In mathematics , the Minkowski–Steiner formula is a formula relating the surface area and volume of compact subsets of Euclidean space . More precisely, it defines the surface area as the "derivative" of enclosed volume in an appropriate sense.
The Minkowski–Steiner formula is used, together with the Brunn–Minkowski theorem , to prove the isoperimetric inequality . It is named after Hermann Minkowski and Jakob Steiner .
Let n ≥ 2 {\displaystyle n\geq 2} , and let A ⊊ R n {\displaystyle A\subsetneq \mathbb {R} ^{n}} be a compact set. Let μ ( A ) {\displaystyle \mu (A)} denote the Lebesgue measure (volume) of A {\displaystyle A} . Define the quantity λ ( ∂ A ) {\displaystyle \lambda (\partial A)} by the Minkowski–Steiner formula
λ ( ∂ A ) := lim inf δ → 0 μ ( A + B δ ¯ ) − μ ( A ) δ , {\displaystyle \lambda (\partial A):=\liminf _{\delta \to 0}{\frac {\mu \left(A+{\overline {B_{\delta }}}\right)-\mu (A)}{\delta }},} where
B δ ¯ := { x = ( x 1 , … , x n ) ∈ R n | | x | := x 1 2 + ⋯ + x n 2 ≤ δ } {\displaystyle {\overline {B_{\delta }}}:=\left\{x=(x_{1},\dots ,x_{n})\in \mathbb {R} ^{n}\left||x|:={\sqrt {x_{1}^{2}+\dots +x_{n}^{2}}}\leq \delta \right.\right\}} denotes the closed ball of radius δ > 0 {\displaystyle \delta >0} , and
A + B δ ¯ := { a + b ∈ R n | a ∈ A , b ∈ B δ ¯ } {\displaystyle A+{\overline {B_{\delta }}}:=\left\{a+b\in \mathbb {R} ^{n}\left|a\in A,b\in {\overline {B_{\delta }}}\right.\right\}} is the Minkowski sum of A {\displaystyle A} and B δ ¯ {\displaystyle {\overline {B_{\delta }}}} , so that
A + B δ ¯ = { x ∈ R n | | x − a | ≤ δ for some a ∈ A } . {\displaystyle A+{\overline {B_{\delta }}}=\left\{x\in \mathbb {R} ^{n}{\mathrel {|}}\ {\mathopen {|}}x-a{\mathclose {|}}\leq \delta {\mbox{ for some }}a\in A\right\}.} For "sufficiently regular" sets A {\displaystyle A} , the quantity λ ( ∂ A ) {\displaystyle \lambda (\partial A)} does indeed correspond with the ( n − 1 ) {\displaystyle (n-1)} -dimensional measure of the boundary ∂ A {\displaystyle \partial A} of A {\displaystyle A} . See Federer (1969) for a full treatment of this problem.
When the set A {\displaystyle A} is a convex set , the lim-inf above is a true limit , and one can show that
μ ( A + B δ ¯ ) = μ ( A ) + λ ( ∂ A ) δ + ∑ i = 2 n − 1 λ i ( A ) δ i + ω n δ n , {\displaystyle \mu \left(A+{\overline {B_{\delta }}}\right)=\mu (A)+\lambda (\partial A)\delta +\sum _{i=2}^{n-1}\lambda _{i}(A)\delta ^{i}+\omega _{n}\delta ^{n},} where the λ i {\displaystyle \lambda _{i}} are some continuous functions of A {\displaystyle A} (see quermassintegrals ) and ω n {\displaystyle \omega _{n}} denotes the measure (volume) of the unit ball in R n {\displaystyle \mathbb {R} ^{n}} :
ω n = 2 π n / 2 n Γ ( n / 2 ) , {\displaystyle \omega _{n}={\frac {2\pi ^{n/2}}{n\Gamma (n/2)}},} where Γ {\displaystyle \Gamma } denotes the Gamma function .
Example: volume and surface area of a ball [ edit ] Taking A = B R ¯ {\displaystyle A={\overline {B_{R}}}} gives the following well-known formula for the surface area of the sphere of radius R {\displaystyle R} , S R := ∂ B R {\displaystyle S_{R}:=\partial B_{R}} :
λ ( S R ) = lim δ → 0 μ ( B R ¯ + B δ ¯ ) − μ ( B R ¯ ) δ {\displaystyle \lambda (S_{R})=\lim _{\delta \to 0}{\frac {\mu \left({\overline {B_{R}}}+{\overline {B_{\delta }}}\right)-\mu \left({\overline {B_{R}}}\right)}{\delta }}} = lim δ → 0 [ ( R + δ ) n − R n ] ω n δ {\displaystyle =\lim _{\delta \to 0}{\frac {[(R+\delta )^{n}-R^{n}]\omega _{n}}{\delta }}} = n R n − 1 ω n , {\displaystyle =nR^{n-1}\omega _{n},} where ω n {\displaystyle \omega _{n}} is as above.