by Jeremy Batterson

Archimedes’ discovery of the method of determination of the volume of a sphere was a discovery of such beauty and with such astonishing implications, that Archimedes, before his death, instructed that it be engraved upon his tombstone. And, yet, almost none, in our day, have ever worked through its proof, although it stands as a precursor of Leibniz’s later idea of the integral, as well as, it seems to me, hinting at the existence of the unseen domain which Gauss and others would later investigate. I had thought, myself, that this proof must have involved enormously complicated calculations, when, in fact, it is very easily accessible to any who desires to do a bit of mental work, so that, once having worked it through, we realize how totally laughable it is that every person who has EVER studied geometry has not worked through this and other problems of the ancient Greeks. It is so ridiculous that it would be as if you were to go to a modern university to study the subject of economics and not even study LaRouche’s works. What would YOU think about such a laughable thing?

Beginning with a sphere, Archimedes circumscribed it with a cylinder whose height was equal to the diameter of the sphere it circumscribed, but whose diameter was twice the diameter of this sphere, with the two solids based on a common center, namely, the center of the base of the cylinder. Finally, at this same common center, he placed the apex of a cone, whose height was equal to that of the sphere and cylinder, and with base equal in diameter to that of the cylinder. Thus, all three solids had the same height and the same common central axis, with the cylinder having a constant diameter all up and down this common axis, but the sphere and cone having constantly CHANGING diameters along its length. To draw a cross section of this construction, which will be needed for our further elaboration, draw a circle, and denote two opposite poles of this circle as “A” and “B”, so that line AB is the diameter D of this circle. Next, draw a rectangle whose shorter side is D and whose longer side is 2D, such that the circle is exactly centered within it, with point A lying on the center of one of the longer sides of the rectangle. Now, from point A, draw the two lines of maximum possible length from A to the opposite side of the rectangle, which will cause these lines to terminate at the far corners of the rectangle, points D and E, producing a triangle ADE, which is half of a square. Let the corner of the rectangle above A be point C and the corner below A be point F. This diagram represents a cross section of the sphere (circle), cone, (triangle) and cylinder (rectangle).

And now, Archimedes did something surprising! (Remember it was Archimedes who said: “Give me a fulcrum large enough and I will move the earth!”) If you ever played on a see-saw, as a kid, you may remember the principle of the fulcrum: A far lighter weight can lift a heavier weight, across a fulcrum, if the length of the balancing board between the two weights is longer on the side of the lighter weight. Thus, a toddler can hold his heavier teenage sibling up in the air this way, by simply placing the fulcrum far away from his end of the balancing board, and much closer to the end at which his sibling is sitting. Think of the fulcrum as being the “sun”, and the distance of balancing board between it and the end of the shorter length of board concerned as being the “perihelion” of an orbit, with the longer distance being the “aphelion.” The principle is that the heavier weight, of weight X times that of the lighter weight, equally balances the lighter weight when its distance from the fulcrum is 1/X the distance of the lighter weight from the fulcrum. Thus, for example, were the heavier weight twice the weight of the lighter, it would have to be 1/2 the distance from the fulcrum as this lighter. Or put otherwise, the product of the heavier weight and its distance from the fulcrum must equal the product of the lighter weight with its distance from the fulcrum, if they are to balance out. (X times 1/X = 1/X times X.) Behold, now, how Archimedes used the principle of the fulcrum in his discovery:

First, imagine the cut, line M2M, in our diagram, passing horizontally down the exact center of the construction, from the center of DC to the center of EF. This line passes exactly through the center O of the sphere. and also intersects both the sphere and cone at two points M3 and M4. Line OM3 is the radius of both the cone and the sphere at this particular point and is also the radius R of the sphere. If R is one, than the AREAS of the two corresponding circular slices of the cone and sphere will both equal 2?. The CYLINDRICAL cut, however, creates a circle of radius 2R, and hence, an area of 4?. Archimedes now asks, if the SUM of areas of the conical and spherical cuts is 2?, thus exactly HALF the area of the cylindrical cut, and if we treat the relative areas as WEIGHTS, where will these differing weights have to be to balance each other out? Leaving the cylindrical cut where it is, at distance R from A, he places a fulcrum at point F, and moves the areas of the conical and spherical cuts TWICE R, or D, to the other side of the fulcrum, to point G!

Now, project from any ARBITRARY point X along axis AB a perpendicular which extends upwards to line DC, intersecting this line at point X2. This line intersects the sphere at Xs and the Cone at Xc. We know, from Jonathan Tennenbaum’s pedagogical on Archytus that the line XXs, which is the radius of the circular cut of the sphere at this point, is the geometric MEAN between line XA and line XB. Thus, it follows that length XA/XXs = XXs/XB. Now, imagine point X, on the axis AB, as it travels from A to B. As its distance from A widens, the cylinder’s radius XX2, and, hence, corresponding WEIGHT of the cylindrical cut, remains constant, since all cylindrical cuts in the construction have the same radius. However, its DISTANCE XA FROM A is increasing, so that the weight needed to balance it out must either increase, or pass further and further away from the fulcrum. Let us now look more closely, then, at the SUM of the areas of the circular and conical cuts, the counterbalancing weight concerned:

Since all conical cuts in the construction result in icosoles right triangles, it follows that, for any such cut, the corresponding vertical radius (the line XXc in our diagram) will always equal this cut’s corresponding distance XA from A, and, hence, its lateral distance from F. Henceforth, let us call the line XA, which is the same as the radius of the conical cut, “MINOR.” Similarly, we will call the line XB “MAJOR” and the line XXs the geometric “MEAN” of these two extremes. (Don’t be confused by the fact that MINOR becomes longer than MAJOR once X crosses point O, but, rather, think of it as being the first of the two extremes. We could also call the two “origin” and “destination,” for example.) The sum of proportional areas of the conical and spherical cuts, will thus be, respectively, (MINOR times MINOR) + (MAJOR times MINOR), the spherical cut’s proportional area being such because MEAN times MEAN will always equal MAJOR times MINOR. Since the two proportional areas have a common factor, namely, MINOR, and since the sum of lines MINOR and MAJOR is the DIAMETER D of the sphere, the sum of the proportional AREAS of the conical and spherical cut must, thus, always be D times MINOR. Meanwhile, the proportional area of the cylindrical cuts, as we noted, remains D times D, since D is the radius of all cylindrical cuts.

As we recall, for the two weights to balance each other, the weight of the cylindrical cut times its distance from A (hence its lateral distance from F) must equal the sum of weights of the conical and spherical cuts times their distance from A. So, where does any particular pair of conical and spherical cuts balance out their corresponding cylindrical cut, assuming that we leave the cylindrical cut in its original position of distance MINOR from A? We know that DD times MINOR must equal some UNKNOWN DISTANCE times (D times MINOR.) What is that unknown distance? Indeed, it is D, since we can clearly see that

DD times MINOR = D times (D times MINOR)! And, this must be true for ALL cases, since the sum of proportional areas of the spherical and conical cuts is ALWAYS (D times MINOR), as we showed above! Thus, FOR ALL CASES, THE SUM OF WEIGHTS OF THE CONICAL AND SPHERICAL CUTS WILL BALANCE OUT THEIR CORRESPONDING CYLINDRICAL CUT’S WEIGHT AT A DISTANCE D FROM THE FULCRUM!!

Now, since this be true, it follows that, were we to make all possible infinite numbers of cuts through the construction, and thus encompass its entire VOLUME, we would end up with the cylinder remaining exactly in its original position, with all its weight focused at its center, namely the point directly beneath the origin of the circle, or point M on our diagram. This point is at R, or (1/2)D distance from the fulcrum. Meanwhile, the entire volume of both the cone and sphere would be squashed together into a PLANE, a circle balanced at point G, which is at distance D from the fulcrum. This balancing ratio of D:(1/2)D tells us that the weight of the SUM OF VOLUMES of the cone and sphere, contained within this squashed up plane, must be exactly HALF that of the Cylinder! But we are trying to find the weight, and, hence, the VOLUME of the sphere. Archimedes already knew (from Eudoxus, I believe) that the volume of the cone was 1/3 that of the cylinder encompassing it. Thus, If the total volume of the cylinder were 1, then the volume of the cone would have to be 1/3, while the volume of the sphere would have to be that which, when added to 1/3, yields 1/2 of 1. Thus, the volume of the sphere must be 1/6 that of the cylinder.

Archimedes took this one step further, by noting that the sphere which exactly encompasses the sphere would be of diameter D, instead of 2D, but have the same HEIGHT, and, hence, would have a volume 1/4 that of the larger cylinder he had used in his construction. Thus, the sphere would have a volume of FOUR times 1/6, which is 4/6 or 2/3 that of the sphere which encompassed it. However, since the volume of a cylinder is the area of its base, times its height, and since the height of the cylinder encompassing the sphere is 2R, this cylinder’s volume would be (?R^{2})2R. Since the sphere must be 2/3 of THIS volume, it is (4/3)?R^{3}. Hence the famous solution for the volume of a sphere.”

Now, ask yourself this: Did Archimedes figure this out by “tinkering,” by playing around with different shapes, until the right arrangement popped out, “by magic,” or, rather, did he find this particular construction BECAUSE he was proceeding from a principle? For example, imagine the following possibility: Since he knew that MEAN times MEAN was equal to MAJOR times MINOR, might he have, then, asked which form would create the circumstance wherein its cut would equal MINOR times MINOR, and seen instantly that this must be a isosceles right triangle, hence a section of the particular cone he used in the construction? Similarly, might he have then asked which Cylinder, when coupled with this arrangement, led to a certain lawful result? Or might it have been something similar to this, perhaps far more elaborated?

Now, is that beautiful, or what?