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Hoping a philomath can help - I'm searching how Archimedes calculates the centroid of a hemisphere without calculus as behind the below sentence in the wikipedia entry on "the Method of Mechanical Theorems":

"This type of method* [leverage argument - please see wikipedia] "can be used to find the area of an arbitrary section of a parabola, and similar arguments can be used to find the integral of any power of x, although higher powers become complicated without algebra. Archimedes only went as far as the integral of x3, which he used to find the center of mass of a hemisphere, and in other work, the center of mass of a parabola." ... "Other propositions in the palimpsest "A series of propositions of geometry are proved in the palimpsest by similar arguments. One theorem is that the location of a center of mass of a hemisphere is located 5/8 of the way from the pole to the center of the sphere. This problem is notable, because it is evaluating a cubic integral."

https://en.wikipedia.org/wiki/The_Method_of_Mechanical_Theorems

I found reference to Proposition 12 in Archimedes, the Center of Gravity, and the First Law of Mechanics 2nd edition The Law of the Lever Andre K.T. Assis which is "If any number of straight lines drawn from the origin to meet the spiral make equal angles with one another, the lines will be in arithmetical progression."

I am intrigued how Archimedes could use the integral of a cube by the leverage argument to determine the centre of mass of a hemisphere and a parabola? I know the proofs by calculus which Archimedes did not have full access to - although he seems to have used some of its concepts. Archimedes intuitive and elegant method of leverage has much didactic benefit - but this part escapes me.

Maths stackexchange refers to Pappus but also has footnotes referring to Archimedes but again no explanation:

https://math.stackexchange.com/questions/387640/compute-the-centroid-of-a-semicircle-without-calculus

Rupert

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  • $\begingroup$ See Apostol, A Fresh Look at the Method of Archimedes, p. 506. $\endgroup$
    – Conifold
    Nov 23 '20 at 20:08
  • $\begingroup$ This doesnt explain how Archimedes could use the integral of a cube by the leverage argument shown on wikipedia. $\endgroup$
    – rupert
    Nov 24 '20 at 9:29
  • $\begingroup$ Archimedes did not "use" integral of a cube, or any other, nor did he "find" or "evaluate" any integrals as numbers, despite what Wikipedia says. He only related volumes or centroids of one shape to those of another, in this case hemisphere and prism. And that is done by relating areas or centroids of the corresponding slices, as Apostol explains. $\endgroup$
    – Conifold
    Nov 24 '20 at 16:24
  • $\begingroup$ The Works Of Archimedes : Heath,T.L. shows he did after a bit of work ... $\endgroup$
    – rupert
    Nov 25 '20 at 11:30
  • $\begingroup$ In other words, Heath did it 2000+ years later. Just because we are now used to computing volumes with integrals, and modern commentators explain things accordingly, does not mean that Archimedes did it, especially before integrals were even invented. $\endgroup$
    – Conifold
    Nov 25 '20 at 12:03
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I confirm that Archimedes did get the centroid of a hemisphere. This can be done - without calculus -using the leverage argument in the wikipedia article relying on the integral of a cube which follows from the leverage argument after using Archimedes method for the centroid of a parabolic segment which Archimedes go in Proposition 8 of Book 2 (see The Works Of Archimedes : Heath,T.L.). Take a hemisphere of radius 1. So for the hemisphere, volume of hemisphere is 2/3pi. Here Archimedes proves the centroid of a parabola is 3/4 which shows the integral of the cube is 1/4 - by geometry and the leverage argument! (start with the center of gravity of parabolic segment and go from there). Using leverage this balances the integral of pi.x(1-x^2)=pi(x-x^3)=pi(1/2-1/4)=pi/4 So if centroid of hemisphere is "x" then x.2/3pi =pi/4 so x=3/8 QED Hats off to Archimedes! Eureka!! Wikipedia is right on this!!!

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