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The Method of Mechanical Theorems
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The Method of Mechanical Theorems
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{{italic title}}The Method of Mechanical Theorems (), also referred to as The Method, is considered one of the major surviving works of the ancient Greek polymath Archimedes. The Method takes the form of a letter from Archimedes to Eratosthenes,{{harvnb|Archimedes|1912}} the chief librarian at the Library of Alexandria, and contains the first attested explicit use of indivisibles (sometimes referred to as infinitesimals).{{harvnb|Archimedes|1912}}Netz, Reviel; Saito, Ken; Tchernetska, Natalie: A new reading of Method Proposition 14: preliminary evidence from the Archimedes palimpsest. I. SCIAMVS 2 (2001), 9â€“29. 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 law of the lever, which was first demonstrated by Archimedes, and of the center of mass (or centroid), which he had found for many special shapes.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.- the content below is remote from Wikipedia
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Area of a parabola
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
int_0^1 x^2 , dx = frac{1}{3},
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 x-y plane between the x-axis and y = x2 as x varies from 0 to 1. The triangle is the region in the x-y plane between the x-axis and the line y = x, also as x varies from 0 to 1.Slice the parabola and triangle into vertical slices, one for each value of x. Imagine that the x-axis is a lever, with a fulcrum at x = 0. 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 x, the slice of the triangle at position x has a mass equal to its height x, and is at a distance x from the fulcrum; so it would balance the corresponding slice of the parabola, of height x2, if the latter were moved to x = −1, at a distance of 1 on the other side of the fulcrum.Since each pair of slices balances, moving the whole parabola to x = −1 would balance the whole triangle. This means that if the original uncut parabola is hung by a hook from the point x = −1 (so that the whole mass of the parabola is attached to that point), it will balance the triangle sitting between x = 0 and x = 1.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 E, 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 E, 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 y = x/2, while a second median is the line y = 1 − x. Solving these equations, we see that the intersection of these two medians is above the point x = 2/3, 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 x = 0. 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 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.First proposition in the palimpsest
Consider the parabola in the figure to the right. Pick two points on the parabola and call them A and B.missing image!
- Archie1small.png -
right
Suppose the line segment AC is parallel to the axis of symmetry of the parabola. Further suppose that the line segment BC lies on a line that is tangent to the parabola at B.The first proposition states:
- Archie1small.png -
right
The area of the triangle ABC is exactly three times the area bounded by the parabola and the secant line AB.
Proof:
Let D be the midpoint of AC. Construct a line segment JB through D, where the distance from J to D is equal to the distance from B to D. We will think of the segment JB as a "lever" with D as its fulcrum. As Archimedes had previously shown, the center of mass of the triangle is at the point I on the "lever" where DI :DB = 1:3. Therefore, it suffices to show that if the whole weight of the interior of the triangle rests at I, and the whole weight of the section of the parabola at J, the lever is in equilibrium.Consider an infinitely small cross-section of the triangle given by the segment HE, where the point H lies on BC, the point E lies on AB, and HE is parallel to the axis of symmetry of the parabola. Call the intersection of HE and the parabola F and the intersection of HE and the lever G. If the whole weight of the triangle rests at I, it exerts the same torque on the lever JB as it does on HE. Thus, we wish to show that if the weight of the cross-section HE rests at G and the weight of the cross-section EF of the section of the parabola rests at J, then the lever is in equilibrium. In other words, it suffices to show that EF :GD = EH :JD. But that is a routine consequence of the equation of the parabola. Q.E.D.Volume of a sphere
Again, to illuminate the mechanical method, it is convenient to use a little bit of coordinate geometry. If a sphere of radius 1 is placed with its center at x = 1, the vertical cross sectional radius rho_S at any x between 0 and 2 is given by the following formula:
rho_S(x) = sqrt{x(2-x)}.
pi rho_S(x)^2 = 2pi x - pi x^2.
rho_C(x) = x
pi rho_C^2 = pi x^2.
M(x) = 2pi x.
V_S = 4pi - {8over 3}pi = {4over 3}pi.
Surface area of a sphere
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 scriptstyle Sr/3, which must equal the volume of the sphere: scriptstyle 4pi r^3/3. Therefore, the surface area of the sphere must be 4pi r^2, or "four times its largest circle". Archimedes proves this rigorously in On the Sphere and Cylinder.Curvilinear shapes with rational volumes
One of the remarkable things about the Method is that Archimedes finds two shapes defined by sections of cylinders, whose volume does not involve π, despite the shapes having curvilinear boundaries. This is a central point of the investigation—certain curvilinear shapes could be rectified by ruler and compass, so that there are nontrivial rational relations between the volumes defined by the intersections of geometrical solids.Archimedes emphasizes this in the beginning of the treatise, and invites the reader to try to reproduce the results by some other method. Unlike the other examples, the volume of these shapes is not rigorously computed in any of his other works. From fragments in the palimpsest, it appears that Archimedes did inscribe and circumscribe shapes to prove rigorous bounds for the volume, although the details have not been preserved.The two shapes he considers are the intersection of two cylinders at right angles, which is the region of (x, y, z) obeying:
(2Cyl) x^2 + y^2
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