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.