Size of a mathematical 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 formula_1 where formula_2 is the volume of the unit n-ball, the n-ball of radius 1.
The real number formula_2 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 formula_4, the area of the unit n-sphere.
Formulas.
The first volumes are as follows:
Closed form.
The "n"-dimensional volume of a Euclidean ball of radius "R" in "n"-dimensional Euclidean space is:
formula_5
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" + ) 
 ("n" − ) · ("n" − ) · … · · π1/2 if "n" 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 "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:
formula_6
This allows computation of "V""n"("R") in approximately "n" / 2 steps.
Alternative forms.
The volume can also be expressed in terms of an ("n" − 1)-ball using the one-dimension recurrence relation:
formula_7
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:
formula_8
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 k, these are:
formula_9
The volume can also be expressed in terms of double factorials. For a positive odd integer 2"k" + 1, the double factorial is defined by
formula_10
The volume of an odd-dimensional ball is
formula_11
There are multiple conventions for double factorials of even integers. Under the convention in which the double factorial satisfies
formula_12
the volume of an "n"-dimensional ball is, regardless of whether "n" is even or odd,
formula_13
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:
formula_14
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.
formula_15
formula_16
In particular, for any fixed value of "R" the volume tends to a limiting value of 0 as "n" goes to infinity. Which value of n maximizes "V""n"("R") depends upon the value of R; 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.
Also, there is an asymptotic formula for the surface areaformula_17
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:
formula_18
Thus, "A""n" − 1("R") inherits formulas and recursion relationships from "V""n"("R"), such as
formula_19
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 "n"th 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 "R""n":
formula_20
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 "R""n" times the volume of "K". This is a direct consequence of the change of variables formula:
formula_21
where "dx" = "dx"1…"dxn" 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:
formula_22
it is possible by the inductive hypothesis to remove a factor of "R" from the radius of the ("n" − 1)-ball to get:
formula_23
Making the change of variables "t" 
 "" leads to:
formula_24
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 . 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:
formula_25
The azimuthal coordinate can be immediately integrated out. Applying the proportionality relation shows that the volume equals
formula_26
The integral can be evaluated by making the substitution "u" 
 1 − ("") to get
formula_27
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() 
 · Γ() 
 . 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:
formula_28
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" 
 (""). This transforms the expression into
formula_29
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
formula_30
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
formula_31
Using the value gives the one-dimension recursion formula:
formula_32
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 formula_33 can be computed by integrating the volume element in spherical coordinates. The spherical coordinate system has a radial coordinate "r" and angular coordinates "φ"1, …, "φ""n" − 1, where the domain of each "φ" except "φ""n" − 1 is , and the domain of "φ""n" − 1 is . The spherical volume element is:
formula_34
and the volume is the integral of this quantity over "r" between 0 and "R" and all possible angles:
formula_35
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:
formula_36
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 [0, ]:
formula_37
Each of the remaining integrals is now a particular value of the beta function:
formula_38
The beta functions can be rewritten in terms of gamma functions:
formula_39
This product telescopes. Combining this with the values and Γ(1) 
 1 and the functional equation "z"Γ("z") 
 Γ("z" + 1) leads to
formula_40
Gaussian integrals.
The volume formula can be proven directly using Gaussian integrals. Consider the function:
formula_41
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:
formula_42
where "dV" is the "n"-dimensional volume element. Using rotational invariance, the same integral can be computed in spherical coordinates:
formula_43
where "S""n" − 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 "A""n" − 1("r") is the surface area of an ("n" − 1)-sphere of radius "r", then:
formula_44
Applying this to the above integral gives the expression
formula_45
Substituting "t" 
formula_46
The integral on the right is the gamma function evaluated at .
Combining the two results shows that
formula_47
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 "z"Γ("z") 
 Γ("z" + 1):
formula_48
Geometric proof.
The relations formula_49 and formula_50 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 formula_51 is obtained from a unit ball formula_52 by rescaling all directions in formula_51 times, formula_33 is proportional to formula_55, which implies formula_56. Also, formula_57 because a ball is a union of concentric spheres and increasing radius by "ε" corresponds to a shell of thickness "ε". Thus, formula_58; equivalently, formula_49.
formula_50 follows from existence of a volume-preserving bijection between the unit sphere formula_61 and formula_62:
 formula_63
(formula_64 is an "n"-tuple; formula_65; 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 formula_66 times in the direction of constant formula_67) that exactly matches the compression in the direction of the gradient of formula_68 on formula_69 (the relevant angles being equal). For formula_70, 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" 
 ("x"1, …, "x""n") in R"n" is
formula_71
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
formula_72
These volumes satisfy recurrence relations similar to those for "p" 
 2:
formula_73
and 
formula_74
which can be written more concisely using a generalized binomial coefficient,
formula_75
For "p" 
 2, one recovers the recurrence for the volume of a Euclidean ball because .
For example, in the cases "p" 
 1 (taxicab norm) and "p" 
 ∞ (max norm), the volumes are:
formula_76
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 formula_77 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" 
 ∞,