A standard question in vector calculus is to calculate the volume of the shape carved out by the intersection of $2$ or $3$ perpendicular cylinders of radius $1$ in three dimensional space. Such shapes are known as Steinmetz Solids, and for two and three cylinders we have $$V_2=4\cdot \frac{4}{3},$$ $$ V_3=6(2-\sqrt{2})\cdot \frac{4}{3}.$$

Here, and throughout, when we say that the cylinders intersect, we mean that the centers of the cylinders (which are lines) all intersect at a single point. Moreton Moore wrote an article where he calculates the volume of the intersection of $4$ and $6$ cylinders which arise from identifying faces of the octahedron and dodecahedron respectively. In this case, we have $$V_4=\frac{4}{3}\cdot9(2\sqrt{2}-\sqrt{6}),$$ $$V_6=\frac{4}{3} 4\left(3+2\sqrt{3}-4\sqrt{2}\right).$$

Define $$K_n=\inf\{V: V \text{ is the volume of the intersection of }n\text{ cylinders of radius }1\}.$$ Since we are assuming that the centers of the cylinders share a common point, the sphere will be contained in any configuration of cylinders. The maximum volume is infinite, and the limiting volume will be the sphere, that is $$K_\infty=\pi\frac{4}{3}.$$ I believe that for reasons of symmetry, for $n=2,3,4,6$ we have $K_n=V_n$. Naturally, this leads to the question:

What is the optimal configuration of cylinders for each $n$, and what is the resulting volume?

While this may be a difficult problem, I am interested in the more approachable problem of evaluating the rate of convergence to the sphere. Can we identify how fast $K_n$ approaches $K_\infty$?

Question:Let $$f(n)=\frac{K_n}{4/3}-\pi.$$ Then $f(n)$ is monotonically decreasing to $0$, and $f(n)$ measures how fast $K_n$ approaches the volume of the sphere. At what rate is $f(n)$ decreasing to $0$? Can we find the order of magnitude of $f(n)$?