Sphere on a grid So, this is a little tricky kind of a question and I'm not totally sure if it's a mathematic question or a more programming one, but I nevertheless hope to find answers. 
I want to find out the error of a volume and surface calculation of a 3-D sampled sphere. That means, the sphere is lying in a 3-D grid and is downsampled to/by that grid. That means each voxel is either one or zero, depending on how much of the sphere is touching that voxel (to make it easy, let's assume 50%).
The question now is, how much the Volume of the sphere differs from that of the cubes.
Another way of asking would be: How many equal sized cubes that share one surface can be placed in a sphere?
 A: Suppose you have a sphere with a radius $R$, and you're filling it with cubes of size $a$, where $a<R$ (and probably, $a \ll R$). The volume of the sphere is $\frac{4}{3} \pi R^3$. The volume of each cube is $a^3$. The question is what to do with the points on the surface of the sphere—some of which are counted as "inside" the sphere, and others of which are "outside" the sphere. 
Method #1
We could estimate the error by thinking of the largest possible volume and the smallest possible volume. Consider two more spheres: a smaller sphere with radius $R-a$ and a larger sphere with radius $R+a$. The smaller sphere would contain only cubes that were already counted in the original sphere of radius $R$ (but would not count all of them). The larger sphere would contain all the cubes in the original sphere, but would also contain all the cubes we were uncertain about. If we calculate the differences in volume ($\Delta V$) between the largest sphere and the smallest sphere, and divide that volume up into cubes, then that's a rough upper estimate to the number of cubes that would be on the boundary:
$$
\Delta V \equiv V_{large} - V_{small} \\
\Delta V = \frac{4}{3} \pi (R+a)^3 - \frac{4}{3} \pi (R-a)^3 \\
\Delta V = \frac{4}{3} \pi \left[ (R+a)^3 - (R-a)^3 \right] \\
\Delta V = \frac{4}{3} \pi \left[ 2 a^3 + 6 a R^2 \right] \\
\Delta V = \frac{8}{3} \pi \left[ a^3 + 3 a R^2 \right] \\
$$
$$
N_{cubes} \approx \frac{\Delta V}{a^3} \\
N_{cubes} \approx \frac{8}{3} \pi \left[ 1 + 3 \left(\frac{R}{a}\right)^2 \right] \\
$$
The actual number of cubes that will be inside or outside this boundary should be quite a bit less. (If I had to guess, I would guess the number of cubes "mostly outside" the original sphere would be closer to half this amount, and the other half would be "mostly inside".) Note that for spheres much larger than the cubes ($R \gg a$), this reduces to the following: 
$$
N_{cubes} \approx 8 \pi \left(\frac{R}{a}\right)^2 \\
$$
Method #2
We'll again divide a small $\Delta V$ by the number of cubes, but this time we'll calculate $\Delta V$ as the surface area of the original sphere times the height of each cube. Visually, for a sphere much larger than the cubes ($R \gg a$), this is sort of like placing a cube on the surface of the sphere and counting them up:
$$
\Delta V \equiv A_{sph} a \\
\Delta V = 4 \pi R^2 a 
$$
As before, we divide this fractional volume by the volume of a cube to get the number of cubes that would be in this "border region": 
$$
N_{cubes} \approx \frac{\Delta V}{a^3} \\
N_{cubes} \approx 4 \pi \left(\frac{R}{a}\right)^2 \\
$$
Conclusions
Comparing the methods, we see that they pretty much agree in the case of large spheres in terms of both the number of cubes in the edge region and in terms of the volume of this edge region. 
Honestly, the best way to figure this out is probably to write a quick little program to calculate this for you. It will probably give you something very similar to my equations, but probably off by some coefficient that depends on your cutoff value (the 50% you mentioned). 
I hope this is a correct interpretation of your question! 
