# 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 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 $R^{n}V_{n},$ where $V_{n}$ is the volume of the unit n-ball, the n-ball of radius 1.
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.

## Formulas

The first volumes are as follows:
| Dimension | Volume of a ball of radius R | Radius of a ball of volume V |
| 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 | V$_n$(R) | R$_n$(V) |

### Closed form

The n-dimensional volume of a Euclidean ball of radius R in n-dimensional Euclidean space is:
    $V_{n}(R)={\frac {\pi ^{n/2}}{\Gamma {\bigl{\tfrac {n}{2}}+1{\bigr )}}}R^{n},$
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) = (n − ⁠1/2⁠) · (n − ⁠3/2⁠) · … · ⁠1/2⁠ · π$^{1/2}$ 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:
    $V_{n}(R)={\begin{cases}1&{\text{if }}n=0,\\[0.5ex]2R&{\text{if }}n=1,\\[0.5ex]{\dfrac {2\pi }{n}}R^{2}\times V_{n-2}(R)&{\text{otherwise}}.\end{cases}}$
This allows computation of 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:
    ${\begin{aligned}V_{0}(R)&=1,\\V_{n}(R)&={\frac {\Gamma {\bigl{\tfrac {n}{2}}+{\tfrac {1}{2}}{\bigr )}{\sqrt {\pi }}}{\Gamma {\bigl{\tfrac {n}{2}}+1{\bigr )}}}R\,V_{n-1}(R).\end{aligned}}$
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:
    ${\begin{aligned}R_{n}(V)&={\bigl{\tfrac {1}{2}}n{\bigr )}^{1/n}\left(\Gamma {\bigl{\tfrac {n}{2}}{\bigr )}V\right)^{-2/(n(n-2))}R_{n-2}(V),\\R_{n}(V)&={\frac {\Gamma {\bigl{\tfrac {n}{2}}+1{\bigr )}^{1/n}}{\Gamma {\bigl{\tfrac {n}{2}}+{\tfrac {1}{2}}{\bigr )}^{1/(n-1)}}}V^{-1/(n(n-1))}R_{n-1}(V).\end{aligned}}$
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. For non-negative integer k, these are:
    ${\begin{aligned}V_{2k}(R)&={\frac {\pi ^{k}}{k!}}R^{2k},\\V_{2k+1}(R)&={\frac {2(k!)(4\pi )^{k}}{(2k+1)!}}R^{2k+1}.\end{aligned}}$
The volume can also be expressed in terms of double factorials. For a positive odd integer 2k + 1, 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
    $V_{2k+1}(R)={\frac {2(2\pi )^{k}}{(2k+1)!!}}R^{2k+1}.$
There are multiple conventions for double factorials of even integers. Under the convention in which the double factorial satisfies
    $(2k)!!=(2k)\cdot (2k-2)\dotsm 4\cdot 2\cdot {\sqrt {2/\pi }}=2^{k}\cdot k!\cdot {\sqrt {2/\pi }},$
the volume of an n-dimensional ball is, regardless of whether n is even or odd,
    $V_{n}(R)={\frac {2(2\pi )^{(n-1)/2}}{n!!}}R^{n}.$
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:
    ${\begin{aligned}R_{n}(V)&={\frac {\Gamma {\bigl{\tfrac {n}{2}}+1{\bigr )}^{1/n}}{\sqrt {\pi }}}V^{1/n}\\&=\left({\frac {n!!V}{2(2\pi )^{(n-1)/2}}}\right)^{1/n}\\R_{2k}(V)&={\frac {(k!V)^{1/(2k)}}{\sqrt {\pi }}},\\R_{2k+1}(V)&=\left({\frac {(2k+1)!V}{2(k!)(4\pi )^{k}}}\right)^{1/(2k+1)}.\end{aligned}}$

### 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.
    $V_{n}(R)\sim {\frac {1}{\sqrt {n\pi }}}\left({\frac {2\pi e}{n}}\right)^{n/2}R^{n}.$
    $R_{n}(V)\sim (\pi n)^{1/(2n)}{\sqrt {\frac {n}{2\pi e}}}V^{1/n}.$
In particular, for any fixed value of 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 area$\lim _{n}{\frac {1}{n}}\ln A_{n-1}({\sqrt {n}})={\frac {1}{2}}(\ln(2\pi )+1)$

### 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:
    $A_{n-1}(R)={\frac {d}{dR}}V_{n}(R)={\frac {n}{R}}V_{n}(R).$
Thus, A$_{n−1}$(R) inherits formulas and recursion relationships from V$_n$(R), such as
    $A_{n-1}(R)={\frac {2\pi ^{n/2}}{\Gamma {\bigl{\tfrac {n}{2}}{\bigr )}}}R^{n-1}.$
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 R$^n$:
    $V_{n}(R)\propto R^{n}.$
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 R$^n$ times the volume of K. This is a direct consequence of the change of variables formula:
    $V(RK)=\int _{RK}dx=\int _{K}R^{n}\,dy=R^{n}V(K)$
where dx = dx$_1$…dx$_n$ 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:
    $V_{n}(R)=\int _{-R}^{R}V_{n-1}\!\left({\sqrt {R^{2}-x^{2}}}\right)dx,$
it is possible by the inductive hypothesis to remove a factor of R from the radius of the (n − 1)-ball to get:
    $V_{n}(R)=R^{n-1}\!\int _{-R}^{R}V_{n-1}\!\left({\sqrt {1-\left({\frac {x}{R}}\right)^{2}}}\right)dx.$
Making the change of variables t = ⁠x/R⁠ leads to:
    $V_{n}(R)=R^{n}\!\int _{-1}^{1}V_{n-1}\!\left({\sqrt {1-t^{2}}}\right)dt=R^{n}V_{n}(1),$
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 √R$^2$ − r$^2$. 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:
    $V_{n}(R)=\int _{0}^{2\pi }\int _{0}^{R}V_{n-2}\!\left({\sqrt {R^{2}-r^{2}}}\right)r\,dr\,d\theta ,$
The azimuthal coordinate can be immediately integrated out. Applying the proportionality relation shows that the volume equals
    $V_{n}(R)=2\pi V_{n-2}(R)\int _{0}^{R}\left(1-\left({\frac {r}{R}}\right)^{2}\right)^{(n-2)/2}\,r\,dr.$
The integral can be evaluated by making the substitution u = 1 − (⁠r/R⁠)$^2$ to get
    ${\begin{aligned}V_{n}(R)&=2\pi V_{n-2}(R)\cdot \left[-{\frac {R^{2}}{n}}\left(1-\left({\frac {r}{R}}\right)^{2}\right)^{n/2}\right]_{r=0}^{r=R}\\&={\frac {2\pi R^{2}}{n}}V_{n-2}(R),\end{aligned}}$
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:
    $V_{n}(R)=V_{n-1}(R)\int _{-R}^{R}\left(1-\left({\frac {x}{R}}\right)^{2}\right)^{(n-1)/2}\,dx.$
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
    $V_{n-1}(R)\cdot R\cdot \int _{0}^{1}(1-u)^{(n-1)/2}u^{-{\frac {1}{2}}}\,du$
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
    $V_{n}(R)=V_{n-1}(R)\cdot R\cdot \mathrm {B} \left({\tfrac {n+1}{2}},{\tfrac {1}{2}}\right).$
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
    $V_{n}(R)=V_{n-1}(R)\cdot R\cdot {\frac {\Gamma {\bigl{\tfrac {n}{2}}+{\tfrac {1}{2}}{\bigr )}\Gamma {\bigl{\tfrac {1}{2}}{\bigr )}}{\Gamma {\bigl{\tfrac {n}{2}}+1{\bigr )}}}.$
Using the value Γ(⁠1/2⁠) = √π gives the one-dimension recursion formula:
    $V_{n}(R)=R{\sqrt {\pi }}{\frac {\Gamma {\bigl{\tfrac {n}{2}}+{\tfrac {1}{2}}{\bigr )}}{\Gamma {\bigl{\tfrac {n}{2}}+1{\bigr )}}}V_{n-1}(R).$
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 $V_{n}(R)$ can be computed by integrating the volume element in spherical coordinates. The spherical coordinate system has a radial coordinate r and angular coordinates φ$_1$, …, φ$_{n−1}$, where the domain of each φ except φ$_{n−1}$ is [0, π), and the domain of φ$_{n−1}$ is [0, 2π). The spherical volume element is:
    $dV=r^{n-1}\sin ^{n-2}(\varphi _{1})\sin ^{n-3}(\varphi _{2})\cdots \sin(\varphi _{n-2})\,dr\,d\varphi _{1}\,d\varphi _{2}\cdots d\varphi _{n-1},$
and the volume is the integral of this quantity over r between 0 and R and all possible angles:
    $V_{n}(R)=\int _{0}^{R}\int _{0}^{\pi }\cdots \int _{0}^{2\pi }r^{n-1}\sin ^{n-2}(\varphi _{1})\cdots \sin(\varphi _{n-2})\,d\varphi _{n-1}\cdots d\varphi _{1}\,dr.$
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:
    $V_{n}(R)=\left(\int _{0}^{R}r^{n-1}\,dr\right)\!\left(\int _{0}^{\pi }\sin ^{n-2}(\varphi _{1})\,d\varphi _{1}\right)\cdots \left(\int _{0}^{2\pi }d\varphi _{n-1}\right).$
The integral over the radius is ⁠R$^n$/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]:
    $V_{n}(R)={\frac {R^{n}}{n}}\left(2\int _{0}^{\pi /2}\sin ^{n-2}(\varphi _{1})\,d\varphi _{1}\right)\cdots \left(4\int _{0}^{\pi /2}d\varphi _{n-1}\right).$
Each of the remaining integrals is now a particular value of the beta function:
    $V_{n}(R)={\frac {R^{n}}{n}}\mathrm {B} {\bigl{\tfrac {n-1}{2}},{\tfrac {1}{2}}{\bigr )}\mathrm {B} {\bigl{\tfrac {n-2}{2}},{\tfrac {1}{2}}{\bigr )}\cdots \mathrm {B} {\bigl1,{\tfrac {1}{2}}{\bigr )}\cdot 2\,\mathrm {B} {\bigl{\tfrac {1}{2}},{\tfrac {1}{2}}{\bigr )}.$
The beta functions can be rewritten in terms of gamma functions:
    $V_{n}(R)={\frac {R^{n}}{n}}\cdot {\frac {\Gamma {\bigl{\tfrac {n}{2}}-{\tfrac {1}{2}}{\bigr )}\Gamma {\bigl{\tfrac {1}{2}}{\bigr )}}{\Gamma {\bigl{\tfrac {n}{2}}{\bigr )}}}\cdot {\frac {\Gamma {\bigl{\tfrac {n}{2}}-1{\bigr )}\Gamma {\bigl{\tfrac {1}{2}}{\bigr )}}{\Gamma {\bigl{\tfrac {n}{2}}-{\tfrac {1}{2}}{\bigr )}}}\cdots {\frac {\Gamma (1)\Gamma {\bigl{\tfrac {1}{2}}{\bigr )}}{\Gamma {\bigl{\tfrac {3}{2}}{\bigr )}}}\cdot 2{\frac {\Gamma {\bigl{\tfrac {1}{2}}{\bigr )}\Gamma {\bigl{\tfrac {1}{2}}{\bigr )}}{\Gamma (1)}}.$
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
    $V_{n}(R)={\frac {2\pi ^{n/2}R^{n}}{n\,\Gamma {\bigl{\tfrac {n}{2}}{\bigr )}}}={\frac {\pi ^{n/2}R^{n}}{\Gamma {\bigl{\tfrac {n}{2}}+1{\bigr )}}}.$

### Gaussian integrals

The volume formula can be proven directly using Gaussian integrals. Consider the function:
    $f(x_{1},\ldots ,x_{n})=\exp {\biggl{-{\tfrac {1}{2}}\sum _{i=1}^{n}x_{i}^{2}}{\biggr )}.$
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:
    $\int _{\mathbf {R} ^{n}}f\,dV=\prod _{i=1}^{n}\left(\int _{-\infty }^{\infty }\exp \left(-{\tfrac {1}{2}}x_{i}^{2}\right)\,dx_{i}\right)=(2\pi )^{n/2},$
where dV is the n-dimensional volume element. Using rotational invariance, the same integral can be computed in spherical coordinates:
    $\int _{\mathbf {R} ^{n}}f\,dV=\int _{0}^{\infty }\int _{S^{n-1}(r)}\exp \left(-{\tfrac {1}{2}}r^{2}\right)\,dA\,dr,$
where 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:
    $A_{n-1}(r)=r^{n-1}A_{n-1}(1).$
Applying this to the above integral gives the expression
    $(2\pi )^{n/2}=\int _{0}^{\infty }\int _{S^{n-1}(r)}\exp \left(-{\tfrac {1}{2}}r^{2}\right)\,dA\,dr=A_{n-1}(1)\int _{0}^{\infty }\exp \left(-{\tfrac {1}{2}}r^{2}\right)\,r^{n-1}\,dr.$
Substituting t = ⁠r$^2$/2⁠:
    $\int _{0}^{\infty }\exp \left(-{\tfrac {1}{2}}r^{2}\right)\,r^{n-1}\,dr=2^{(n-2)/2}\int _{0}^{\infty }e^{-t}t^{(n-2)/2}\,dt.$
The integral on the right is the gamma function evaluated at ⁠n/2⁠.
Combining the two results shows that
    $A_{n-1}(1)={\frac {2\pi ^{n/2}}{\Gamma {\bigl{\tfrac {n}{2}}{\bigr )}}}.$
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 ≤ r ≤ R and apply the functional equation zΓ(z) = Γ(z + 1):
    $V_{n}(R)=\int _{0}^{R}{\frac {2\pi ^{n/2}}{\Gamma {\bigl{\tfrac {n}{2}}{\bigr )}}}\,r^{n-1}\,dr={\frac {2\pi ^{n/2}}{n\,\Gamma {\bigl{\tfrac {n}{2}}{\bigr )}}}R^{n}={\frac {\pi ^{n/2}}{\Gamma {\bigl{\tfrac {n}{2}}+1{\bigr )}}}R^{n}.$

### Geometric proof

The relations $V_{n+1}(R)={\frac {R}{n+1}}A_{n}(R)$ 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 ${\frac {dV_{n}(R)}{dR}}={\frac {n}{R}}V_{n}(R)$. Also, $A_{n-1}(R)={\frac {dV_{n}(R)}{dR}}$ because a ball is a union of concentric spheres and increasing radius by ε corresponds to a shell of thickness ε. Thus, $V_{n}(R)={\frac {R}{n}}A_{n-1}(R)$; 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 $S_{1}\times B_{n}$:
    $(x,y,{\vec {z}})\mapsto \left({\frac {x}{\sqrt {x^{2}+y^{2}}}},{\frac {y}{\sqrt {x^{2}+y^{2}}}},{\vec {z}}\right)$
(${\vec {z}}$ is an n-tuple; $|(x,y,{\vec {z}})|=1$; 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 1/√(x$^2$+y$^2$){\textstyle 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 $|{\vec {z}}|$ 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 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
    $\|x\|_{p}={\biggl\sum _{i=1}^{n}|x_{i}|^{p}{\biggr )}^{\!1/p},$
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
    $V_{n}^{p}(R)={\frac {{\Bigl2\,\Gamma {\bigl{\tfrac {1}{p}}+1{\bigr )}{\Bigr )}^{n}}{\Gamma {\bigl{\tfrac {n}{p}}+1{\bigr )}}}R^{n}.$
These volumes satisfy recurrence relations similar to those for p = 2:
    $V_{n}^{p}(R)={\frac {{\Bigl2\,\Gamma {\bigl{\tfrac {1}{p}}+1{\bigr )}{\Bigr )}^{p}p}{n}}R^{p}\,V_{n-p}^{p}(R)$
and
    $V_{n}^{p}(R)=2{\frac {\Gamma {\bigl{\tfrac {1}{p}}+1{\bigr )}\Gamma {\bigl{\tfrac {n-1}{p}}+1{\bigr )}}{\Gamma {\bigl{\tfrac {n}{p}}+1{\bigr )}}}R\,V_{n-1}^{p}(R),$
which can be written more concisely using a generalized binomial coefficient,
    $V_{n}^{p}(R)={\frac {2}{\dbinom {n/p}{1/p}}}R\,V_{n-1}^{p}(R).$
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:
    ${\begin{aligned}V_{n}^{1}(R)&={\frac {2^{n}}{n!}}R^{n},\\V_{n}^{\infty }(R)&=2^{n}R^{n}.\end{aligned}}$
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 $A_{n-1}^{p}(R)$ 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 = ∞, this factor is one. However, if p = 1 then the correction factor is √n: the surface area of an L$^1$ sphere of radius R in R$^n$ is √n times the derivative of the volume of an L$^1$ ball. This can be seen most simply by applying the divergence theorem to the vector field F(x) = x to get
    $nV_{n}^{1}(R)=$$\iiint _{V}\left(\mathbf {\nabla } \cdot \mathbf {F} \right)\,dV=$ $\scriptstyle S$ $(\mathbf {F} \cdot \mathbf {n} )\,dS$ $=$ $\scriptstyle S$ ${\frac {1}{\sqrt {n}}}(|x_{1}|+\cdots +|x_{n}|)\,dS$ $={\frac {R}{\sqrt {n}}}$ $\scriptstyle S$ $\,dS$ $={\frac {R}{\sqrt {n}}}A_{n-1}^{1}(R).$
For other values of p, the constant is a complicated integral.

### Generalizations

The volume formula can be generalized even further. For positive real numbers p$_1$, …, p$_n$, define the (p$_1$, …, p$_n$) ball with limit L ≥ 0 to be
    $B_{p_{1},\ldots ,p_{n}}(L)=\left\{x=(x_{1},\ldots ,x_{n})\in \mathbf {R} ^{n}:\vert x_{1}\vert ^{p_{1}}+\cdots +\vert x_{n}\vert ^{p_{n}}\leq L\right\}.$
The volume of this ball has been known since the time of Dirichlet:
    $V{\biglB_{p_{1},\ldots ,p_{n}}(L){\bigr )}={\frac {2^{n}\Gamma {\bigl{\tfrac {1}{p_{1}}}+1{\bigr )}\cdots \Gamma {\bigl{\tfrac {1}{p_{n}}}+1{\bigr )}}{\Gamma {\bigl{\tfrac {1}{p_{1}}}+\cdots +{\tfrac {1}{p_{n}}}+1{\bigr )}}}L^{{\tfrac {1}{p_{1}}}+\cdots +{\tfrac {1}{p_{n}}}}.$

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