Intrinsic volume

Remarks

Equivalently, with $B$ the closed unit ball, the Steiner formula reads $$\operatorname{vol}(K + tB) = \sum_{j=0}^n \kappa_{n-j} V_j(K)\, t^{n-j} \quad (t \ge 0).$$ Special values: $V_n(K) = \operatorname{vol}(K)$, $V_{n-1}(K)$ is half the surface area, and $V_0(K) = 1$. If $K$ lies in a $k$-dimensional affine subspace, then $V_j(K)$ is the same whether computed in $\R^n$ or in that subspace, and $V_j(K) = 0$ for $j > k$.

Example: the unit square $K = [0,1]^2$ in $\R^2$. Its parallel body has area $1 + 4t + \pi t^2$, and the formula above gives $V_2(K) + 2 V_1(K)\, t + \pi V_0(K)\, t^2$. So $V_2(K) = 1$, $V_1(K) = 2$ (half the perimeter) and $V_0(K) = 1$.

Non-example: $K = \set{0, 1} \subseteq \R$ is compact but not convex. The length of $K + tB$ is $4t$ for $t \le \frac12$ and $1 + 2t$ for $t \ge \frac12$, which is not one polynomial, so there are no intrinsic volumes of $K$ in this sense.