Архимедтің механикалық теоремалар әдісі

Mathematical treatise by Archimedes

The Method of Mechanical Theorems (Περὶ μηχανικῶν θεωρημάτων πρὸς Ἐρατοσθένη ἔφοδος), also referred to as The Method, is one of the major surviving works of the ancient Greek polymath Archimedes. The Method takes the form of a letter from Archimedes to Eratosthenes, the chief librarian at the Library of Alexandria, and contains the first attested explicit use of indivisibles (indivisibles are geometric versions of infinitesimals). The work was originally thought to be lost, but in 1906 was rediscovered in the celebrated Archimedes Palimpsest. The palimpsest includes Archimedes' account of the "mechanical method", so called because it relies on the center of weights of figures (centroid) and the law of the lever, which were demonstrated by Archimedes in On the Equilibrium of Planes. Archimedes did not admit the method of indivisibles as part of rigorous mathematics, and therefore did not publish his method in the formal treatises that contain the results. In these treatises, he proves the same theorems by exhaustion, finding rigorous upper and lower bounds which both converge to the answer required. Nevertheless, the mechanical method was what he used to discover the relations for which he later gave rigorous proofs.

To explain Archimedes' method today, it is convenient to make use of a little bit of Cartesian geometry, although this of course was unavailable at the time. His idea is to use the law of the lever to determine the areas of figures from the known center of mass of other figures. The simplest example in modern language is the area of the parabola. Archimedes uses a more elegant method, but in Cartesian language, his method is calculating the integral

which can easily be checked nowadays using elementary integral calculus. The idea is to mechanically balance the parabola (the curved region being integrated above) with a certain triangle that is made of the same material. The parabola is the region in the plane between the axis and the curve as varies from 0 to 1. The triangle is the region in the same plane between the axis and the line , also as varies from 0 to 1. Slice the parabola and triangle into vertical slices, one for each value of Imagine that the axis is a lever, with a fulcrum at The law of the lever states that two objects on opposite sides of the fulcrum will balance if each has the same torque, where an object's torque equals its weight times its distance to the fulcrum. For each value of , the slice of the triangle at position has a mass equal to its height , and is at a distance from the fulcrum; so it would balance the corresponding slice of the parabola, of height , if the latter were moved to , at a distance of 1 on the other side of the fulcrum. Since each pair of slices balances, moving the whole parabola to would balance the whole triangle. This means that if the original uncut parabola is hung by a hook from the point (so that the whole mass of the parabola is attached to that point), it will balance the triangle sitting between and
The center of mass of a triangle can be easily found by the following method, also due to Archimedes. If a median line is drawn from any one of the vertices of a triangle to the opposite edge , the triangle will balance on the median, considered as a fulcrum. The reason is that if the triangle is divided into infinitesimal line segments parallel to , each segment has equal length on opposite sides of the median, so balance follows by symmetry. This argument can be easily made rigorous by exhaustion by using little rectangles instead of infinitesimal lines, and this is what Archimedes does in On the Equilibrium of Planes. So the center of mass of a triangle must be at the intersection point of the medians. For the triangle in question, one median is the line , while a second median is the line Solving these equations, we see that the intersection of these two medians is above the point , so that the total effect of the triangle on the lever is as if the total mass of the triangle were pushing down on (or hanging from) this point. The total torque exerted by the triangle is its area, 1/2, times the distance 2/3 of its center of mass from the fulcrum at This torque of 1/3 balances the parabola, which is at a distance 1 from the fulcrum. Hence, the area of the parabola must be 1/3 to give it the opposite torque. This type of method 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 , although higher powers become complicated without algebra. Archimedes only went as far as the integral of , which he used to find the center of mass of a hemisphere, and in other work, the center of mass of a parabola.

To find the surface area of the sphere, Archimedes argued that just as the area of the circle could be thought of as infinitely many infinitesimal right triangles going around the circumference (see Measurement of the Circle), the volume of the sphere could be thought of as divided into many cones with height equal to the radius and base on the surface. The cones all have the same height, so their volume is 1/3 the base area times the height. Archimedes states that the total volume of the sphere is equal to the volume of a cone whose base has the same surface area as the sphere and whose height is the radius. There are no details given for the argument, but the obvious reason is that the cone can be divided into infinitesimal cones by splitting the base area up, and the each cone makes a contribution according to its base area, just the same as in the sphere. Let the surface of the sphere be S. The volume of the cone with base area S and height r is , which must equal the volume of the sphere: Therefore, the surface area of the sphere must be , or "four times its largest circle". Archimedes proves this rigorously in On the Sphere and Cylinder.

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.