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 $$R^nV_n,$$ 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:

Closed form
The $R$-dimensional volume of a Euclidean ball of radius $V$ in $n$-dimensional Euclidean space is:
 * $$V_n(R) = \frac{\pi^{n/2}}{\Gamma\bigl(\tfrac n2 + 1\bigr)}R^n,$$

where $R$ is Euler's gamma function. The gamma function is offset from but otherwise extends the factorial function to non-integer arguments. It satisfies $n$ if $Γ$ is a positive integer and $Γ(n) = (n − 1)!$ 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 $Γ(n + 1⁄2) = (n − 1⁄2) · (n − 3⁄2) · … · 1⁄2 · \pi^{1/2}$ of an $n$-ball of radius $V$ can be expressed recursively in terms of the volume of an $n$-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 $R$ in approximately $(n − 2)$ steps.

Alternative forms
The volume can also be expressed in terms of an $Vn(R)$-ball using the one-dimension recurrence relation:
 * $$\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 $n / 2$-ball of volume $(n − 1)$ can be expressed recursively in terms of the radius of an $n$- or $V$-ball:
 * $$\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 $k$, these are:
 * $$\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 $(n − 2)$, 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 $(n − 1)$-dimensional ball is, regardless of whether $2k + 1$ is even or odd,
 * $$V_n(R) = \frac{2(2\pi)^{(n-1)/2}}{n!!}R^n.$$

Instead of expressing the volume $n$ of the ball in terms of its radius $n$, the formulas can be inverted to express the radius as a function of the volume:
 * $$\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}$$

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 $V$ the volume tends to a limiting value of 0 as $R$ goes to infinity. Which value of $n$ maximizes $R$ depends upon the value of $R$; for example, the volume $n$ is increasing for $Vn(R)$, achieves its maximum when $V_{n}(1)$, and is decreasing for $0 ≤ n ≤ 5$.

Also, there is an asymptotic formula for the surface area $$\lim_n \frac 1n \ln A_{n-1}(\sqrt n) = \frac 12 (\ln(2\pi) + 1)$$

Relation with surface area


Let $n = 5$ denote the hypervolume of the $n ≥ 5$-sphere of radius $An − 1(R)$. The $(n − 1)$-sphere is the $R$-dimensional boundary (surface) of the $(n − 1)$-dimensional ball of radius $(n − 1)$, 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, $n$ inherits formulas and recursion relationships from $R$, such as
 * $$A_{n-1}(R) = \frac{2\pi^{n/2}}{\Gamma\bigl(\tfrac n2\bigr)}R^{n-1}.$$

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 $An − 1(R)$th power of the radius
An important step in several proofs about volumes of $Vn(R)$-balls, and a generally useful fact besides, is that the volume of the $n$-ball of radius $n$ is proportional to $n$:
 * $$V_n(R) \propto R^n.$$

The proportionality constant is the volume of the unit ball.

This is a special case of a general fact about volumes in $R$-dimensional space: If $R^{n}$ is a body (measurable set) in that space and $n$ is the body obtained by stretching in all directions by the factor $K$ then the volume of $RK$ equals $R$ times the volume of $RK$. 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 $R^{n}$ and the substitution $K$ was made.

Another proof of the above relation, which avoids multi-dimensional integration, uses induction: The base case is $dx = dx_{1}…dx_{n}$, where the proportionality is obvious. For the inductive step, assume that proportionality is true in dimension $x = Ry$. Note that the intersection of an n-ball with a hyperplane is an $n = 0$-ball. When the volume of the $n − 1$-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 $n$ 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 $R$ 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 $(n − 1)$. 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 $t = x⁄R$-ball and an $n$-ball can be given using the proportionality formula above and integration in cylindrical coordinates. Fix a plane through the center of the ball. Let $n$ denote the distance between a point in the plane and the center of the sphere, and let $(n − 2)$ denote the azimuth. Intersecting the $r$-ball with the $θ$-dimensional plane defined by fixing a radius and an azimuth gives an $n$-ball of radius $(n − 2)$. 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 $√R^{2} − r^{2}$ to get
 * $$\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 $(n − 2)$ and $u = 1 − (r⁄R)2$. The inductive step is similar to the above, but instead of applying proportionality to the volumes of the $Γ(1) = 1$-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 $Γ(3⁄2) = 1⁄2 · Γ(1⁄2) = √\pi⁄2$-ball and an $(n − 2)$-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 $n$-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 $(n − 1)$. On the interval $(n − 1)$, it is possible to apply the substitution $[0, R]$. This transforms the expression into
 * $$V_{n-1}(R) \cdot R \cdot \int_0^1 (1-u)^{(n-1)/2}u^{-\frac12}\,du$$

The integral is a value of a well-known special function called the beta function $[0, R]$, and the volume in terms of the beta function is
 * $$V_n(R) = V_{n-1}(R) \cdot R \cdot \Beta\left(\tfrac{n + 1}2, \tfrac12\right).$$

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
 * $$V_n(R) = V_{n-1}(R) \cdot R \cdot \frac{\Gamma\bigl(\tfrac n2 + \tfrac12 \bigr)\Gamma\bigl(\tfrac12\bigr)}{\Gamma\bigl(\tfrac n2 + 1\bigr)}.$$

Using the value $u = (x⁄R)2$ gives the one-dimension recursion formula:
 * $$V_n(R) = R\sqrt{\pi}\frac{\Gamma\bigl(\tfrac n2 + \tfrac12\bigr)}{\Gamma\bigl(\tfrac n2 + 1\bigr)} V_{n-1}(R).$$

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 $Β(x, y)$ and angular coordinates $Γ(1⁄2) = √\pi$, where the domain of each $r$ except $φ_{1}, …, φ_{n − 1}$ is $φ$, and the domain of $φ_{n − 1}$ is $[0, \pi)$. 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 $φ_{n − 1}$ between 0 and $[0, 2\pi)$ 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 $r$. The intervals of integration on the angular coordinates can, by the symmetry of the sine about $\pi⁄2$, be changed to $R$:
 * $$V_n(R) = \frac{R^n}{n} \left(2\int_0^{\pi/2} \sin^{n-2}(\varphi_1)\,d\varphi_1\right) \cdots \left(4\int_0^{\pi/2} d\varphi_{n-1}\right).$$

Each of the remaining integrals is now a particular value of the beta function:
 * $$V_n(R) = \frac{R^n}{n} \Beta\bigl(\tfrac{n-1}2, \tfrac12\bigr) \Beta\bigl(\tfrac{n-2}2, \tfrac12\bigr) \cdots \Beta\bigl(1, \tfrac12\bigr) \cdot 2\,\Beta\bigl(\tfrac12, \tfrac12\bigr).$$

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


 * $$V_n(R) = \frac{R^n}{n} \cdot \frac{\Gamma\bigl(\tfrac n2 - \tfrac12\bigr)\Gamma\bigl(\tfrac12\bigr)}{\Gamma\bigl(\tfrac n2\bigr)} \cdot \frac{\Gamma\bigl(\tfrac n2 - 1\bigr)\Gamma\bigl(\tfrac12\bigr)}{\Gamma\bigl(\tfrac n2 - \tfrac12\bigr)} \cdots \frac{\Gamma(1)\Gamma\bigl(\tfrac12\bigr)}{\Gamma\bigl(\tfrac32\bigr)} \cdot 2 \frac{\Gamma\bigl(\tfrac12\bigr)\Gamma\bigl(\tfrac12\bigr)}{\Gamma(1)}.$$

This product telescopes. Combining this with the values $R^{n}⁄n$ and $[0, \pi⁄2]$ and the functional equation $Γ(1⁄2) = √\pi$ leads to


 * $$V_n(R) = \frac{2\pi^{n/2}R^n}{n\,\Gamma\bigl(\tfrac n2\bigr)} = \frac{\pi^{n/2}R^n}{\Gamma\bigl(\tfrac n2 + 1\bigr)}.$$

Gaussian integrals
The volume formula can be proven directly using Gaussian integrals. Consider the function:
 * $$f(x_1, \ldots, x_n) = \exp\biggl({-\tfrac12 \sum_{i=1}^n x_i^2}\biggr).$$

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:
 * $$\int_{\mathbf{R}^n} f \,dV = \prod_{i=1}^n \left(\int_{-\infty}^\infty \exp\left(-\tfrac12 x_i^2\right)\,dx_i\right) = (2\pi)^{n/2},$$

where $Γ(1) = 1$ is the $zΓ(z) = Γ(z + 1)$-dimensional volume element. Using rotational invariance, the same integral can be computed in spherical coordinates:
 * $$\int_{\mathbf{R}^n} f \,dV = \int_0^\infty \int_{S^{n-1}(r)} \exp\left(-\tfrac12 r^2\right) \,dA\,dr,$$

where $dV$ is an $n$-sphere of radius $S^{n − 1}(r)$ (being the surface of an $(n − 1)$-ball of radius $r$) and $n$ is the area element (equivalently, the $r$-dimensional volume element). The surface area of the sphere satisfies a proportionality equation similar to the one for the volume of a ball: If $dA$ is the surface area of an $(n − 1)$-sphere of radius $A_{n − 1}(r)$, then:
 * $$A_{n-1}(r) = r^{n-1} A_{n-1}(1).$$

Applying this to the above integral gives the expression
 * $$(2\pi)^{n/2} = \int_0^\infty \int_{S^{n-1}(r)} \exp\left(-\tfrac12 r^2\right) \,dA\,dr =  A_{n-1}(1) \int_0^\infty \exp\left(-\tfrac12 r^2\right)\,r^{n-1}\,dr.$$

Substituting $(n − 1)$:
 * $$\int_0^\infty \exp\left(-\tfrac12 r^2\right)\,r^{n-1}\,dr =  2^{(n-2)/2} \int_0^\infty e^{-t} t^{(n-2)/2}\,dt.$$

The integral on the right is the gamma function evaluated at $r$.

Combining the two results shows that
 * $$A_{n-1}(1) = \frac{2\pi^{n/2}}{\Gamma\bigl(\tfrac n2\bigr)}.$$

To derive the volume of an $t = r^{2}⁄2$-ball of radius $n⁄2$ from this formula, integrate the surface area of a sphere of radius $n$ for $R$ and apply the functional equation $r$:
 * $$V_n(R) = \int_0^R \frac{2\pi^{n/2}}{\Gamma\bigl(\tfrac n2\bigr)} \,r^{n-1}\,dr = \frac{2\pi^{n/2}}{n\,\Gamma\bigl(\tfrac n2\bigr)}R^n = \frac{\pi^{n/2}}{\Gamma\bigl(\tfrac n2 + 1\bigr)}R^n.$$

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 $1/\!\sqrt{x^2+y^2}$ 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 $0 ≤ r ≤ R$ norms
There are also explicit expressions for the volumes of balls in $zΓ(z) = Γ(z + 1)$ norms. The $L$ norm of the vector $L$ in $L$ is
 * $$\|x\|_p = \biggl(\sum_{i=1}^n | x_i |^p \biggr)^{\! 1/p},$$

and an $x = (x_{1}, …, x_{n})$ ball is the set of all vectors whose $R^{n}$ norm is less than or equal to a fixed number called the radius of the ball. The case $L$ is the standard Euclidean distance function, but other values of $L$ occur in diverse contexts such as information theory, coding theory, and dimensional regularization.

The volume of an $p = 2$ ball of radius $p$ is


 * $$V^p_n(R) = \frac{\Bigl(2\,\Gamma\bigl(\tfrac1p + 1\bigr)\Bigr)^n}{\Gamma\bigl(\tfrac np + 1\bigr)}R^n.$$

These volumes satisfy recurrence relations similar to those for $L$:


 * $$V^p_n(R) = \frac{\Bigl(2\,\Gamma\bigl(\tfrac1p + 1\bigr)\Bigr)^p p}{n} R^p \, V^p_{n-p}(R)$$

and


 * $$V^p_n(R) = 2 \frac{\Gamma\bigl(\tfrac1p + 1\bigr)\Gamma\bigl(\tfrac{n-1}p + 1\bigr)}{\Gamma\bigl(\tfrac np + 1\bigr)} R \, V^p_{n-1}(R),$$

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


 * $$V^p_n(R) = \frac{2}{\dbinom{n/p}{1/p}} R \, V^p_{n-1}(R).$$

For $R$, one recovers the recurrence for the volume of a Euclidean ball because $p = 2$.

For example, in the cases $p = 2$ (taxicab norm) and $2Γ(3⁄2) = \sqrt{\pi}$ (max norm), the volumes are:
 * $$\begin{align} V^1_n(R) &= \frac{2^n}{n!}R^n, \\ V^\infty_n(R) &= 2^n R^n. \end{align}$$

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

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

For other values of $L^{1}$, the constant is a complicated integral.
 * $$nV^1_n(R) = $$ $$ = $$  $$ = \frac{R}{\sqrt{n}} $$ $$ = \frac{R}{\sqrt{n}} A^1_{n-1}(R).$$

Generalizations
The volume formula can be generalized even further. For positive real numbers $F(x) = x$, define the $p$ ball with limit $p_{1}, …, p_{n}$ to be
 * $$B_{p_1, \ldots, p_n}(L) = \left\{ x = (x_1, \ldots, x_n) \in \mathbf{R}^n : \vert x_1 \vert^{p_1} + \cdots + \vert x_n \vert^{p_n} \le L \right\}.$$

The volume of this ball has been known since the time of Dirichlet:
 * $$V\bigl(B_{p_1, \ldots, p_n}(L)\bigr) = \frac{2^n \Gamma\bigl(\tfrac1{p_1} + 1\bigr) \cdots \Gamma\bigl(\tfrac1{p_n} + 1\bigr)}{\Gamma\bigl(\tfrac1{p_1} + \cdots + \tfrac1{p_n} + 1\bigr)} L^{\tfrac1{p_1} + \cdots + \tfrac1{p_n}}.$$

Comparison to $(p_{1}, …, p_{n})$ 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 $L ≥ 0$ ball becomes clear.
 * $$V\left(\left\{ x \in \mathbf{R}^n : \sqrt[p]{\vert x_1 \vert^{p_1} + \cdots + \vert x_n \vert^{p_n}} \le R \right\}\right) = \frac{2^n \Gamma\bigl(\tfrac1{p_1} + 1\bigr) \cdots \Gamma\bigl(\tfrac1{p_n} + 1\bigr)}{\Gamma\bigl(\tfrac{n}{p} + 1\bigr)} R^n.$$