a ball with fixed positive radius tends to zero as n → ∞, the maximum volume is achieved for some value of n. The dimension in which this happens depends on the radius R. Volume of an n-ball_sentence_46

When x is not a positive integer, this function has no obvious geometric interpretation. Volume of an n-ball_sentence_47

However, it is smooth, so the techniques of calculus can be used to find maxima. Volume of an n-ball_sentence_48

where ψ is the digamma function, the logarithmic derivative of the gamma function. Volume of an n-ball_sentence_49

The critical points of V(x, R) therefore occur at the solutions of Volume of an n-ball_sentence_50

More explicit, though less precise, estimates may be derived by bounding the digamma function. Volume of an n-ball_sentence_51

For y > 1, the digamma function satisfies: Volume of an n-ball_sentence_52

where γ is the Euler–Mascheroni constant. Volume of an n-ball_sentence_53

Applying these bounds with y = x0/2 + 1 yields Volume of an n-ball_sentence_54

whence Volume of an n-ball_sentence_55

Therefore the maximum of Vn(R) is achieved for some integer n such that Volume of an n-ball_sentence_56

## Proofs Volume of an n-ball_section_8

There are many proofs of the above formulas. Volume of an n-ball_sentence_57

### The volume is proportional to the nth power of the radius Volume of an n-ball_section_9

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 R: Volume of an n-ball_sentence_58

The proportionality constant is the volume of the unit ball. Volume of an n-ball_sentence_59

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 R times the volume of K. This is a direct consequence of the change of variables formula: Volume of an n-ball_sentence_60

where dx = dx1…dxn and the substitution x = Ry was made. Volume of an n-ball_sentence_61

Another proof of the above relation, which avoids multi-dimensional integration, uses induction: The base case is n = 0, where the proportionality is obvious. Volume of an n-ball_sentence_62

For the inductive case, assume that proportionality is true in dimension n − 1. Volume of an n-ball_sentence_63

Note that the intersection of an n-ball with a hyperplane is an (n − 1)-ball. Volume of an n-ball_sentence_64

When the volume of the n-ball is written as an integral of volumes of (n − 1)-balls: Volume of an n-ball_sentence_65

it is possible by the inductive assumption to remove a factor of R from the radius of the (n − 1)-ball to get: Volume of an n-ball_sentence_66

Making the change of variables t = x/R leads to: Volume of an n-ball_sentence_67

which demonstrates the proportionality relation in dimension n. By induction, the proportionality relation is true in all dimensions. Volume of an n-ball_sentence_68

### The two-dimension recursion formula Volume of an n-ball_section_10

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. Volume of an n-ball_sentence_69

Fix a plane through the center of the ball. Volume of an n-ball_sentence_70

Let r denote the distance between a point in the plane and the center of the sphere, and let θ denote the azimuth. Volume of an n-ball_sentence_71

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 √R − r. 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: Volume of an n-ball_sentence_72

The azimuthal coordinate can be immediately integrated out. Volume of an n-ball_sentence_73

Applying the proportionality relation shows that the volume equals: Volume of an n-ball_sentence_74

The integral can be evaluated by making the substitution u = 1 − (r/R) to get: Volume of an n-ball_sentence_75

which is the two-dimension recursion formula. Volume of an n-ball_sentence_76

The same technique can be used to give an inductive proof of the volume formula. Volume of an n-ball_sentence_77

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. Volume of an n-ball_sentence_78

The inductive step is similar to the above, but instead of applying proportionality to the volumes of the (n − 2)-balls, the inductive assumption is applied instead. Volume of an n-ball_sentence_79

### The one-dimension recursion formula Volume of an n-ball_section_11

The proportionality relation can also be used to prove the recursion formula relating the volumes of an n-ball and an (n − 1)-ball. Volume of an n-ball_sentence_80

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. Volume of an n-ball_sentence_81

Instead of making a substitution, however, the proportionality relation can be applied to the volumes of the (n − 1)-balls in the integrand: Volume of an n-ball_sentence_82

The integrand is an even function, so by symmetry the interval of integration can be restricted to [0, R]. Volume of an n-ball_sentence_83

On the interval [0, R], it is possible to apply the substitution u = (x/R) . Volume of an n-ball_sentence_84

This transforms the expression into: Volume of an n-ball_sentence_85

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: Volume of an n-ball_sentence_86

The beta function can be expressed in terms of the gamma function in much the same way that factorials are related to binomial coefficients. Volume of an n-ball_sentence_87

Applying this relationship gives: Volume of an n-ball_sentence_88

Using the value Γ(1/2) = √π gives the one-dimension recursion formula: Volume of an n-ball_sentence_89

As with the two-dimension recursive formula, the same technique can be used to give an inductive proof of the volume formula. Volume of an n-ball_sentence_90

### Direct integration in spherical coordinates Volume of an n-ball_section_12

and the volume is the integral of this quantity over r between 0 and R and all possible angles: Volume of an n-ball_sentence_91

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: Volume of an n-ball_sentence_92

The integral over the radius is R/n. Volume of an n-ball_sentence_93

The intervals of integration on the angular coordinates can, by symmetry, be changed to [0, π/2]: Volume of an n-ball_sentence_94

Each of the remaining integrals is now a particular value of the beta function: Volume of an n-ball_sentence_95

The beta functions can be rewritten in terms of gamma functions: Volume of an n-ball_sentence_96

This product telescopes. Volume of an n-ball_sentence_97

Combining this with the values Γ(1/2) = √π and Γ(1) = 1 and the functional equation zΓ(z) = Γ(z + 1) leads to: Volume of an n-ball_sentence_98

### Gaussian integrals Volume of an n-ball_section_13

The volume formula can be proven directly using Gaussian integrals. Volume of an n-ball_sentence_99

Consider the function: Volume of an n-ball_sentence_100

This function is both rotationally invariant and a product of functions of one variable each. Volume of an n-ball_sentence_101

Using the fact that it is a product and the formula for the Gaussian integral gives: Volume of an n-ball_sentence_102

where dV is the n-dimensional volume element. Volume of an n-ball_sentence_103

Using rotational invariance, the same integral can be computed in spherical coordinates: Volume of an n-ball_sentence_104

where S(r) is an (n − 1)-sphere of radius r and dA is the area element (equivalently, the (n − 1)-dimensional volume element). Volume of an n-ball_sentence_105

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: Volume of an n-ball_sentence_106

Applying this to the above integral gives the expression: Volume of an n-ball_sentence_107

By substituting t = r/2, the expression is transformed into: Volume of an n-ball_sentence_108

This is the gamma function evaluated at n/2. Volume of an n-ball_sentence_109

Combining the two integrations shows that: Volume of an n-ball_sentence_110

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): Volume of an n-ball_sentence_111

## Balls in Lp norms Volume of an n-ball_section_15

There are also explicit expressions for the volumes of balls in L norms. Volume of an n-ball_sentence_112

The L norm of the vector x = (x1, …, xn) in R is: Volume of an n-ball_sentence_113

and an L ball is the set of all vectors whose L norm is less than or equal to a fixed number called the radius of the ball. Volume of an n-ball_sentence_114

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. Volume of an n-ball_sentence_115

The volume of an L ball of radius R is: Volume of an n-ball_sentence_116

These volumes satisfy a recurrence relation similar to the one dimension recurrence for p = 2: Volume of an n-ball_sentence_117

For p = 2, one recovers the recurrence for the volume of a Euclidean ball because 2Γ(3/2) = √π. Volume of an n-ball_sentence_118

For example, in the cases p = 1 (taxicab norm) and p = ∞ (max norm), the volumes are: Volume of an n-ball_sentence_119

These agree with elementary calculations of the volumes of cross-polytopes and hypercubes. Volume of an n-ball_sentence_120

### Relation with surface area Volume of an n-ball_section_16

For other values of p, the constant is a complicated integral. Volume of an n-ball_sentence_121

### Generalizations Volume of an n-ball_section_17

The volume formula can be generalized even further. Volume of an n-ball_sentence_122

For positive real numbers p1, …, pn, define the unit (p1, …, pn) ball to be: Volume of an n-ball_sentence_123

The volume of this ball has been known since the time of Dirichlet: Volume of an n-ball_sentence_124