Volume of an n-ball

{{See also|N-sphere#Volume and area|Unit sphere#Volume and area}}

The first volumes are as follows:

The {{math|n}}-dimensional volume of a Euclidean ball of radius {{math|R}} in {{math|n}}-dimensional Euclidean space is:Equation 5.19.4, NIST Digital Library of Mathematical Functions. http://dlmf.nist.gov/5.19#E4, Release 1.0.6 of 2013-05-06.

:$V\_n(R)\; =\; \backslash frac\{\backslash pi^\{n/2\}\}\{\backslash Gamma\backslash bigl(\backslash tfrac\; n2\; +\; 1\backslash bigr)\}R^n,$

where {{math|Γ}} is Euler's gamma function. The gamma function is offset from but otherwise extends the factorial function to non-integer arguments. It satisfies {{math|Γ(n) {{=}} (n − 1)!}} if {{math|n}} is a positive integer and {{math|Γ(n + {{sfrac|1|2}}) {{=}} (n − {{sfrac|1|2}}) · (n − {{sfrac|3|2}}) · … · {{sfrac|1|2}} · {{pi}}^1/2}} if {{math|n}} is a non-negative integer.

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 {{math|V}} of an {{math|n}}-ball of radius {{math|R}} can be expressed recursively in terms of the volume of an {{math|(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 {{math|V{{sub|n}}(R)}} in approximately {{math|n / 2}} steps.

The volume can also be expressed in terms of an {{math|(n − 1)}}-ball using the one-dimension recurrence relation:

:$\backslash begin\{align\}$

V_0(R) &= 1, \\

V_n(R) &= \frac{\Gamma\bigl(\tfrac n2 + \tfrac12 \bigr)\sqrt\pi}{\Gamma\bigl(\tfrac n2 + 1\bigr)}R\, V_{n-1}(R).

\end{align}

Inverting the above, the radius of an {{math|n}}-ball of volume {{math|V}} can be expressed recursively in terms of the radius of an {{math|(n − 2)}}- or {{math|(n − 1)}}-ball:

:$\backslash begin\{align\}$

R_n(V) &= \bigl(\tfrac12n\bigr)^{1/n}\left(\Gamma\bigl(\tfrac n2\bigr) V\right)^{-2/(n(n-2))}R_{n-2}(V), \\

R_n(V) &= \frac{\Gamma\bigl(\tfrac n2 + 1\bigr)^{1/n}}{\Gamma\bigl(\tfrac n2 + \tfrac12 \bigr)^{1/(n-1)}}V^{-1/(n(n-1))}R_{n-1}(V).

\end{align}

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 {{mvar|k}}, these are:

:$\backslash begin\{align\}$

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{align}

The volume can also be expressed in terms of double factorials. For a positive odd integer {{math|2k + 1}}, the double factorial is defined by

:$(2k\; +\; 1)!!\; =\; (2k\; +\; 1)\; \backslash cdot\; (2k\; -\; 1)\; \backslash dotsm\; 5\; \backslash cdot\; 3\; \backslash cdot\; 1.$

The volume of an odd-dimensional ball is

:$V\_\{2k+1\}(R)\; =\; \backslash frac\{2(2\backslash pi)^k\}\{(2k\; +\; 1)!!\}R^\{2k+1\}.$

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

:$(2k)!!\; =\; (2k)\; \backslash cdot\; (2k\; -\; 2)\; \backslash dotsm\; 4\; \backslash cdot\; 2\; \backslash cdot\; \backslash sqrt\{2/\backslash pi\}\; =\; 2^k\; \backslash cdot\; k!\; \backslash cdot\; \backslash sqrt\{2/\backslash pi\},$

the volume of an {{math|n}}-dimensional ball is, regardless of whether {{math|n}} is even or odd,

:$V\_n(R)\; =\; \backslash frac\{2(2\backslash pi)^\{(n-1)/2\}\}\{n!!\}R^n.$

Instead of expressing the volume {{math|V}} of the ball in terms of its radius {{math|R}}, the formulas can be inverted to express the radius as a function of the volume:

:$\backslash begin\{align\}$

R_n(V) &= \frac{\Gamma\bigl(\tfrac n2 + 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{align}

Stirling's approximation for the gamma function can be used to approximate the volume when the number of dimensions is high.

:$V\_n(R)\; \backslash sim\; \backslash frac\{1\}\{\backslash sqrt\{n\backslash pi\}\}\backslash left(\backslash frac\{2\backslash pi\; e\}\{n\}\backslash right)^\{n/2\}R^n.$

:$R\_n(V)\; \backslash sim\; (\backslash pi\; n)^\{1/(2n)\}\backslash sqrt\{\backslash frac\{n\}\{2\backslash pi\; e\}\}\; V^\{1/n\}.$

In particular, for any fixed value of {{math|R}} the volume tends to a limiting value of 0 as {{math|n}} goes to infinity. Which value of {{mvar|n}} maximizes {{math|V{{sub|n}}(R)}} depends upon the value of {{mvar|R}}; for example, the volume {{math|V_n(1)}} is increasing for {{math|0 ≤ n ≤ 5}}, achieves its maximum when {{math|n {{=}} 5}}, and is decreasing for {{math|n ≥ 5}}.Smith, David J. and Vamanamurthy, Mavina K., "How Small Is a Unit Ball?", Mathematics Magazine, Volume 62, Issue 2, 1989, pp. 101–107, https://doi.org/10.1080/0025570X.1989.11977419.

Also, there is an asymptotic formula for the surface area{{Cite web | url=https://www.stat.berkeley.edu/~songmei/Teaching/STAT260_Spring2021/Lecture_notes/scribe_lecture7.pdf | title=Lecture 7: Concentration Inequalities and Field theoretic calculations | date=2021-02-10 | author=Song Mei | website=www.stat.berkeley.edu}}$$\backslash lim\_n\; \backslash frac\; 1n\; \backslash ln\; A\_\{n-1\}(\backslash sqrt\; n)\; =\; \backslash frac\; 12\; (\backslash ln(2\backslash pi)\; +\; 1)$$

File:Surface areas of unit hyperspheres.svg

Let {{math|A{{sub|n − 1}}(R)}} denote the hypervolume of the n-sphere of radius {{math|R}}. The {{math|(n − 1)}}-sphere is the {{math|(n − 1)}}-dimensional boundary (surface) of the {{math|n}}-dimensional ball of radius {{math|R}}, and the sphere's hypervolume and the ball's hypervolume are related by:

:$A\_\{n-1\}(R)\; =\; \backslash frac\{d\}\{dR\}\; V\_\{n\}(R)\; =\; \backslash frac\{n\}\{R\}V\_\{n\}(R).$

Thus, {{math|A{{sub|n − 1}}(R)}} inherits formulas and recursion relationships from {{math|V{{sub|n}}(R)}}, such as

:$A\_\{n-1\}(R)\; =\; \backslash frac\{2\backslash pi^\{n/2\}\}\{\backslash Gamma\backslash bigl(\backslash tfrac\; n2\backslash bigr)\}R^\{n-1\}.$

There are also formulas in terms of factorials and double factorials.

There are many proofs of the above formulas.

An important step in several proofs about volumes of {{math|n}}-balls, and a generally useful fact besides, is that the volume of the {{math|n}}-ball of radius {{math|R}} is proportional to {{math|Rⁿ}}:

:$V\_n(R)\; \backslash propto\; R^n.$

The proportionality constant is the volume of the unit ball.

This is a special case of a general fact about volumes in {{math|n}}-dimensional space: If {{math|K}}

is a body (measurable set) in that space and {{math|RK}} is the body obtained by stretching in all directions by the factor {{math|R}} then the volume of {{math|RK}} equals {{math|Rⁿ}} times the volume of {{math|K}}. This is a direct consequence of the change of variables formula:

:$V(RK)\; =\; \backslash int\_\{RK\}\; dx\; =\; \backslash int\_K\; R^n\backslash ,\; dy\; =\; R^n\; V(K)$

where {{math|1=dx = dx₁…dx_n}} and the substitution {{math|1=x = Ry}} was made.

Another proof of the above relation, which avoids multi-dimensional integration, uses induction: The base case is {{math|n {{=}} 0}}, where the proportionality is obvious. For the inductive step, assume that proportionality is true in dimension {{math|n − 1}}. Note that the intersection of an n-ball with a hyperplane is an {{math|(n − 1)}}-ball. When the volume of the {{math|n}}-ball is written as an integral of volumes of {{math|(n − 1)}}-balls:

:$V\_n(R)\; =\; \backslash int\_\{-R\}^R\; V\_\{n-1\}\backslash !\backslash left(\backslash sqrt\{R^2\; -\; x^2\}\backslash right)\; dx,$

it is possible by the inductive hypothesis to remove a factor of {{math|R}} from the radius of the {{math|(n − 1)}}-ball to get:

:$V\_n(R)\; =\; R^\{n-1\}\backslash !\; \backslash int\_\{-R\}^R\; V\_\{n-1\}\backslash !\backslash left(\backslash sqrt\{1\; -\; \backslash left(\backslash frac\{x\}\{R\}\backslash right)^2\}\backslash right)\; dx.$

Making the change of variables {{math|t {{=}} {{sfrac|x|R}}}} leads to:

:$V\_n(R)\; =\; R^n\backslash !\; \backslash int\_\{-1\}^1\; V\_\{n-1\}\backslash !\backslash left(\backslash sqrt\{1\; -\; t^2\}\backslash right)\; dt\; =\; R^n\; V\_n(1),$

which demonstrates the proportionality relation in dimension {{math|n}}. By induction, the proportionality relation is true in all dimensions.

A proof of the recursion formula relating the volume of the {{math|n}}-ball and an {{math|(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 {{math|r}} denote the distance between a point in the plane and the center of the sphere, and let {{math|θ}} denote the azimuth. Intersecting the {{math|n}}-ball with the {{math|(n − 2)}}-dimensional plane defined by fixing a radius and an azimuth gives an {{math|(n − 2)}}-ball of radius {{math|{{sqrt|R² − r²}}}}. The volume of the ball can therefore be written as an iterated integral of the volumes of the {{math|(n − 2)}}-balls over the possible radii and azimuths:

:$V\_n(R)\; =\; \backslash int\_0^\{2\backslash pi\}\; \backslash int\_0^R\; V\_\{n-2\}\backslash !\backslash left(\backslash sqrt\{R^2\; -\; r^2\}\backslash right)\; r\backslash ,dr\backslash ,d\backslash theta,$

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

:$V\_n(R)\; =\; 2\backslash pi\; V\_\{n-2\}(R)\; \backslash int\_0^R\; \backslash left(1\; -\; \backslash left(\backslash frac\{r\}\{R\}\backslash right)^2\backslash right)^\{(n-2)/2\}\backslash ,r\backslash ,dr.$

The integral can be evaluated by making the substitution {{math|u {{=}} 1 − ({{sfrac|r|R}}){{su|p=2}}}} to get

:$\backslash begin\{align\}$

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{align}

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 {{math|Γ(1) {{=}} 1}} and {{math|Γ({{sfrac|3|2}}) {{=}} {{sfrac|1|2}} · Γ({{sfrac|1|2}}) {{=}} {{sfrac|{{sqrt|{{pi}}}}|2}}}}. The inductive step is similar to the above, but instead of applying proportionality to the volumes of the {{math|(n − 2)}}-balls, the inductive hypothesis is applied instead.

The proportionality relation can also be used to prove the recursion formula relating the volumes of an {{math|n}}-ball and an {{math|(n − 1)}}-ball. As in the proof of the proportionality formula, the volume of an {{math|n}}-ball can be written as an integral over the volumes of {{math|(n − 1)}}-balls. Instead of making a substitution, however, the proportionality relation can be applied to the volumes of the {{math|(n − 1)}}-balls in the integrand:

:$V\_n(R)\; =\; V\_\{n-1\}(R)\; \backslash int\_\{-R\}^R\; \backslash left(1\; -\; \backslash left(\backslash frac\{x\}\{R\}\backslash right)^2\backslash right)^\{(n-1)/2\}\; \backslash ,dx.$

The integrand is an even function, so by symmetry the interval of integration can be restricted to {{math|[0, R]}}. On the interval {{math|[0, R]}}, it is possible to apply the substitution {{math|u {{=}} ({{sfrac|x|R}}){{su|p=2}}}}. This transforms the expression into

:$V\_\{n-1\}(R)\; \backslash cdot\; R\; \backslash cdot\; \backslash int\_0^1\; (1-u)^\{(n-1)/2\}u^\{-\backslash frac12\}\backslash ,du$

The integral is a value of a well-known special function called the beta function {{math|Β(x, y)}}, and the volume in terms of the beta function is

:$V\_n(R)\; =\; V\_\{n-1\}(R)\; \backslash cdot\; R\; \backslash cdot\; \backslash Beta\backslash left(\backslash tfrac\{n\; +\; 1\}2,\; \backslash tfrac12\backslash right).$

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

:$V\_n(R)\; =\; V\_\{n-1\}(R)\; \backslash cdot\; R\; \backslash cdot\; \backslash frac\{\backslash Gamma\backslash bigl(\backslash tfrac\; n2\; +\; \backslash tfrac12\; \backslash bigr)\backslash Gamma\backslash bigl(\backslash tfrac12\backslash bigr)\}\{\backslash Gamma\backslash bigl(\backslash tfrac\; n2\; +\; 1\backslash bigr)\}.$

Using the value {{math|Γ({{sfrac|1|2}}) {{=}} {{sqrt|{{pi}}}}}} gives the one-dimension recursion formula:

:$V\_n(R)\; =\; R\backslash sqrt\{\backslash pi\}\backslash frac\{\backslash Gamma\backslash bigl(\backslash tfrac\; n2\; +\; \backslash tfrac12\backslash bigr)\}\{\backslash Gamma\backslash bigl(\backslash tfrac\; n2\; +\; 1\backslash bigr)\}\; V\_\{n-1\}(R).$

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

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 {{math|r}} and angular coordinates {{math|φ₁, …, φ_{n − 1}}}, where the domain of each {{math|φ}} except {{math|φ_{n − 1}}} is {{math|[0, {{pi}})}}, and the domain of {{math|φ_{n − 1}}} is {{math|[0, 2{{pi}})}}. The spherical volume element is:

:$dV\; =\; r^\{n-1\}\backslash sin^\{n-2\}(\backslash varphi\_1)\backslash sin^\{n-3\}(\backslash varphi\_2)\; \backslash cdots\; \backslash sin(\backslash varphi\_\{n-2\})\backslash ,dr\backslash ,d\backslash varphi\_1\backslash ,d\backslash varphi\_2\; \backslash cdots\; d\backslash varphi\_\{n-1\},$

and the volume is the integral of this quantity over {{math|r}} between 0 and {{math|R}} and all possible angles:

:$V\_n(R)\; =\; \backslash int\_0^R\; \backslash int\_0^\backslash pi\; \backslash cdots\; \backslash int\_0^\{2\backslash pi\}\; r^\{n-1\}\backslash sin^\{n-2\}(\backslash varphi\_1)\; \backslash cdots\; \backslash sin(\backslash varphi\_\{n-2\})\backslash ,d\backslash varphi\_\{n-1\}\; \backslash cdots\; d\backslash varphi\_1\backslash ,dr.$

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:

:$V\_n(R)\; =\; \backslash left(\backslash int\_0^R\; r^\{n-1\}\backslash ,dr\backslash right)\backslash !\backslash left(\backslash int\_0^\backslash pi\; \backslash sin^\{n-2\}(\backslash varphi\_1)\backslash ,d\backslash varphi\_1\backslash right)\backslash cdots\backslash left(\backslash int\_0^\{2\backslash pi\}\; d\backslash varphi\_\{n-1\}\backslash right).$

The integral over the radius is {{math|{{sfrac|Rⁿ|n}}}}. The intervals of integration on the angular coordinates can, by the symmetry of the sine about {{sfrac|{{pi}}|2}}, be changed to {{math|[0, {{sfrac|{{pi}}|2}}]}}:

:$V\_n(R)\; =\; \backslash frac\{R^n\}\{n\}\; \backslash left(2\backslash int\_0^\{\backslash pi/2\}\; \backslash sin^\{n-2\}(\backslash varphi\_1)\backslash ,d\backslash varphi\_1\backslash right)\; \backslash cdots\; \backslash left(4\backslash int\_0^\{\backslash pi/2\}\; d\backslash varphi\_\{n-1\}\backslash right).$

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

:$V\_n(R)\; =\; \backslash frac\{R^n\}\{n\}\; \backslash Beta\backslash bigl(\backslash tfrac\{n-1\}2,\; \backslash tfrac12\backslash bigr)\; \backslash Beta\backslash bigl(\backslash tfrac\{n-2\}2,\; \backslash tfrac12\backslash bigr)\; \backslash cdots\; \backslash Beta\backslash bigl(1,\; \backslash tfrac12\backslash bigr)\; \backslash cdot\; 2\backslash ,\backslash Beta\backslash bigl(\backslash tfrac12,\; \backslash tfrac12\backslash bigr).$

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

:$V\_n(R)\; =\; \backslash frac\{R^n\}\{n\}\; \backslash cdot\; \backslash frac\{\backslash Gamma\backslash bigl(\backslash tfrac\; n2\; -\; \backslash tfrac12\backslash bigr)\backslash Gamma\backslash bigl(\backslash tfrac12\backslash bigr)\}\{\backslash Gamma\backslash bigl(\backslash tfrac\; n2\backslash bigr)\}\; \backslash cdot\; \backslash frac\{\backslash Gamma\backslash bigl(\backslash tfrac\; n2\; -\; 1\backslash bigr)\backslash Gamma\backslash bigl(\backslash tfrac12\backslash bigr)\}\{\backslash Gamma\backslash bigl(\backslash tfrac\; n2\; -\; \backslash tfrac12\backslash bigr)\}\; \backslash cdots\; \backslash frac\{\backslash Gamma(1)\backslash Gamma\backslash bigl(\backslash tfrac12\backslash bigr)\}\{\backslash Gamma\backslash bigl(\backslash tfrac32\backslash bigr)\}\; \backslash cdot\; 2\; \backslash frac\{\backslash Gamma\backslash bigl(\backslash tfrac12\backslash bigr)\backslash Gamma\backslash bigl(\backslash tfrac12\backslash bigr)\}\{\backslash Gamma(1)\}.$

This product telescopes. Combining this with the values {{math|Γ({{sfrac|1|2}}) {{=}} {{sqrt|{{pi}}}}}} and {{math|Γ(1) {{=}} 1}} and the functional equation {{math|zΓ(z) {{=}} Γ(z + 1)}} leads to

:$V\_n(R)\; =\; \backslash frac\{2\backslash pi^\{n/2\}R^n\}\{n\backslash ,\backslash Gamma\backslash bigl(\backslash tfrac\; n2\backslash bigr)\}\; =\; \backslash frac\{\backslash pi^\{n/2\}R^n\}\{\backslash Gamma\backslash bigl(\backslash tfrac\; n2\; +\; 1\backslash bigr)\}.$

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

:$f(x\_1,\; \backslash ldots,\; x\_n)\; =\; \backslash exp\backslash biggl(\{-\backslash tfrac12\; \backslash sum\_\{i=1\}^n\; x\_i^2\}\backslash biggr).$

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:

:$\backslash int\_\{\backslash mathbf\{R\}^n\}\; f\; \backslash ,dV\; =\; \backslash prod\_\{i=1\}^n\; \backslash left(\backslash int\_\{-\backslash infty\}^\backslash infty\; \backslash exp\backslash left(-\backslash tfrac12\; x\_i^2\backslash right)\backslash ,dx\_i\backslash right)\; =\; (2\backslash pi)^\{n/2\},$

where {{math|dV}} is the {{math|n}}-dimensional volume element. Using rotational invariance, the same integral can be computed in spherical coordinates:

:$\backslash int\_\{\backslash mathbf\{R\}^n\}\; f\; \backslash ,dV\; =\; \backslash int\_0^\backslash infty\; \backslash int\_\{S^\{n-1\}(r)\}\; \backslash exp\backslash left(-\backslash tfrac12\; r^2\backslash right)\; \backslash ,dA\backslash ,dr,$

where {{math|S^{n − 1}(r)}} is an {{math|(n − 1)}}-sphere of radius {{math|r}} (being the surface of an {{math|n}}-ball of radius {{math|r}}) and {{math|dA}} is the area element (equivalently, the {{math|(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 {{math|A_{n − 1}(r)}} is the surface area of an {{math|(n − 1)}}-sphere of radius {{math|r}}, then:

:$A\_\{n-1\}(r)\; =\; r^\{n-1\}\; A\_\{n-1\}(1).$

Applying this to the above integral gives the expression

:$(2\backslash pi)^\{n/2\}\; =\; \backslash int\_0^\backslash infty\; \backslash int\_\{S^\{n-1\}(r)\}\; \backslash exp\backslash left(-\backslash tfrac12\; r^2\backslash right)\; \backslash ,dA\backslash ,dr\; =\; A\_\{n-1\}(1)\; \backslash int\_0^\backslash infty\; \backslash exp\backslash left(-\backslash tfrac12\; r^2\backslash right)\backslash ,r^\{n-1\}\backslash ,dr.$

Substituting {{math|t {{=}} {{sfrac|r²|2}}}}:

:$\backslash int\_0^\backslash infty\; \backslash exp\backslash left(-\backslash tfrac12\; r^2\backslash right)\backslash ,r^\{n-1\}\backslash ,dr\; =\; 2^\{(n-2)/2\}\; \backslash int\_0^\backslash infty\; e^\{-t\}\; t^\{(n-2)/2\}\backslash ,dt.$

The integral on the right is the gamma function evaluated at {{math|{{sfrac|n|2}}}}.

Combining the two results shows that

:$A\_\{n-1\}(1)\; =\; \backslash frac\{2\backslash pi^\{n/2\}\}\{\backslash Gamma\backslash bigl(\backslash tfrac\; n2\backslash bigr)\}.$

To derive the volume of an {{math|n}}-ball of radius {{math|R}} from this formula, integrate the surface area of a sphere of radius {{math|r}} for {{math|0 ≤ r ≤ R}} and apply the functional equation {{math|zΓ(z) {{=}} Γ(z + 1)}}:

:$V\_n(R)\; =\; \backslash int\_0^R\; \backslash frac\{2\backslash pi^\{n/2\}\}\{\backslash Gamma\backslash bigl(\backslash tfrac\; n2\backslash bigr)\}\; \backslash ,r^\{n-1\}\backslash ,dr\; =\; \backslash frac\{2\backslash pi^\{n/2\}\}\{n\backslash ,\backslash Gamma\backslash bigl(\backslash tfrac\; n2\backslash bigr)\}R^n\; =\; \backslash frac\{\backslash pi^\{n/2\}\}\{\backslash Gamma\backslash bigl(\backslash tfrac\; n2\; +\; 1\backslash bigr)\}R^n.$

The relations $V\_\{n+1\}(R)\; =\; \backslash frac\{R\}\{n+1\}A\_n(R)$ and $A\_\{n+1\}(R)\; =\; (2\backslash pi\; R)V\_n(R)$ 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 $\backslash frac\{dV\_n(R)\}\{dR\}\; =\; \backslash frac\{n\}\{R\}\; V\_n(R)$. Also, $A\_\{n-1\}(R)\; =\; \backslash frac\{dV\_n(R)\}\{dR\}$ because a ball is a union of concentric spheres and increasing radius by ε corresponds to a shell of thickness ε. Thus, $V\_\{n\}(R)\; =\; \backslash frac\{R\}\{n\}A\_\{n-1\}(R)$; equivalently, $V\_\{n+1\}(R)\; =\; \backslash frac\{R\}\{n+1\}A\_n(R)$.

$A\_\{n+1\}(R)\; =\; (2\backslash pi\; R)V\_n(R)$ follows from existence of a volume-preserving bijection between the unit sphere $S\_\{n+1\}$ and $S\_1\; \backslash times\; B\_n$:

: $(x,y,\backslash vec\{z\})\; \backslash mapsto\; \backslash left(\backslash frac\{x\}\{\backslash sqrt\{x^2+y^2\}\},\backslash frac\{y\}\{\backslash sqrt\{x^2+y^2\}\},\backslash vec\{z\}\backslash right)$

($\backslash vec\{z\}$ is an n-tuple; $|(x,y,\backslash vec\{z\})|=1$; 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 $1/\backslash !\backslash sqrt\{x^2+y^2\}$ times in the direction of constant $x^2+y^2$) that exactly matches the compression in the direction of the gradient of $|\backslash vec\{z\}|$ 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.

There are also explicit expressions for the volumes of balls in Lp spaces. The {{math|L{{isup|p}}}} norm of the vector {{math|x {{=}} (x₁, …, x_n)}} in {{math|Rⁿ}} is

:$\backslash |x\backslash |\_p\; =\; \backslash biggl(\backslash sum\_\{i=1\}^n\; |\; x\_i\; |^p\; \backslash biggr)^\{\backslash !\; 1/p\},$

and an {{math|L{{isup|p}}}} ball is the set of all vectors whose {{math|L{{isup|p}}}} norm is less than or equal to a fixed number called the radius of the ball. The case {{math|p {{=}} 2}} is the standard Euclidean distance function, but other values of {{math|p}} occur in diverse contexts such as information theory, coding theory, and dimensional regularization.

The volume of an {{math|L{{isup|p}}}} ball of radius {{math|R}} is

:$V^p\_n(R)\; =\; \backslash frac\{\backslash Bigl(2\backslash ,\backslash Gamma\backslash bigl(\backslash tfrac1p\; +\; 1\backslash bigr)\backslash Bigr)^n\}\{\backslash Gamma\backslash bigl(\backslash tfrac\; np\; +\; 1\backslash bigr)\}R^n.$

These volumes satisfy recurrence relations similar to those for {{math|p {{=}} 2}}:

:$V^p\_n(R)\; =\; \backslash frac\{\backslash Bigl(2\backslash ,\backslash Gamma\backslash bigl(\backslash tfrac1p\; +\; 1\backslash bigr)\backslash Bigr)^p\; p\}\{n\}\; R^p\; \backslash ,\; V^p\_\{n-p\}(R)$

and

:$V^p\_n(R)\; =\; 2\; \backslash frac\{\backslash Gamma\backslash bigl(\backslash tfrac1p\; +\; 1\backslash bigr)\backslash Gamma\backslash bigl(\backslash tfrac\{n-1\}p\; +\; 1\backslash bigr)\}\{\backslash Gamma\backslash bigl(\backslash tfrac\; np\; +\; 1\backslash bigr)\}\; R\; \backslash ,\; V^p\_\{n-1\}(R),$

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

:$V^p\_n(R)\; =\; \backslash frac\{2\}\{\backslash dbinom\{n/p\}\{1/p\}\}\; R\; \backslash ,\; V^p\_\{n-1\}(R).$

For {{math|p {{=}} 2}}, one recovers the recurrence for the volume of a Euclidean ball because {{math|2Γ({{sfrac|3|2}}) {{=}} {{radic|{{pi}}}}}}.

For example, in the cases {{math|p {{=}} 1}} (taxicab norm) and {{math|p {{=}} ∞}} (max norm), the volumes are:

:

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

For most values of {{math|p}}, the surface area $A\_\{n-1\}^p(R)$ of an {{math|L{{isup|p}}}} sphere of radius {{math|R}} (the boundary of an {{math|L{{isup|p}}}} {{math|n}}-ball of radius {{math|R}}) cannot be calculated by differentiating the volume of an {{math|L{{isup|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 {{math|p}}-norm varies from point to point. For {{math|p {{=}} 2}} and {{math|p {{=}} ∞}}, this factor is one. However, if {{math|p {{=}} 1}} then the correction factor is {{math|{{sqrt|n}}}}: the surface area of an {{math|L¹}} sphere of radius {{math|R}} in {{math|Rⁿ}} is {{math|{{sqrt|n}}}} times the derivative of the volume of an {{math|L¹}} ball. This can be seen most simply by applying the divergence theorem to the vector field {{math|F(x) {{=}} x}} to get

:$nV^1\_n(R)\; =${{oiint

| preintegral = $\backslash iiint\_V\backslash left(\backslash mathbf\{\backslash nabla\}\backslash cdot\backslash mathbf\{F\}\backslash right)\backslash ,dV=$

| intsubscpt = $\backslash scriptstyle\; S$

| integrand = $(\backslash mathbf\{F\}\backslash cdot\backslash mathbf\{n\})\backslash ,dS$

}} $=$ {{oiint

| intsubscpt = $\backslash scriptstyle\; S$

| integrand = $\backslash frac\{1\}\{\backslash sqrt\{n\}\}(|x\_1|+\backslash cdots+|x\_n|)\backslash ,dS$

}} $=\; \backslash frac\{R\}\{\backslash sqrt\{n\}\}${{oiint

| intsubscpt = $\backslash scriptstyle\; S$

| integrand = $\backslash ,dS$

}} $=\; \backslash frac\{R\}\{\backslash sqrt\{n\}\}\; A^1\_\{n-1\}(R).$

For other values of {{math|p}}, the constant is a complicated integral.

The volume formula can be generalized even further. For positive real numbers {{math|p₁, …, p_n}}, define the {{math|(p₁, …, p_n)}} ball with limit {{math|L ≥ 0}} to be

:$B\_\{p\_1,\; \backslash ldots,\; p\_n\}(L)\; =\; \backslash left\backslash \{\; x\; =\; (x\_1,\; \backslash ldots,\; x\_n)\; \backslash in\; \backslash mathbf\{R\}^n\; :\; \backslash vert\; x\_1\; \backslash vert^\{p\_1\}\; +\; \backslash cdots\; +\; \backslash vert\; x\_n\; \backslash vert^\{p\_n\}\; \backslash le\; L\; \backslash right\backslash \}.$

The volume of this ball has been known since the time of Dirichlet:{{cite journal|last=Dirichlet|first=P. G. Lejeune|title=Sur une nouvelle méthode pour la détermination des intégrales multiples|trans-title=On a novel method for determining multiple integrals|journal=Journal de Mathématiques Pures et Appliquées|volume=4|date=1839|pages=164–168}}

:$V\backslash bigl(B\_\{p\_1,\; \backslash ldots,\; p\_n\}(L)\backslash bigr)\; =\; \backslash frac\{2^n\; \backslash Gamma\backslash bigl(\backslash tfrac1\{p\_1\}\; +\; 1\backslash bigr)\; \backslash cdots\; \backslash Gamma\backslash bigl(\backslash tfrac1\{p\_n\}\; +\; 1\backslash bigr)\}\{\backslash Gamma\backslash bigl(\backslash tfrac1\{p\_1\}\; +\; \backslash cdots\; +\; \backslash tfrac1\{p\_n\}\; +\; 1\backslash bigr)\}\; L^\{\backslash tfrac1\{p\_1\}\; +\; \backslash cdots\; +\; \backslash tfrac1\{p\_n\}\}.$

Using the harmonic mean $p\; =\; \backslash frac\{n\}\{\backslash frac\{1\}\{p\_1\}\; +\; \backslash cdots\; \backslash frac\{1\}\{p\_n\}\}$ and defining $R\; =\; \backslash sqrt[p]\{L\}$, the similarity to the volume formula for the {{math|L{{isup|p}}}} ball becomes clear.

:$V\backslash left(\backslash left\backslash \{\; x\; \backslash in\; \backslash mathbf\{R\}^n\; :\; \backslash sqrt[p]\{\backslash vert\; x\_1\; \backslash vert^\{p\_1\}\; +\; \backslash cdots\; +\; \backslash vert\; x\_n\; \backslash vert^\{p\_n\}\}\; \backslash le\; R\; \backslash right\backslash \}\backslash right)\; =\; \backslash frac\{2^n\; \backslash Gamma\backslash bigl(\backslash tfrac1\{p\_1\}\; +\; 1\backslash bigr)\; \backslash cdots\; \backslash Gamma\backslash bigl(\backslash tfrac1\{p\_n\}\; +\; 1\backslash bigr)\}\{\backslash Gamma\backslash bigl(\backslash tfrac\{n\}\{p\}\; +\; 1\backslash bigr)\}\; R^n.$

• n-sphere
• Sphere packing
• Hamming bound

{{Reflist}}

• {{Cite journal |last=Hayes |first=Brian| author-link=Brian Hayes (scientist) |title=An Adventure in the Nth Dimension |url=https://www.americanscientist.org/article/an-adventure-in-the-nth-dimension |access-date=October 24, 2024 |journal=American Scientist |volume=99 |issue=9| date=November–December 2011| page=442 |doi=10.1511/2011.93.442}}, bibliography, accessible to layman.
• Updated version: {{Cite book |last=Hayes |first=Brian |author-link=Brian Hayes (scientist) |chapter=Playing Ball in the nth Dimension |title=Foolproof, and Other Mathematical Meditations |publisher=The MIT Press |publication-date=September 22, 2017 |isbn=9780262036863}}, bibliography, accessible to layman.

• [http://www.brouty.fr/Maths/sphere.html Derivation in hyperspherical coordinates] {{in lang|fr}}
• [http://mathworld.wolfram.com/Hypersphere.html Hypersphere] on Wolfram MathWorld
• [http://www.mathreference.com/ca-int,hsp.html Volume of the Hypersphere] at Math Reference

Category:Multi-dimensional geometry

Category:Size