Volumes of Spheres

When we first learn geometry, spheres feel completely intuitive. A circle in the plane. A ball in three-dimensional space. Everything is visual. High-dimensional geometry behaves in ways that feel almost paradoxical. Volumes shrink, surfaces dominate interiors, and many familiar formulas suddenly depend on special functions like the Gamma function.

Understanding why requires stepping away from visual geometry and entering the world of analysis, integrals, and functional identities.

An $\displaystyle (n-1)$ –sphere of radius $\displaystyle R$ is defined as the set of points in $\displaystyle \mathbb{R}^n$ satisfying $\displaystyle x_1^2 + x_2^2 + \dots + x_n^2 = R^2$ . The interior region is called the $\displaystyle n$ –ball: $\displaystyle B_n(R) = \{x \in \mathbb{R}^n : x_1^2 + \dots + x_n^2 \le R^2\}$ . We denote the volume of the $\displaystyle n$ –ball by $\displaystyle V_n(R)$ . Dimensional scaling implies $\displaystyle V_n(R) = C_n R^n$ where $\displaystyle C_n$ is the volume of the unit ball. Thus the central problem becomes determining the constant $\displaystyle C_n$ .

The Gaussian Integral Method
A powerful analytic approach to compute $\displaystyle C_n$ uses the classical Gaussian integral

$\displaystyle \int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$ .

Consider now the integral $\displaystyle I_n = \int_{\mathbb{R}^n} e^{-|x|^2} d^n x$ . Since $\displaystyle e^{-|x|^2} = e^{-(x_1^2 + \dots + x_n^2)} = \prod_{i=1}^{n} e^{-x_i^2}$ , the integral separates into $\displaystyle I_n = \prod_{i=1}^{n} \int_{-\infty}^{\infty} e^{-x_i^2} dx_i$ . Each integral contributes $\displaystyle \sqrt{\pi}$ , giving $\displaystyle I_n = (\sqrt{\pi})^n = \pi^{n/2}$ .

We can evaluate the same integral using radial symmetry. Let $\displaystyle r = |x|$ . Then $\displaystyle I_n = \int_0^\infty S_{n-1}(r) e^{-r^2} dr$ where $\displaystyle S_{n-1}(r)$ is the surface area of the $\displaystyle (n-1)$ –sphere. Since surface area scales with radius, $\displaystyle S_{n-1}(r) = n C_n r^{n-1}$ . Therefore $\displaystyle I_n = \int_0^\infty n C_n r^{n-1} e^{-r^2} dr$ .

Introduce the substitution $\displaystyle u = r^2$ . Then $\displaystyle du = 2r\,dr$ . The integral becomes $\displaystyle I_n = \frac{nC_n}{2} \int_0^\infty u^{n/2 - 1} e^{-u} du$ . But the integral $\displaystyle \Gamma(z) = \int_0^\infty t^{z-1} e^{-t} dt$ defines the Euler Gamma function. Thus $\displaystyle I_n = \frac{nC_n}{2} \Gamma(n/2)$ . Equating with the earlier result $\displaystyle \pi^{n/2} = \frac{nC_n}{2} \Gamma(n/2)$ gives $\displaystyle C_n = \frac{\pi^{n/2}}{\Gamma(n/2+1)}$ .

Hence the volume of the $\displaystyle n$ –ball is $\displaystyle V_n(R) = \frac{\pi^{n/2}}{\Gamma(n/2+1)} R^n$ .

The surface area becomes $\displaystyle S_{n-1}(R) = \frac{2\pi^{n/2}}{\Gamma(n/2)} R^{n-1}$ .

Evaluating the formula yields familiar results.
For $\displaystyle n=1$ : $\displaystyle V_1(R) = 2R$ . For $\displaystyle n=2$ : $\displaystyle V_2(R) = \pi R^2$ . For $\displaystyle n=3$ : $\displaystyle V_3(R) = \frac{4}{3}\pi R^3$ . For $\displaystyle n=4$ : $\displaystyle V_4(R) = \frac{\pi^2}{2}R^4$ .

Before interpreting the constant geometrically, it is useful to recall the basic identity $\displaystyle \Gamma(k)=(k-1)!$ for integers $\displaystyle k$ . In particular, when the real dimension $\displaystyle 2n$ is even we have $\displaystyle \Gamma(n+1)=n!$ , so the Euclidean ball volume formula $\displaystyle V_{2n}(1)=\frac{\pi^n}{\Gamma(n+1)}$ reduces simply to $\displaystyle V_{2n}(1)=\frac{\pi^n}{n!}.$

For the unit ball, $\displaystyle V_n(1) = \frac{\pi^{n/2}}{\Gamma(n/2+1)}$ . Using Stirling’s approximation $\displaystyle \Gamma(z+1) \approx \sqrt{2\pi z}\left(\frac{z}{e}\right)^z$ we obtain $\displaystyle V_n(1) \sim \left(\frac{2\pi e}{n}\right)^{n/2}$ . Because the factor $\displaystyle \frac{2\pi e}{n}$ becomes smaller than one for large $\displaystyle n$ , we find $\displaystyle V_n(1) \rightarrow 0$ as $\displaystyle n \to \infty$ . Thus extremely high dimensional spheres contain almost no volume (compared to a unit high dimensional cube).

Another striking consequence of high dimension is that most of the volume of the ball lies very close to its boundary. In fact the probability that $\displaystyle r\le 1-\varepsilon$ is approximately $\displaystyle (1-\varepsilon)^n\approx e^{-n\varepsilon}$ , so almost all of the volume lies in a thin outer shell of width about $\displaystyle O(1/n)$ near the boundary of the ball.

The Gaussian method for computing sphere volumes can also be interpreted through probabilistic sampling. If we draw a vector $\displaystyle G=(G_1,\dots,G_n)$ whose coordinates are independent standard Gaussians $\displaystyle G_i\sim\mathcal N(0,1)$ , the joint density $\displaystyle e^{-|G|^2}$ depends only on the radius $\displaystyle |G|$ , so the distribution is rotationally symmetric. Consequently the direction $\displaystyle U=\frac{G}{|G|}$ is uniformly distributed on the sphere $\displaystyle S^{n-1}$ , while the radius $\displaystyle r=|G|$ is an independent random variable with density proportional to $\displaystyle r^{\,n-1}e^{-r^2/2}$ . Thus sampling independent Gaussians is equivalent to sampling a random radius $\displaystyle r$ together with a uniform direction $\displaystyle U$ and writing $\displaystyle G=rU$ . The Gaussian integral computation then corresponds to integrating over all directions on the sphere while integrating the radial variable according to this Gaussian–induced distribution, which gives the Gamma functions.

Recursive Geometric Constructions
Another perspective arises from the recurrence relation $\displaystyle C_n = \frac{2\pi}{n} C_{n-2}$ . To derive it, consider slicing the ball along a two–dimensional plane. Fix polar coordinates $\displaystyle (r,\theta)$ . The remaining coordinates form an $\displaystyle (n-2)$ –ball with radius $\displaystyle \sqrt{R^2-r^2}$ . Thus $\displaystyle V_n(R) = \int_0^{2\pi}\int_0^R V_{n-2}(\sqrt{R^2-r^2})\,r\,dr\,d\theta$ . Substituting $\displaystyle V_{n-2}(R) = C_{n-2}R^{n-2}$ gives $\displaystyle V_n(R) = 2\pi C_{n-2} \int_0^R (R^2-r^2)^{(n-2)/2} r\,dr$ . Evaluating the integral produces $\displaystyle C_n = \frac{2\pi}{n}C_{n-2}$ .

Another useful recurrence relation for $\displaystyle V_n(R)$ arises by slicing an $\displaystyle n$ –ball along a single coordinate axis. Consider the $\displaystyle n$ –ball $\displaystyle x_1^2 + x_2^2 + \dots + x_n^2 \le R^2$ . Fix a value of $\displaystyle x_1 = x$ . The remaining coordinates satisfy $\displaystyle x_2^2 + \dots + x_n^2 \le R^2 - x^2$ , which describes an $\displaystyle (n-1)$ –ball of radius $\displaystyle \sqrt{R^2 - x^2}$ . Thus the volume of the $\displaystyle n$ –ball can be obtained by stacking these slices: $\displaystyle V_n(R) = \int_{-R}^{R} V_{n-1}(\sqrt{R^2-x^2})\,dx.$ Since $\displaystyle V_{n-1}(r)=C_{n-1}r^{\,n-1}$ , we obtain $\displaystyle V_n(R) = C_{n-1} \int_{-R}^{R} (R^2-x^2)^{(n-1)/2}dx.$ . Let $\displaystyle x = R\sin\theta$ . Then $\displaystyle dx = R\cos\theta\,d\theta$ and $\displaystyle R^2-x^2 = R^2\cos^2\theta$ . The limits become $\displaystyle -\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$ . Substituting gives $\displaystyle V_n(R) = C_{n-1}R^n \int_{-\pi/2}^{\pi/2} \cos^n\theta\,d\theta.$ . The remaining integral $\int_{-\pi/2}^{\pi/2}\cos^n\theta\,d\theta$ is a classical Wallis integral. Using symmetry, $\int_{-\pi/2}^{\pi/2}\cos^n\theta\,d\theta = 2\int_0^{\pi/2}\cos^n\theta\,d\theta.$ This can be expressed using the Beta function: $\int_0^{\pi/2}\cos^n\theta\,d\theta = \frac12 B\!\left(\frac12,\frac{n+1}{2}\right).$ Using the identity $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)},$ we obtain $\int_{-\pi/2}^{\pi/2}\cos^n\theta\,d\theta = \sqrt{\pi} \frac{\Gamma((n+1)/2)}{\Gamma(n/2+1)}.$ .

Substituting back gives $\displaystyle V_n(R) = C_{n-1}R^n \sqrt{\pi} \frac{\Gamma((n+1)/2)}{\Gamma(n/2+1)}.$ Since $\displaystyle V_n(R)=C_nR^n$ , we obtain the recurrence $\displaystyle C_n = C_{n-1} \sqrt{\pi} \frac{\Gamma((n+1)/2)}{\Gamma(n/2+1)}.$

The previous recurrences arise from slicing the $\displaystyle n$ –ball along either one or two coordinates. This idea extends naturally to slicing along a $\displaystyle k$ –dimensional subspace. In general, we obtain $\displaystyle C_n = C_{n-k} \frac{\pi^{k/2}\Gamma\!\left(\frac{n-k}{2}+1\right)}{\Gamma\!\left(\frac{n}{2}+1\right)}.$

The recurrence $\displaystyle C_n = C_{n-1}\sqrt{\pi}\,\frac{\Gamma((n+1)/2)}{\Gamma(n/2+1)}$ also reveals a key geometric feature of high dimensions. Using the asymptotic behavior of the Gamma function, $\displaystyle \frac{\Gamma((n+1)/2)}{\Gamma(n/2+1)} \sim \sqrt{\frac{2}{n}},$ the effective region contributing to the slice integral $\displaystyle V_n(R)=\int_{-R}^{R} V_{n-1}(\sqrt{R^2-x^2})\,dx$ has width $\displaystyle dx = O\left(\frac{\Gamma((n+1)/2)}{\Gamma(n/2+1)}\right) \approx O\left(\frac{1}{\sqrt{n}}\right).$ Thus most of the volume of a high–dimensional ball lies in a thin shell near the boundary.

Archimedes’ Area–Preserving Projection
One of Archimedes’ most elegant discoveries is that the surface of a sphere can be projected onto the lateral surface of its circumscribed cylinder in a way that preserves area. Consider a sphere of radius $\displaystyle R$ : $\displaystyle x^2+y^2+z^2=R^2$ . Surround the sphere with a cylinder of the same radius and height $\displaystyle 2R$ : $\displaystyle x^2+y^2=R^2, \quad -R\le z\le R$ . Archimedes observed that if points on the sphere are projected radially onto the cylinder, then small surface patches keep the same area.

At height $\displaystyle z$ , the sphere intersects the horizontal plane in a circle of radius $\displaystyle \sqrt{R^2-z^2}$ . A thin spherical band between $\displaystyle z$ and $\displaystyle z+dz$ forms a curved strip around the sphere. The circumference of the sphere at that height is $\displaystyle 2\pi \sqrt{R^2-z^2}$ . However the slope of the sphere causes the band to stretch so that its true surface width becomes $\displaystyle \frac{R}{\sqrt{R^2-z^2}}\,dz$ . Multiplying circumference by width gives the band area $\displaystyle dA_{\text{sphere}} = 2\pi R\,dz.$ But the cylindrical strip between the same heights has area $\displaystyle dA_{\text{cyl}} = (2\pi R)\,dz.$ Thus $\displaystyle dA_{\text{sphere}} = dA_{\text{cyl}}$ . Since corresponding horizontal strips have equal area, the projection preserves area globally. Integrating from $\displaystyle -R$ to $\displaystyle R$ gives $\displaystyle A_{\text{sphere}} = \int_{-R}^{R} 2\pi R\,dz = 4\pi R^2,$ which matches the lateral surface area of the cylinder.

This generalizes naturally to all dimensions by splitting the coordinates of $\displaystyle \mathbb{R}^n$ into a $\displaystyle 2$ –dimensional plane and the remaining $\displaystyle (n-2)$ coordinates. Write a point as $\displaystyle (u,w)\in\mathbb{R}^2\times\mathbb{R}^{n-2}$ , where the $\displaystyle (n-1)$ –sphere satisfies $\displaystyle |u|^2+|w|^2=R^2$ . For each fixed $\displaystyle w$ with $\displaystyle |w|<R$ , the sphere intersects the plane $\displaystyle \mathbb{R}^2\times{w}$ in a circle of radius $\displaystyle \rho=\sqrt{R^2-|w|^2}$ , which can be parameterized by an angle $\displaystyle \theta$ . Thus a convenient parameterization of the sphere is $\displaystyle F(\theta,w)=(\rho\cos\theta,\rho\sin\theta,w)$ . Archimedes’ projection sends this point radially outward in the $\displaystyle u$ –plane to the cylinder of radius $\displaystyle R$ while leaving $\displaystyle w$ unchanged, giving $\displaystyle (R\cos\theta,R\sin\theta,w)$ . Computing the surface Jacobian shows that the spherical area element becomes $\displaystyle dA=R\,d\theta\,dw$ , exactly the same expression obtained from the natural parameterization of the cylinder, so the projection preserves $(n-1)$ –dimensional surface measure. Integrating therefore gives the total surface area
$\displaystyle S_{n-1}(R)=\int_{|w|<R}\int_0^{2\pi}R\,d\theta\,dw=(2\pi R) {V}(B_{n-2}(R))$ , which for $\displaystyle n=3$ reduces to Archimedes’ famous result $\displaystyle 4\pi R^2$ and in general yields the recurrence $\displaystyle C_n=\frac{2\pi}{n}C_{n-2}$ for the volume constants of $\displaystyle n$ –balls.

Lp Spaces

The Euclidean ball is only one member of a much larger family of “balls’’ defined by norms. For $\displaystyle p>0$ define the $\displaystyle L^p$ norm on $\displaystyle \mathbb{R}^n$ by $\displaystyle |x|_p=\left(\sum_{i=1}^n |x_i|^p\right)^{1/p}$ ,
and the $\displaystyle L^p$ ball of radius $\displaystyle R$ by $\displaystyle B_{n,p}(R)=\{x:|x|_p\le R\}$ . For $\displaystyle p\ge 1$ this is a centrally symmetric convex body (for $\displaystyle 0<p<1$ it is still a star-shaped set but not convex). The remarkable point is that its volume can be computed in closed form using the same Gamma–function technology that appears in the Euclidean case.

A convenient way to compute the volume of an $\displaystyle L^p$ ball is to evaluate the integral $\displaystyle I_{n,p}=\int_{\mathbb{R}^n}e^{-|x|_p^p}dx$ in two ways. Because the exponent splits as $\displaystyle \sum |x_i|^p$ , the integral factorizes into $\displaystyle n$ one–dimensional integrals, while an $\displaystyle L^p$ radial decomposition expresses it as an integral over radii involving the unknown volume constant; equating the two gives $\displaystyle V_{n,p}(1)=\frac{(2\Gamma(1+1/p))^n}{\Gamma(1+n/p)}$ and hence $\displaystyle V_{n,p}(R)=\frac{(2\Gamma(1+1/p))^n}{\Gamma(1+n/p)}R^n$ .

Euclidean case $\displaystyle p=2$ . Here $\displaystyle \Gamma(1+1/2)=\Gamma(3/2)=\frac{\sqrt{\pi}}{2}$ , so $\displaystyle 2\Gamma(1+1/2)=\sqrt{\pi}$ , and the formula becomes $\displaystyle V_{n,2}(1)=\frac{\pi^{n/2}}{\Gamma(1+n/2)},$ the usual Euclidean constant. Cross-polytope case $\displaystyle p=1$ . Here $\displaystyle \Gamma(2)=1$ , so $\displaystyle 2\Gamma(2)=2$ , and $\displaystyle V_{n,1}(1)=\frac{2^n} {\Gamma(1+n)}=\frac{2^n}{n!},$ the volume of the $\displaystyle n$ –dimensional octahedron (diamond). Hypercube limit $\displaystyle p\to\infty$ .
As $\displaystyle p\to\infty$ , we have $\displaystyle \Gamma(1+1/p)\to\Gamma(1)=1$ and $\displaystyle \Gamma(1+n/p)\to\Gamma(1)=1$ , so $\displaystyle V_{n,p}(1)\to 2^n,$ the volume of the cube $\displaystyle [-1,1]^n$ .

For fixed $\displaystyle p$ , the expression $\displaystyle V_{n,p}(1)=\frac{\left(2\Gamma(1+1/p)\right)^n}{\Gamma(1+n/p)}$ shows the same basic competition as in the Euclidean case: an exponential numerator versus a Gamma function in the denominator. Using Stirling-type asymptotics for $\displaystyle \Gamma(1+n/p)$ , one finds that for any fixed $\displaystyle 1\le p<\infty$ , the unit $\displaystyle L^p$ ball volume eventually decreases to $\displaystyle 0$ as $\displaystyle n\to\infty$ .

Categorification: Polydisk Mapping and Symplectic Volume

The appearance of the Gamma function in the analytic derivations is, in these dimensions, nothing more mysterious than the factorial coming from repeated integration. What is interesting is that this factorial also admits a geometric interpretation: it reflects a hidden $\displaystyle S_n$ symmetry in the coordinates, which becomes visible when one compares the complex ball with the polydisk.

The analytic computations above show that the complex ball $\displaystyle B=\{z\in\mathbb C^n:\sum |z_j|^2\le1\}$ has volume $\displaystyle \mathrm{Vol}(B)=\pi^n/n!$ . A useful geometric way to understand the factor $\displaystyle 1/n!$ is to compare the ball with the unit polydisk $\displaystyle D_n=\{(z_1,\dots,z_n)\in\mathbb C^n:|z_j|\le1\}$ , whose volume is simply $\displaystyle \mathrm{Vol}(D_n)=\pi^n$ because it is the Cartesian product of $\displaystyle n$ unit disks. The symmetric group $\displaystyle S_n$ of permutations acts on $\displaystyle D_n$ by permuting coordinates, so the quotient space $\displaystyle D_n/S_n$ has volume $\displaystyle \pi^n/n!$ . Thus the ball volume formula suggests that the ball should be geometrically
related to this quotient.

Write each coordinate in polar form $\displaystyle z_j=r_j e^{i\theta_j}$ and set $\displaystyle \rho_j=r_j^2$ . The volume form becomes $\displaystyle dV=\frac{1}{2^n}\,d\rho_1\cdots d\rho_n\,d\theta_1\cdots d\theta_n$ . After integrating over the angles, the geometry reduces entirely to the variables $\displaystyle (\rho_1,\dots,\rho_n)$ . For the polydisk the constraints $\displaystyle |z_j|\le1$ give $\displaystyle 0\le\rho_j\le1$ , so $\displaystyle (\rho_1,\dots,\rho_n)$ ranges over the cube $\displaystyle [0,1]^n$ . For the ball the constraint $\displaystyle \sum |z_j|^2\le1$ becomes $\displaystyle \rho_1+\cdots+\rho_n\le1$ , so the variables range over the simplex $\displaystyle \Delta_n=\{\rho_j\ge0,\ \rho_1+\cdots+\rho_n\le1\}$ .

The cube $\displaystyle [0,1]^n$ naturally decomposes into $\displaystyle n!$ regions according to the orderings of the coordinates $\displaystyle \rho_{\sigma(1)}\le\cdots\le\rho_{\sigma(n)}$ . Each such region has the same volume, and each is affinely equivalent to the simplex $\displaystyle \Delta_n$ . Thus the simplex occupies exactly $\displaystyle 1/n!$ of the cube, which explains why $\displaystyle \mathrm{Vol}(B)=\mathrm{Vol}(D_n)/n!$ .

The above computation shows that the polydisk is a cube in the rho coordinates and in the same coordinates the disk is a simplex, the factorial comes from the fact the simplex is a equivalent to another simplex that is obtained by restricting to a certain fixed ordering of the coordinates.

What Blass–Schanuel show is that this relationship is not merely a combinatorial coincidence but reflects a deeper geometric equivalence. Write each coordinate in polar form $\displaystyle z_j=r_j e^{i\theta_j}$ and first permute the coordinates so that $\displaystyle r_1\ge r_2\ge\cdots\ge r_n.$ On this ordered chamber define new coordinates $\displaystyle w_1,\dots,w_n$ by $\displaystyle |w_1|^2=r_1^2-r_2^2,\quad |w_2|^2=r_2^2-r_3^2,\ \dots,\|w_n|^2=r_n^2,$ with phases twisted triangularly, $\displaystyle \arg(w_k)=\theta_1+\cdots+\theta_k.$ The radii telescope so that $\displaystyle \sum_{k=1}^n |w_k|^2=r_1^2\le1,$ hence the image lies inside the ball $\displaystyle B_{2n}$ . The triangular mixing of the angles ensures that the standard symplectic form $\displaystyle \omega=\sum_{j=1}^n dx_j\wedge dy_j =\sum_{j=1}^n r_j\,dr_j\wedge d\theta_j$ is preserved under the map. Since symplectic maps preserve the top-degree form $\displaystyle \omega^n/n!$ , they preserve volume locally.

In this way the ball can be viewed as a symplectic rearrangement of the polydisk modulo the $\displaystyle S_n$ symmetry. The factor $\displaystyle 1/n!$ in the volume formula therefore reflects the $\displaystyle n!$ different orderings of the radii, which the map collapses into a single canonical configuration.

Coordinate free calculations

A clean geometric way to relate the volume of the ball to the surface area of the sphere uses the Euler (radial) vector field. Let $\displaystyle dV = dx_1 \wedge \dots \wedge dx_n$ be the standard volume form on $\displaystyle \mathbb{R}^n$ , and let the radial vector field be $\displaystyle X=\sum_{i=1}^n x_i \partial_{x_i}.$ Contracting the volume form with $\displaystyle X$ gives the $(n-1)$ –form $\displaystyle \omega=\iota_X dV =\sum_{j=1}^n (-1)^{j-1}x_j\,dx_1\wedge\cdots\wedge\widehat{dx_j}\wedge\cdots\wedge dx_n.$ One may write $\displaystyle \omega = r\, *dr,$ in terms of hodge operator. Geometrically, this projects the volume form onto directions tangent to the sphere. On the unit sphere $\displaystyle S^{n-1}$ the vector $\displaystyle X$ is the outward normal, so $\displaystyle \omega$ restricts to the natural surface area form. To relate this to the interior volume, compute $\displaystyle d\omega=d(\iota_X dV)=\mathcal{L}_X dV,$ using Cartan’s formula. The Lie derivative of the volume form is $\displaystyle \mathcal{L}_X dV=(\mathrm{div}\,X)\,dV=n\,dV,$ since $\displaystyle \mathrm{div}(X)=n$ . Applying Stokes’ theorem over the unit ball $\displaystyle B_n$ gives $\displaystyle \int_{S^{n-1}}\omega=\int_{B_n} d\omega =\int_{B_n} n\,dV=n\,\mathrm{Vol}(B_n).$ Thus $\displaystyle \mathrm{Vol}(B_n)=\frac{1}{n}\mathrm{Area}(S^{n-1}).$ Geometrically this expresses the ball as a continuum of cones with apex at the origin and base element $\displaystyle dA$ on the sphere; each cone has volume $\displaystyle \frac{1}{n}dA$ , and integrating these contributions over the boundary yields the total volume.