Surface area and volume of a sphere in finite Euclidean space
2025-04-06·10 min read·
A derivation of the volume of the Euclidean unit ball and the surface area of the unit sphere via spherical coordinates, induced metrics, and the Gaussian integral.
Motivation
The computation of the area and volume of a sphere is one of the classical
problems of geometry. In dimension two, the corresponding formulas are already
familiar:
Length(∂B2(0,r))=2πr,Area(B2(0,r))=πr2.
In dimension three, the standard formulas are
Area(Sr2)=4πr2,Vol(B3(0,r))=34πr3.
Historically, the three-dimensional case goes back to Archimedes. In "On the
Sphere and Cylinder", Archimedes established the relation between a sphere and
its circumscribed cylinder: the volume of the sphere is two-thirds the volume of
the circumscribed cylinder, and the surface area of the sphere is equal to the
lateral area of that cylinder.
Thus, in low dimensions, the formulas look elementary and geometric. However,
once we pass from R2 and R3 to arbitrary
Rn, the pattern is no longer obvious. One might expect powers of
π, but the exact dependence on n is more subtle. The answer involves
the Gamma function:
Vol(Bn)=Γ(n/2+1)πn/2,Area(Sn−1)=Γ(n/2)2πn/2.
This already suggests that the general n-dimensional problem is not merely
a problem of elementary geometry, it belongs more naturally to analysis. The
Gamma function itself arose from the problem of extending the factorial beyond
integer values. This problem appeared already in the work of Wallis on
quadrature, and was developed much further in the 18th century by Euler
and Stirling. Euler's work on the so-called Eulerian integrals made it possible
to treat expressions such as
Γ(s)=∫0∞ts−1e−tdt
as a continuous analogue of the factorial.
There is also a natural connection with the theory of multiple integrals. In the
nineteenth century, Dirichlet studied methods for determining multiple
integrals, and this analytic tradition is exactly the setting in which the
volume of an n-dimensional ball becomes a clean calculation. Instead of
trying to visualize a high-dimensional sphere, one evaluates an integral in two
different ways: first by Fubini's theorem, and then by changing to spherical
coordinates.
The key example is the Gaussian integral
∫Rne−∥x∥2dx.
On the one hand, Fubini's theorem reduces it to the one-dimensional Gaussian
integral:
∫Rne−∥x∥2dx=(∫−∞∞e−x2dx)n=πn/2.
On the other hand, spherical coordinates express the same integral as
∫Rne−∥x∥2dx=∫0∞∫Sn−1e−r2rn−1dσdr.
The radial part then becomes a Gamma integral. Comparing the two evaluations
reveals the surface area of the unit sphere, and integrating once more in the
radial variable gives the volume of the unit ball.
Therefore, the goal of this article is to derive the formulas
An−1=Area(Sn−1)=Γ(n/2)2πn/2
and
Vn=Vol(Bn)=Γ(n/2+1)πn/2
Conventions
Throughout this article, we denote
Bn=B(0,1)={x∈Rn∣∥x∥<1}andSn−1={x∈Rn∣∥x∥=1}.
The volume of the unit ball is defined by the ordinary coordinate integral
Vn=Vol(Bn)=∫Bndx1⋯dxn.
And the surface area of the unit sphere is defined by integrating the induced area element
An−1=Area(Sn−1)=∫Sn−1dσ.
The goal is to compute Vn and An−1 explicitly.
Necessary Tools
Two elementary analytic tools are used repeatedly.
Lemma (Gaussian Integral)
∫−∞∞e−x2dx=π.
Proof.
Let I=∫−∞∞e−x2dx. Since e−x2>0, I>0. Then
I2=(∫−∞∞e−x2dx)(∫−∞∞e−y2dy)=∫R2e−(x2+y2)dxdy.
Use polar coordinates x=rcosθ, y=rsinθ, whose Jacobian is r:
I2=∫02π∫0∞e−r2rdrdθ=2π∫0∞re−r2dr.
With t=r2 and dt=2rdr,
I2=2π⋅21∫0∞e−tdt=π⋅[−e−t]0∞=π.
Since I>0, I=π.
■
Lemma (Properties of the Gamma Function)
For s>0, the Gamma function defined by
Γ(s)=∫0∞ts−1e−tdt.
satisfies the following properties:
Γ(1)=1.
Γ(s+1)=sΓ(s) for all s>0.
Consequently, for n∈N, Γ(n)=(n−1)!
Γ(21)=π.
Proof.
1. Direct computation: Γ(1)=∫0∞e−tdt=[−e−t]0∞=1.
2. Integration by parts with u=ts and dv=e−tdt:
Γ(s+1)=∫0∞tse−tdt=[−tse−t]0∞+s∫0∞ts−1e−tdt.
The boundary term vanishes since tse−t→0 as t→0+ and t→∞. Therefore Γ(s+1)=sΓ(s).
3. From 1 and 2, induction gives Γ(n+1)=n!, i.e., Γ(n)=(n−1)!.
4. Substituting t=x2 in the definition,
Γ(21)=∫0∞t−1/2e−tdt=2∫0∞e−x2dx=∫−∞∞e−x2dx=π.
■
Computing Vn
We will start by changing variables into polar coordinates. Consider the equation x12+⋯+xn2=1. Since −1≤x1≤1, write x1=cosθ1. Then x22+⋯+xn2=sin2θ1. Next write x2=sinθ1cosθ2, so x32+⋯+xn2=sin2θ1sin2θ2. Iterating this substitution gives
The remaining determinant is evaluated through its Gram matrix. Set
A=(σ,∂θ1∂σ,…,∂θn−1∂σ).
Then ∣detA∣2=det(ATA).
Induced Metric and Area Element
The geometric origin of the induced metric is classical. For a surface
parametrized by X(u,v)⊆R3, Gauss studied the quadratic
expression
ds2=Edu2+2Fdudv+Gdv2,
where
E=⟨Xu,Xu⟩,F=⟨Xu,Xv⟩,G=⟨Xv,Xv⟩.
This is the first fundamental form of the surface. In modern language, it is
the metric induced on the surface by the Euclidean inner product of the ambient
space. Thus the basic idea of an induced metric goes back to Gauss's theory of
surfaces.
Riemann later abstracted this idea. Instead of studying only surfaces embedded
in Euclidean space, he considered n-dimensional manifolds equipped with an
intrinsic line element. This is the origin of Riemannian geometry: a manifold is
given a smoothly varying inner product on its tangent spaces, and geometric
quantities such as length, angle, area, and volume are derived from this metric.
Theorem (Riemannian Volume Form)
Let (Σm,g) be an oriented Riemannian manifold. In any positively oriented smooth coordinates (xi),
ωg=det(gij)dx1∧⋯∧dxm.
Proof.
Recall that in local coordinates (xi),
dxi1∧⋯∧dxik(∂xj1,…,∂xjk)=δj1⋯jki1⋯ik
for all multi-indices I=(i1,…,ik) and J=(j1,…,jk).
In positively oriented smooth coordinates, write
ωg=fdx1∧⋯∧dxm.
Since (∂xi∣p) is a basis for TpΣ, Gram-Schmidt gives a positively oriented local orthonormal frame (Ek) and a transition matrix C=[Cik] such that
∂xi=k=1∑mCikEk.
Let (εk) be the dual coframe of (Ek), so εk(El)=δlk. Since ωg is multilinear and alternating, and ωg(E1,…,Em)=1,
Since both (∂xi) and (Ek) are positively oriented, det(C)>0. Hence f=det(G).
■
Historically, this formula belongs to the Riemannian tradition, since the metric
determines infinitesimal volume. However, the modern expression as a top-degree
differential form belongs to the language of differential forms, developed
systematically by Cartan and later by de Rham.
The area element used above comes from the Riemannian volume form of the metric induced on the hypersurface.
Let Mm,Nn be smooth manifolds and let g be a Riemannian metric on N. For a smooth map F:M→N, the pullback F∗g on M defined by
(F∗g)p(X,Y):=gF(p)(dFp(X),dFp(Y)),X,Y∈TpM.
It's not hard to prove that F∗g is also a Riemannian metric on M.
Let Σ⊆M be a Riemannian submanifold with the metric induced from (Mm+1,g), and let F:U⊆Rm→Σ be a local parametrization. For the coordinate frame (∂xi) on U,
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