Mean Width of a Planar Compact Set with More than 1 component

This question relates to the mean width of a planar compact set that is itself a finite union of planar compact sets. Let $K$ be a compact set in $\Bbb R^n$ with nonempty interior. The mean width of $K$ is defined by $$b(K)=\frac{2}{\text{vol}_{n-1}(\partial B_2^n)}\int_{\Bbb S^{n-1}} h_K(u) du$$ where $\text{vol}_{n-1}(\partial B_2^n)$ is the surface area of the Euclidean unit ball and $h_K(u)=\sup_{x\in K}\langle x,u\rangle$ is the support function of $K$ at $u\in\Bbb S^{n-1}$.

If $K$ is a compact set in $\Bbb R^2$, it is known that the mean width $b(K)$ and perimeter $p(K)$ are related by $$b(K)=\frac{p(K)}{\pi}.$$ My question is: Does this formula hold even when $K$ consists of a finite number of connected components? More specifically, if $K=B_2^2\sqcup(B_2^2+(3,0))$ is the (disjoint) union of the unit balls centered at $(0,0)$ and $(3,0)$, does it also hold that $b(K)=\frac{p(K)}{\pi}=\frac{2(2\pi)}{\pi}=4$?

It seems to me that this statement is not true, even for compact sets with only one connected component. For example, consider the unit square $[0,1]\times[0,1]$ in $\Bbb R^2$. Choose points on the square $[-1,1]\times[-1,1]$ and consider the compact set $C$ with spikes emanating from all sides of the unit square with vertices on $[-1,1]^2$. Then as the number of spikes on each side of $C$ approaches $\infty$, the perimeter of $C$ tends to $\infty$, whereas the mean width of $C$ tends to the finite quantity $b(C)\leq b([-1,1]^2)<\infty$. Thus, for the equation $b(K)=p(K)/\pi$ to hold, $K$ needs to be convex and compact.