Three holes are drilled through the centre of a solid sphere in a way that the holes are orthogonal to each...?

We start by calculating the volume of the sphere of radius R remaining after a hole of radius r has been drilled vertically down through the centre O at the origin of the cordinate system. The height h of the hole above the origin is given by Pythagoras

h² = R² - r²

At a distance z along the vertical axis, the area of the remaining annulus is

π.(R² - z² - r²) and so the volume of the sphere remaining is

V1 = π.∫ (R² - z² - r²).dz over the range from -h to +h ie

V1 = π.[(R² - r²).z - z³/3] from -h to +h = 2.π.[(R² - r²).h - h³/3]

which, using the first equation above, gives V1 = (4.π/3).h³

The amount of material removed by the drilling the first hole is therefore

V2 = (4.π/3).(R³ - h³)

When drilling the second hole, V2 overestimates the amount of material removed, because of the presence of the first hole at the centre of the sphere. However, the overlap of the orthogonal cylindrical holes at the centre forms a bicylinder, a Steinmetz solid, whose volume can be calculated to provide a correction, so that

V3 = (4.π/3).(R³ - h³) - (16/3).r³

And similarly, when the third hole is drilled, the overlap region at the centre forms a tricylinder, another Steinmetz solid whose volume has been calculated (see reference) to correct for the material missing at the centre

V4 = (4.π/3).(R³ - h³) - (16 - 8√2).r³

Hence the total amount of material removed is given by V2 + V3 + V4, and this is required to be one-half of the original volume of the sphere, assuming that it was of uniform density, giving

(1/2).(4.π/3).R³ = 3.(4.π/3).(R³ - h³) - (16/3 +16 - 8√2).r³

By substituting for h, this provides a complex equation relating R and r. It is not, as far as I can tell, analytically soluble but numerical analysis shows that it has a single solution given by r = 0.382.R. Note that r is the radius of the hole, so the diameter of the drill required would be twice this.

This solution agrees reasonably well with my initial estimate but, like Yahoo, I give no guarantees on this one.

Note added: Thanks; I.m not sure what you mean by the 'extra h terms', but certainly the difficult part is estimating the material common to each of the holes drilled. I had two solutions - the one given and an alternative which subtracted twice the tricylinder volume given in V4 only. The problem was that I was uncertain of the validity of the expression for that, which is smaller than the bicylinder volume in V3 whereas intuitively I thought it should be larger. I had calculated this in a previous answer in this forum, so I was confident in that expression.

In fact the difference produced in the result is small; the ratio r/R given above is 0.382, for the alternative it's 0.379. I'm tending towards the latter after thinking about it, but validating the V4 term might take a significant amount of time.