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?

Edit: the squares (not cubes because we are in two dimensions) need to have sides parallel to the $x$ and $y$ axes. Maybe this changes things a little...

  • $\begingroup$ It will be like many thin curves taking a circuit tour and visiting the four cardinal points: it will take 8R squares, perhaps with a few more squares for cases where the path interior contains a lattice point. You might read up on the Gauss circle problem for more info. Gerhard "Sort Of Like Manhattan Distance" Paseman, 2017.03.24. $\endgroup$ – Gerhard Paseman Mar 24 '17 at 19:36
  • $\begingroup$ Do the squares (I presume you mean that, rather than cubes) need a fixed orientation? If not, the number will be approximately $2 \pi R/\sqrt{2}$. $\endgroup$ – Robert Israel Mar 24 '17 at 19:40
  • $\begingroup$ About $\pi\sqrt{2E}$: look how they cover a big circle. $\endgroup$ – Fedor Petrov Mar 24 '17 at 19:40
  • $\begingroup$ @RobertIsrael I want the squares (yes I mean squares not cubes) to have sides parallel to the coordinate axes. $\endgroup$ – Michela Mar 24 '17 at 20:01
  • $\begingroup$ If the squares have integral coordinates for vertices, then think Manhattan; the reason it won't always be exactly 8R is because sometimes the annulus contains a neighborhood of a lattice point. If the Gauss circle problem applies, current technology will say the true answer is off from 8R by something like R^{2/3}. Gerhard "Think Of Squares, Not Circles" Paseman, 2017.03.24. $\endgroup$ – Gerhard Paseman Mar 24 '17 at 20:30

Gerhard already mentioned what I'm going to say in a comment, but let me make it explicit anyway: To cover a line of slope $0\le m\le 1$ and length $L\gg 1$, you need about $L/\sqrt{1+m^2}$ unit squares with sides parallel to the axes.

So the part of the circle of radius $R$ with angle between $\alpha$ and $\alpha+d\alpha$ requires $R\, d\alpha/\sqrt{1+\tan^2\alpha}$ unit squares, for a total of $$ \simeq 8R \int_0^{\pi/4} \frac{d\alpha}{\sqrt{1+\tan^2\alpha}} = 8R \int_0^{\pi/4}\cos\alpha\, d\alpha = 4\sqrt{2} R $$ squares to cover the whole circle or (more or less equivalently) the very thin annulus.


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