Minkowski–Steiner formula

Let n \geq 2, and let A \subsetneq \mathbb{R}^{n} be a compact set. Let \mu (A) denote the Lebesgue measure (volume) of A. Define the quantity \lambda (\partial A) by the Minkowski–Steiner formula :\lambda (\partial A) := \liminf_{\delta \to 0} \frac{\mu \left( A + \overline{B_{\delta}} \right) - \mu (A)}{\delta}, where :\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 \delta > 0, and :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 and \overline{B_{\delta}}, so that :A + \overline{B_{\delta}} = \left\{ x \in \mathbb{R}^{n} \mathrel|\ \mathopen| x - a \mathclose| \leq \delta \mbox{ for some } a \in A \right\}. ==Remarks==

Surface measure For "sufficiently regular" sets A, the quantity \lambda (\partial A) does indeed correspond with the (n - 1)-dimensional measure of the boundary \partial A of A. See Federer (1969) for a full treatment of this problem. Convex sets When the set A is a convex set, the lim-inf above is a true limit, and one can show that :\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 \lambda_{i} are some continuous functions of A (see quermassintegrals) and \omega_{n} denotes the measure (volume) of the unit ball in \mathbb{R}^{n}: :\omega_{n} = \frac{2 \pi^{n / 2}}{n \Gamma (n / 2)}, where \Gamma denotes the Gamma function. ==Example: volume and surface area of a ball==

Taking A = \overline{B_{R}} gives the following well-known formula for the surface area of the sphere of radius R, S_{R} := \partial B_{R}: :\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_{\delta \to 0} \frac{[ (R + \delta)^{n} - R^{n} ] \omega_{n}}{\delta} ::= n R^{n - 1} \omega_{n}, where \omega_{n} is as above. ==References==