Volume of a "Swiss Cube" Inspired by this video, I wanted to see if I could calculate the volume that remains if a cube is bored through with cylinders on all sides. Here is the final product from the video:

I decided to focus on just one unit cube, with cylinders bored through on all sides, and assumed that the ratio of the volume removed for just one unit cube would be the same as the ratio for the "Swiss Cube" with 25 cylinders bored through on each side.
The trouble I'm running in to is the overlap of the cylinders as you bore through from all directions. I do not know how to calculate the area of overlap between cylinders bored through in even two dimensions, let alone the total three. After one cylinder is bored through, the volume remaining would be:

$V_{\text{remaining}} = s^3 - \pi\left( \frac{1}{2}s \right)^2 s = s^3 \left(\frac{4 - \pi}{4} \right)$,

where s is the unite cube side length. 
I am not sure where to go from here though... How would one figure out how much overlap there is between the cylinder bored through in one dimension and the cylinder bored through in a perpendicular direction? And then, how would one do the same for the third cylinder bored through?
 A: Compute $$V = 2 \int_{x=1/\sqrt{2}}^1 \int_{y=\sqrt{1-x^2}}^x 1 - \sqrt{1-y^2} \, dy \, dx = 1 + \sqrt{2} - \frac{3\pi}{4}.$$  This is one "corner" of a single cube of side length $2$ that remains after drilling out cylinders along each of the three axes, thus a single cube would have remaining volume $$8V = 2(4+4\sqrt{2} - 3\pi).$$  For the given solid shown, the remaining volume is bounded below by $5^3 (8V) = 250(4+4\sqrt{2} - 3\pi)$, with the exact volume being slightly greater due to the fact that adjacent cylindrical bored holes cannot actually touch, thus leaving a slight additional amount of material.  Given that separation distance between adjacent holes, it is trivial to compute a more exact expression for the volume.

The integral is found by considering three mutually intersecting cylinders $$y^2 + z^2 = 1, \quad z^2 + x^2 = 1, \quad x^2 + y^2 = 1,$$ whose axes are the $x$-, $y$-, and $z$-axes, respectively, in the cube with vertices $\{(\pm 2, \pm 2, \pm 2)\}$ where each sign can be chosen independently of the others.  Then in one octant of the cube, say the $(+,+,+)$ octant, there is a twofold symmetry of the region bounded "outside" of all three cylinders and "inside" the cube; let us choose the plane of symmetry to be the plane $y = x$.  The resulting volume, when projected onto the $xy$-plane, comprises the region in the first quadrant bounded by the line $y = x$, the line $x = 1$, and the circle $x^2 + y^2 = 1$.  In this region, the $z$-coordinate of the volume is bounded above by $z = 1$ and below by the cylinder $y^2 + z^2 = 1$.  This is how we obtain the desired integral.  
