|  |     
    Here's a shot ( the idea is good, the manipulation should probably
    be checked ) :
    
    
Problem:
	What is the volume of the solid resulting from the
	intersection of a cube and a cylinder such that the
	axis of the cylinder coincides with the long diagonal
	of the cube.
Assumptions:
	Unit cube
	Cylinder radius (r) from 0 to 1
	Cube vertex labling:
	      F --------------------  B
               /|                  /|
	      / |                 / |
             /  |                /  |
            /   |               /   |
        A  /-------------------- O  |
          |     |              |    |
          |     |              |    |
          |     |              |    |
          |   D ---------------|----
          |    /               |   /
          |   /                |  /
          |  /                 | /
          | /                  |/
          ---------------------
	The clinder's axis will coincide with OD.
Method:
	Consisder cylinder of height Len(OD).  Find volume of clinder not
	in intersection and subtract from total volume.
	Create coordinate system such that cylinder axis coincides with
	Z-axis with O at origin and point F when projected along Z-axis 
	falls on positive X-axis.  Equation of this plane is 
	(1)	Z=Sqr(2)*X.
	Vector to A is
	(2)	( 1/Sqr(6), 1/Sqr(2), 1/Sqr(3) )
	Vector to B is
	(3)	( 1/Sqr(6), -1/Sqr(2), 1/Sqr(3) )
	The projection of the (half) circle X^2 + Y^2 = r^2 onto plane OAFB is
	given by
	(4)	X = Sqr( r^2-y^2)
		Z = Sqr(2) * X
	Proposition:  By symmetry arguments, the volume of the part of the
	cylinder not in the intersection is equal to six times the volume
	of the object bounded by:
		.  Plane Z=0
		.  Plane OAFB
		.  Cylinder X^2 + Y^2 = r^2
		.  Plane Y = Sqr(3) * X
		.  Plane Y = - Sqr(3) * X
	Inspection of the resulting volume integral reveals that this is
	twice the volume of the same object with either of the last two planes
	replaced by the plane Y = 0.  Thus, the boundary conditions on the
	volume integral derived from (1), (2), (3), and (4) are:
		Y = [ -r / Sqr(2), 0 ]
		X = [ -Y / Sqr(3), Sqr( r^2 - Y^2 ) ]
		Z = [ 0, Sqr(2) * X ]
	so
			/ / /
		       | | |
		V = 12 | | | dz dx dy		(with the above boundaries)
		       | | |
		      / / /
	this gives   V = 46 * Sqr( 6 ) * r^3 / 27
	so Vint = ( Sqr(3) * pi * r^2 ) - ( 46 * Sqr( 6 ) * r^3 / 27 )
					Monty
    
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