David Fontaine wrote:

> Wow, how'd you get these? (No, don't answer, my poor little brain can't
> handle it :-) )

Actually it wasn't too bad, even if the answers look complex.  And now
that I've had a good nights sleep I'm ready to type it all up again.

First of all, volume:
The formula for a circle of radius r2, centered at the origin is f(x) =
sqrt(r2^2-x^2)

If you imagine breaking up this circle into rectangles (like a sum
approximation of its area), and then revolving it about the y-axis, each
of these rectangles sweeps out a cylindrical shell.  As the number of
these shells approaches infinity, the volume of each of them becomes
2*pi*x*f(x)*dx

This circle, however, when it is swept out to create the tours, is not
centered at the origin, it is moved r1 units along the x-axis, to
compensate for this, you will need to perform an integral from -r1 to r2.
This also means that you need to change the x in the formula for the
infinitesimal volume of a shell to reflect this by replacing the x with
(x+r1), notice, however, that f(x) remains unchanged.  Also notice that
the volume of the tours is actually twice what would normally be given by
the integral, because f(x) gives the length of a vertical line segment
connecting 0 and the circle (not the two sides of the circle).

The integral to solve for the volume of the tours then is given by:

         /r2
         |
   4*pi* |   (x+r1)*sqrt(r2^2-x^2) dx
         |
         /-r1

          /r2                         /r2
          |                           |
 = 4*pi*( |   x*sqrt(r2^2-x^2) dx  +  |   r1*sqrt(r2^2-x^2) dx )
          |                           |
          /-r1                        /-r1

The first integral is fairly easy to solve with u-substitution, and
evaluates to

-1/3*(r2^2-x^2)^(3/2)

The second integral is a bit more tricky, but fortunately there is a
formula for it.  It evaluates to:

r1/2*(x*sqrt(r2^2-x^2)+r2^2*asin(x/r2))

adding these together, multiplying by the 4*pi and subtracting the value
at x=-r1 from the value at x=r2 gives the answer.


Now for surface area:
The area of the infinitesimal side of the truncated cone formed by
sweeping f(x) around the y axis is 2*pi*f(x)*sqrt(1+f'(x)^2)

Set up the integral in pretty much the same way as was done for finding
the volume, giving:

         /r2
         |
   4*pi* |   (x+r1)*sqrt(1+x^2/(r2^2-x^2)) dx
         |
         /-r1

         /r2
         |
 = 4*pi* |   (x+r1)*sqrt(r2^2/(r2^2-x^2)) dx
         |
         /-r1

         /r2
         |
 = 4*pi* |   (x+r1)*r2/sqrt(r2^2-x^2)) dx
         |
         /-r1

          /r2                            /r2
          |                              |
 = 4*pi*( |   x*r2/sqrt(r2^2-x^2) dx  +  |   r1*r2/sqrt(r2^2-x^2) dx )
          |                              |
          /-r1                           /-r1

Once again, the first integral is easy to solve with u-substitution, and
evaluates to:

-r2*sqrt(r2^2-x^2)

The other side is again, trickier, but in this case has a simple solution:

-r1*r2*acos(x/r2)

If you wanted you could replace the -acos(x/r2) with asin(x/r2) and it
would still be correct, I did it this way so the signs would match up.
Evaluate them at x=r2 and x=-r1, subtract, multiply by 4*pi and again you
should have the answer.

I probably made that a bit more confusing than it really is, but hopefully
it will help you to solve similar problems in the future.

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