I wish i hadn't scrolled down as I glimpsed the solution in the first post which
spoiled my fun of working it out the non-calculus way. However I had in my mind
gone through how to do it using calculus. I haven't looked at the solution but
I believe it will run something as follows.

Let us start by imagining we cut the cored sphere into infinitely thin slices,
then we will have little discs. Then if we add up the area of all the discs we
will get the volume.

If we represent the inner sphere as having a radius of say R and the sphere as
having a radius T then. Now we can write that the height from the axis (say x)
is given by the formula  height^2 = T^2 - R^2

The discs have a radius whose square changes with proportion to the axis of
rotation (say Z), this is because the formula for a circle is x^2+y^2=r^2. Thus
the square of the outer radius of the disc is given by T^2-z^2. Notably at the
origin then we get the radius of the sphere out as z=0. This only works for
values of z between -height and height.

We thus get areas for the disc that are given by: Area = pi*((outside
radius)^2-(inner radius)^2) = pi*(T^2-z^2-R^2) = pi*(height^2-z^2)

Thus to find the volume we must intergrate from -height to height the above area
formula. With a little working we get a mess which equals 2*pi*(2/3*height^3)
or (4*pi*height^3)/3

This I presume is correct given the solution to the non-calculus version is
correct. However potentially this is the proof that the problem is solvable.
Hope noone was too bored by my long mathematical explantion and notably the
calculus involed is not too strenuous! Also thanks Mike as it was a nice little
brain teaser, even to solve the calculus way.

Regards Malcolm

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