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...



How many unit cubes to cover an annulus?
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.


