Let $E>1$ and consider an annulus in $\mathbb{R}^2$ with outer radius $R=\sqrt{E}$ and inner radius $R=\sqrt{E-1}$.
How many unit cubes do I need to cover the annulus?
The area of a $2$-dimensional annulus does not depend on the outer and inner radius, so one could think that the number of needed cubes depends only on the dimension. However, for growing $E$, the width of the annulus becomes thinner and the length of the inner and outer circumference becomes longer. So it looks to me that the number of cubes grows with $\sqrt{E}$, but I don't really understand how. Can anybody help me in this?
First note that $(\sqrt{E}-\sqrt{E-1})^2 = E^2+(E-1)^2-2\sqrt{E}\sqrt{E-1} ≤ E^2+(E-1)^2 < 1$. This means the thickness of the annulus is always less than the width of the unit cube. This means the most efficient way to cover the annulus will be with a "chain". In other words, a collection of cubes that only intersects with the cube before and after it.
As $E$ grows, the thickness tends to $0$ while the curvature also tends to $0$, so the most efficient cover becomes a collection of diagonal cubes as drawn below.
The perimeter will have a length of $2\pi r = 2\pi\sqrt{E}$. The length of $n$ cubes will be $n\sqrt{2}$ (because they have been placed diagonally). Hence the number of cubes needed is about $n=2\pi\sqrt{E}/\sqrt{2} = \sqrt{2}\pi\sqrt{E} \approx 4.44\sqrt{E}$.
Note that this is an approximation, and it holds for large $E$.