I think I can give you an engineer solution using just an integration, but it is not simple. To better visualize our quest for the holy cube volume we will proceed as follows.1)We take four cubes imagine them glued together so they will form another cube which edge will measured two units with.
2)Now we bore all the holes in the four cubes. What it is left supposing the loose pieces fall apart is precisely a body which volume equal to the original holy cube.
3)Analyzing the mentioned body you will found it looks like 3D cross. The hub is a cube and the six arms are like six church steeples formed with four equal pieces of cylindrical roof tangent at the apex. So the volume requested will be the central cube plus the six steeples V2.
Obviously cutting the steeple orthogonally to its symmetry axis you get a square of variable section. Calling r the radius of the cylinders and z the distance measured from the apex. The side L of the square amount to L=2< r- SQRT (r^2 - z^2)>.
The area of the square section, which A(z), values:
A(z) = 4(r- SQRT(r^2-z^2))^2.
Then V2 =INTEG(A (z))
we must integrate z between 0 and SQRT(2)*r/2
That integration renders V2 = ((11*SQRT(2)/3)-2-pi)*r^3. since there are 6 steeples we must multiply V2 later by 6.
4)As for the volume of the central hub cube is obviously.
V1=(2r*(1-SQRT (2)/2))^3 or expanded (20 -14*SQRT(2))*r^3
5)The total volume will be V= V1+6*V2. which after a patient algebra manipulation will give us:
((8+8*SQRT(2))-6pi)*r^3
Which amounts the same result given by JJ since r is the half of x, width of the cube
The numerical value of this expression is 0.058019072 of the cube volume. Just as a verification I ran a Monte Carlo simulation, computing the fraction of randomly points which coordinates situated them simultaneously out of the boundaries of the three cylinders, with respect of the total points randomly considered. Results for several batches over 25.000 points validates our solution.
The only way I believe it can be solve this problem without calculus is either by Monte Carlo simulation or using Guldin theorem provide that in same time we can take from the books the formula for the center of gravity of a portion of circle between the apex and a plane levelled with the top of the cental hub cube... so in my opinion the best answer to solve this case is triple integration.
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