Maximum volume of a 3D shape with 3 identical circular profiles from x,y,z.

[bjshnog]

u = unit of distance.
Take a solid cube of dimensions (1u,1u,1u) with center at (0,0,0).
Cut it straight along x, y and z three times with a circle of diameter 1u parallel to the faces of the cube with the center of the circle at (x,0,0), (0,y,0), (0,0,z) respectively, removing the "shavings" on the outside.
What is the volume of the shape left behind and how do I calculate it?

 

[haruspex]

Take a horizontal slice through it at some height. The slice will be a square.
Let the straight line joining centres of opposite sides of the square subtend an angle 2θ at the centre of the object. Figure out the dimension of the square.
Changing θ by δθ turns the square into a lamina. Find the thickness of the lamina.
Now integrate wrt θ from 0 to π.
(I believe it should involve sin³, for which there's a trick.)

 

[haruspex]

Take a horizontal slice through it at some height. The slice will be a square.
Let the straight line joining centres of opposite sides of the square subtend an angle 2θ at the centre of the object. Figure out the dimension of the square.
Changing θ by δθ turns the square into a lamina. Find the thickness of the lamina.
Now integrate wrt θ from 0 to π.
(I believe it should involve sin³, for which there's a trick.)

Correction:
The horizontal section, being the intersection of only two of the cylinders, will not be a square lamina all the way from 0 to π. As theta increases from 0, at some value it will meet the third cylinder.
What is that angle?
It becomes a square lamina again at a corresponding angle below half way, and on to theta = π.
The same procedure can be applied in the two other dimensions, giving 6 'caps'.
What shape is left after removing these caps?
The final answer I believe to be 16 - 8√2.