A layer through the center of a solid is the region enclosed by a circle
with a radius of r cm. and each plane section perpendicular to a fixed
diameter of the layer is a regular hexagon having a chord of the circle as
a diagonal of the hexagon.  Find the volume of the solid.

I haven't tried it myself yet but just in case I need help I posted it here.

This ventures into the world of slicing for integral calculus.

I don't get it.

[quote author=Yoni link=topic=14577.msg149091#msg149091 date=1143254969]
I don't get it.
[/quote]

Heh, I was thinking the same thing. When I took calc in high school seemed like the hardest part was just trying to figure out what the problem was saying and visualizing the shape.

Any picture of what this would look like? Kind of sounds like a circle inscribed in a hexagon but I'm not too sure about that.

I was going to complain too, but decided to avoid embarassement incase the question actually was properly worded  8).

I think this is what the proper visualization is:

Draw a hexagon, then draw a line that cuts it in half.  Rotate the hexagon
around the line.  Now you have your circular "plane sections".  Of course, finding the answer is now trivial, if this is the proper interpretation.

Modification:  Ok, I won't pull a Fermat :P.  Draw one half of a hexagon, and let it connect to a "x axis."  It will be a continuous non-differentiable piecewise function,
g(x).  Find pi* Integral (g(x)^2  dx ) .

Modification #2:    Alternatively, draw a half circle (f(x) = Sqrt[1-x^2]).  Imagine that this half-circle is the projection of some object dropped through a piece of paper.  Let cross-sections be partial hexagons.  Look up a formula for the area of a hexagon and adjust it to your partial hexagon accordingly.    Relate
the formula for area of a given hexagon to f(x).  Then find the infinitesimal sum
of these areas across the diameter of the half circle, and multiply your result by two.  This sort of logic was used in my solution to the old 4D sphere
problem here. 
This solution (mod#2) does not quite fit with the description of the problem, so I'm sure it is not what your teacher is looking for, but might help you get on the right track.