Is Archimedes's method for computing volumes equivalent to Cavalieri's?

Solution 1:

Archimedes used the optical properties of the parabola to prove that the area of the parabolic segment is $\frac{2}{3}$ of the circumscribed rectangle. In modern terms, the optical property is equivalent to $\frac{d}{dx}x^2 = 2x$, while the area depends on $\int x^2 = \frac{1}{3}x^3$, and the Archimedean argument for the parabolic segment can be viewed as the first attempt to provide an antiderivative for a polynomial.

With such a preliminary result, he proved that the volume of the sphere is $\frac{4\pi}{3}R^3$ by "slicing" it. Since the area of a section is proportional to the square radius of that section, to find the volume of a sphere is essentially the same problem as finding the area of a parabolic segment, since they both depend on $\int x^2$ - and not by chance, the coefficient $\frac{1}{3}$ appears in both of them.

So, in a quite vague (but not so much) sense, we can say that the Archimedean method was a precursor of the Cavalieri's method. And that Archimedes was a true genius.

Solution 2:

The method used by Archimedes in that letter includes Cavalieri's method (and I don't know whether Cavalieri's method is really Cavalieri's), but Archimedes's method also involves centers of gravity. I think Archimedes may have been the first ever to conceive of centers of gravity. Archimedes used his method to show that the center of gravity of the interior of a hemisphere (not a sphere; a hemisphere) is $5/8$ of the way from the north pole (if we were to take it to be the northern hemisphere) to the center of the sphere.