Method 2

Archimedes Mechanical Method with Indivisibles, The Method, Prop. 2©
by Henry Mendell (Cal. State U., L.A.)

Return to Survey of Infinitary Arguments

Introduction to Archimedes Mechanical Method with Indivisibles

(changing diagram or simple diagram) That every sphere is four-times the cone with base equal to the greatest circle of those in the sphere, and height equal to the straight-line from the center of the sphere, and the cylinder with base equal to the greatest circle of those in the sphere, and height equal the diameter of the sphere is half-again the sphere, is here observed in this way.

(diagram 1) For let there be a sphere in which ABGD is a greatest circle, (diagram 2) diameters AG, BD being at right angles to one another, (diagram 3) and let there be a circle in the sphere about diameter BD perpendicular to circle ABGD, (diagram 4) and from this perpendicular circle let there be inscribed a cone with its vertex point A, (diagram 5) and its surface being extended let the cone be cut by a plane through G parallel to the base. (Diagram 6) In fact it will make a circle perpendicular to line AG, (diagram 7) and EZ is its diameter. (Diagram 8) From this circle let there be inscribed a cylinder with axis equal to AG, (diagram 9) and let there be sides of the cylinder EL, ZH. (Diagram 10) And let GA be extended and let AQ be placed equal to it and let GQ be conceived as joined, its middle being A, (diagram 11) and let there be some parallel to BD, namely MN, (diagram 12) and let this cut circle ABGD at X, O, diameter AG at S, straight-line AE at P, AZ at R, and from straight-line let there be erected a plane perpendicular to AG. (Diagram 13) In fact, this will make an intersection in the cylinder that is a circle, whose diameter will be MN, a circle in sphere ABGD whose diameter will be XO, (diagram 14) and a circle in cone AEZ, whose diameter is PR. (diagram 15, all three slices together)

(diagram 16) And since the rectangle by GA, AS is equal to the rectangle by MS, SP, since AG is equal to SM while AS is equal to PS, the square from AX is equal to the rectangle by GA, AS, i.e. the squares from XS, SP, therefore the rectangle from MS, SP is equal to the squares from XS, SP.

(diagram 17) or more simply, diagram 18) Note that ALEG is a square since AK and BK are radii at right angles, i.e., form an isosceles right triangle AKB with AGE similar.
AG = SM & AS = PS (since triangle AKB is isosceles and right and simlar to triangle ASP) => GA*AS = MS*SP.
AX² = GA*AS (since triangle ASX ~ triangle AXG, so that AS : AX = AX : AG)
AX² = XS² + AS² = XS² + PS² (since PS = AS)
Hence, XS² + PS² = GA*AS = MS*SP.

(diagram 16) And since as GA is to AS so is MS to SP, but GA is equal to AQ, therefore as QA is to AS, MS is to SP, that is the square from MS to the rectangle by MS SP. But the squares from XS, SP were proved equal to the rectangle by MS, SP. Therefore, as AQ is to AS, so is the square from MS to the squares from XS, SP.

GA : AS = MS : SP (since MS = EG = AG and AS = SP)
GA = AQ (by the construction of AQ)
AQ : AS = MS : SP = MS² : MS*SP
XS² + PS² = MS*SP (from above)
AQ : AS = MS² : XS² + PS²

But as the square from MS to the squares from XS, SP, so is the square from MN to the squares from XO, PR. (diagram 15, all three slices together) But as the square from MN to the squares from XO, PR so is the circle in the cylinder, whose diameter is MN, to both circles, the one in the cone whose diameter is PR and the one in the sphere whose diameter is XO. Therefore, as QA is to AS, so is the circle in the cylinder to the circles in the sphere and in the cone.

MS² : XS² + PS² = MN² : XO² + PR² (i.e. a² : b² + c² = (2a)² : (2b)² + (2c)²)
MN² : XO² + PR² = circle MN : circle XO + circle PR
Hence, AQ : AS = circle MN : circle XO + circle PR

(diagram 19) And so since as QA is to AS, so is the circle itself in the cylinder when it stays where it is to both circles whose diameters are XO, PR when they are moved over and place thus at Q, so that Q is the center of the weight of each of them, they will balance at point A. (diagram 20) It will be similarly proved that even if another line is drawn in parallelogram LZ parallel to EZ, and a plane is erected from the drawn line perpendicular to AG, that the circle which arises in the cylinder and remaining where it is will balance both circles, the one which arises in the sphere and the one in the cone when they are moved over and placed on the balance beam at Q in such a way that Q is the center of weight of each.

(Diagram 21) And so when the cylinder is filled up by the circles being taken as well as the sphere and the cone, (diagram 22 or more simply diagram 23) the cylinder remaining where it is will balance at point A the sphere and cone moved over and placed on the balance beam at Q, so that Q is the center of weight of each of them. And so since the mentioned solids balance at point A with the cylinder remaining with its center of weight K, the sphere and the cone moved over, as was said, about center of weight Q, it will be that as QA is to AK, so is the cylinder to the sphere and the cone.

(Diagram 22 or more simply diagram 23) But QA is double AK. Therefore the cylinder is also double the sphere and the cone together. But it is three-times the cone. Therefore, three cones are equal to two cones themselves and two spheres. Let two cones in common be taken away. Therefore, one cone with triangle though its axis is AEZ is equal to the mentioned two spheres. (Diagram 24) But the cone whose triangle through its axis is AEZ is equal to eight cones whose triangle through the axis is ABD, since EZ is double BD. Therefore the mentioned eight cones are equal to two spheres. Therefore the sphere, whose greatest circle is ABGD, is four-times the cone whose vertex is point A and whose base is the circle about diameter BD perpendicular to AG.

Cylinder : Sphere+Cone = AQ : A K = 2 : 1 (since the centers of weight are Q for the sphere and cone and K for the cylinder)
Cylinder = 3 Cones = 2 Spheres + 2 Cones
1 Cone = 2 Spheres
1 Cone (AZE) = 8 Cones (ABD) (since AG = 2 AK and ZE = 2 BD, so that the base ZE : base BD = 4 : 1 and the height AG : height AK = 2 : 1)
8 Cones (ABD) = 2 Spheres
4 Cones (ABD) = 1 Sphere

(diagram 25 or more simply diagram 26) Let there be drawn through points B, D in parallelogram LZ parallels, FBC, YDW, to AG. And let there be conceived a cylinder whose bases are the circles about diameters FY, CW, as well as axis AG. And so since the cylinder whose parallelogram through the axis is FW is double the cylinder whose parallelogram through the axis is FD, but this is itself three-times the cone whose triangle through the axis is ABD, as in the Elements, therefore the cylinder whose parallelogram through the axis is FW is six-times the whose whose triangle through the axis is ABD. But it was proved that the sphere whose greatest circle is ABGD is four times the same cone. Therefore, the cylinder is half-again the sphere, which it was required to prove.

Cylinder FW = 2 Cylinder FD = 2*3 Cones ABD = 6 Cones ABD
4 Cones (ABD) = 1 Sphere
Cylinder FW = 1 1/2 Spheres

(diagram 27) When this has been observed, because every sphere is four-times the cone with its base the greatest circle and height equal to the straight-line from the center of the sphere, the conception arises that the surface of every sphere is four-times the greatest circle of those in the sphere. For there was a supposition and (diagram 28) because every circle is equal to a triangle having as its base the circular-arc of the circle and height equal to the straight-line from the center of the circle, and because every sphere is equal to a cone with base the surface of the sphere and height equal to the straight-line from the center of the sphere.

The blue cone has height equal to the radius of the sphere, and the radius of its base is 2-times the radius of the sphere.
The green cones have height and radius of the base equal to the radius of the sphere.
Hence, the base of the blue cone is equal to 4-times the base of a green cone.
Hence, the blue cone = 4 green cones = the sphere.
Archimedes then surmises that the surface area of the sphere is the area of the base of the equal blue cone on the analogy of the triangle with base equal to the circumference and height the radius of a circle.