ahhh. It starts coming back. 
I used x=r1=R as the area center, but it's not the same for a sliced
sircle, of course. Finding the new area center and using that expression
in the revolution integral would have left me with just as much work as
you had there, perhaps even more. (except that I'm cheating; ti89)

sig.


Kevin Wampler wrote:
> 
> 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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