cavalieri and solids of revolution?

Question

i need to apply cavalieri's results to obtain that the volume for the solid is 16(pi)/15. the solid comes from an area bounded by y=1 and y=x^2 revolved around the axis y=1.

this is not an integral problem. i need to use cavalieri's methods. i first found the area:

A=2*sum(from 0 to 1) of (x^0) - 2*sum(from 0 to 1) of (x^2) = 4/3.

so then, cavalieri states that a curve rotated about the x-axis has a volume that can be calculated with:

V = (pi)*sum(from 0 to 1) of (x^(2n)) where n is the power of the curve y=x^n. this is the area under the curve rotated about the x-axis.

how do i go about solving for volume? i need 16(pi)/15, but consistently get 4(pi)/5.... any suggestions? thanks.

Answer 1

(I'll admit I'm not familiar with Cavalieri's methods, but a quick web search makes them look like a proto-integral-calculus. Assuming that's what they are...)

It looks like your main problem here is that you're rotating around y=1, but your calculations are for rotating around y=0 (the x-axis). The volume of revolution of this surface from 0 to 1 around y=0 is indeed 4pi/5, but around y=1 gives 8pi/15. (Since the surface extends to -1, that gives the correct 16pi/15.)

So I suggest you find a way to recast the formula you have to make it account for the different axis of rotation. Alternately, rotate y=1-x^2 around the x-axis.

Notice, in terms of integral calculus:
integrate [0 to 1] pi 1-(x^2)^2 dx = 4pi/5
integrate [0 to 1] pi (1-x^2) dx = 8pi/15 (which is half of the right answer)

The most obvious way is to turn your x^(2n) into a (1-x^n)^2, arguing that the area of a circle is pi r^2 and your r is supposed to be (1-x^n), not x^n. I'm pretty sure this is the right general form but, again, I'm not familiar with Cavalieri's work.

Answer 2

You say this is not an integral problem; what are those sums you refer to?

Check this by using a standard method: First, do an axis transformation so that y=1 becomes the x-axis. Thus f(x) becomes x^2-1. The disc method for finding the volume is

π∫f(x)^2dx


= π∫(x^2 - 1)^2dx = π∫(x^4 - 2x^2 +1)dx (from 0 to -1)

=πª[(x^5)/5 - 2(x^3)/3 + x] (from 0 to -1)

= π[-1/5 + 2/3 - 1] = π(-3 +10 - 15)/15 = -π*8/15 [SEE EDIT BELOW]


EDITED 10/11
EDIT 10/12

NOTE: My error is in the limits of integration which should have been -1 to +1. This gives the result π•16/15