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](http://en.wikipedia.org/wiki/Steinmetz_solid), 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](http://www.jstor.org/discover/10.2307/3615957?uid=2134&uid=2&uid=70&uid=4&sid=21103329679973) 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:



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> What is the optimal configuration of cylinders for each $n$, and what is the resulting volume?
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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 $K\_\infty$?



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> **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 $f(n)$?
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The intersection of $n$ cylinders in $3$-dimensional space
This is a heuristic, suggesting $f(n)=O( 1/n^2)$.


Consider the sphere of radius $r$. Each cylinder intersects this sphere in a great circle. The great circles divide the sphere into a number of regions. By induction we can see that this is at most $n(n-1)+2$. So if the sphere is relatively evenly divided into regions, each region must have diameter asymptotic to $1/n$. The intersection of the $n$ cylinders is a ball, plus a little piece jutting out in each of these regions. The little piece is approximately a cone of radius $O(1/n)$ and slope $O(1/n)$, so it has height $O(1/n^2)$. Since the area of the base is $O(1/n^2)$, it has volume $O(1/n^4)$. So the sum of all the pieces has volume $O(1/n^2)$.


The fact that the cone is on a spherical base and not a flat base cuts its thickness by $1/2$,so is responsible for only a constant factor.


So assuming that:


i. It's possible to choose the great circles to be relatively evenly distributed


ii. Weirdly-shaped regions don't cause any problems


iii. There's not some other error in these heuristic arguments:


the asymptotic is $f(n) = O(1/n^2)$. 


Edit: We can make the upper bound part of this more rigorous using the probabilistic method. Place all the cylinders randomly. Fix a point on the sphere. Consider a ray out from that point - how far does that ray travel in the intersection of the cylinders? It is at most $d^2/2+$-higher order terms, where $d$ is the distance to the nearest great circle corresponding to a cylinder. (In fact, if we express the distance as an angle $\theta$, it is exactly $\sec \theta - 1$.)


Instead of finding the nearest great circle corresponding to a cylinder, we can just look in a random direction and find the nearest great circle corresponding to a cylinder in that direction. So we look along some great circle passing through our point. Each great circle corresponding to a cylinder intersects this one at two random antipodal points. So we need to find the distribution of the distance to the nearest point among $n$ randomly chosen points on the unit circle. This converges rapidly to an exponential distribution of mean $1/n$. Because this distribution decays so rapidly, only the lowest-order term for the radius or the local contribution to volume matters when computing the expected value of the volume. The lowest-order term is proportional to the radius, which is proportional to $d^2$, whose expected value is $1/n^2$. Averaging over all the points, the volume is $O(1/n^2)$


Edit 2: We can make the lower bound rigorous using a simple estimate. The extra radius at a given point is $\sec \theta-1$, where $\theta$ is the distance to the nearest great circle corresponding to a cylinder. For any distance $\epsilon$, the points of distance no more than $\epsilon$ from a single great circle form a thin tube of area at most $ 4 \pi \epsilon$. So the total area of points that are within $\epsilon$ of any great circle form an area of at most $ 4 \pi n \epsilon$. The area of points that are not within $\epsilon$ of any great circle has area at least $4 \pi (1- n \epsilon)$. So if we set $\epsilon=2n/3$, say, then one-third the points of the sphere will extend outwards by at least $\sec \epsilon -1 \ geq \epsilon^2/2 = 2/9n^2$. This gives a lower bound of the form $O(1/n^2)$. 


