Quermassintegral

Remarks

By the Steiner formula, $\operatorname{vol}(K + tB) = \sum_{j=0}^n \binom{n}{j} W_j(K)\, t^j$ for $t \ge 0$. Special values: $W_0(K) = \operatorname{vol}(K)$, $n W_1(K)$ is the surface area of $K$, and $W_n(K) = \kappa_n$, the volume of $B$. Also $W_{n-1}(K) = \frac{\kappa_n}{2} w(K)$, where $w(K)$ is the mean width of $K$. For $1 \le k \le n$, $W_{n-k}(K)$ equals $\kappa_n / \kappa_k$ times the average, over all $k$-dimensional linear subspaces $E$, of the $k$-dimensional volume of the orthogonal projection of $K$ onto $E$; this is Kubota's formula.

Example: $K = rB$, a ball of radius $r \ge 0$. By multilinearity, $W_j(rB) = r^{n-j} V(B, \dots, B) = \kappa_n r^{n-j}$. For $n = 2$ and $r = 1$ this gives $W_1 = \pi$, half the perimeter $2\pi$, as expected.

Non-example: the set $K = \set{0, 1} \subseteq \R$ is not convex, so its mixed volumes are not defined. Indeed the length of $K + tB$ is $4t$ for $t \le \frac12$ and $1 + 2t$ for $t \ge \frac12$, not one polynomial in $t$.