Sphere in a regular grid, area formula Usually, the formula for area of the sphere is $4 \pi r^2$.
My sphere is inside a regular cubic grid, with 1×1×1 unit size cells.
I’d like to know the count of the grid cells that happen to intersect with the surface of my sphere. I only need the approximate answer, for large values of $r$.
Intuitively, this going to be $4 \pi r^2 C$, where $C$ is some constant.
What’s the value of $C$?
 A: The following heuristic makes the value $C={3\over2}$ plausible:
Consider a sphere $S_R$ of radius $R\gg1$. The lattice planes $x=j\in{\mathbb Z}$, $y=k$, or $z=l$ intersect $S_R$ in circles which together partition $S_R$ into a number of cells of various shapes. We have to estimate the number of these cells.
Looking from $(0,0,+\infty)$ down to $S_R$ we see about $\pi R^2$ unit squares. These are projections of quadrangles on $S_R$ produced by the planes $x=j$ and $y=k$, and there are another $\pi R^2$ such squares to be seen on the lower hemisphere.
The latitude circle $z=l$ has radius $r_l=\sqrt{R^2-l^2}$. It intersects about $2r_l$ planes $x=j$ and $2r_l$ planes $y=k$, generating about  $8r_l$ points of intersection. Each arc between two successive such points bisects an "$xy$-quadrangle", hence augments the number of cells by $1$. The total contribution obtained in this way is about
$$\sum_{l=-R}^R 8 r_l\doteq 4\sum_{l=-R}^R 2\sqrt{R^2-l^2}\doteq4\pi R^2\ ,$$
since the last sum is a Riemann sum for the area of a disc of radius $R$. 
It follows that the total number of cells can be estimated as
$$2\pi R^2+4\pi R^2={3\over2}\cdot4\pi R^2\ .$$
