Search results
Results from the WOW.Com Content Network
Archimedes provides the first attested solution to this problem by focusing specifically on the area bounded by a parabola and a chord. [3] Archimedes gives two proofs of the main theorem: one using abstract mechanics and the other one by pure geometry. In the first proof, Archimedes considers a lever in equilibrium under the action of gravity ...
Archimedes' idea is to use the law of the lever to determine the areas of figures from the known center of mass of other figures. [1]: 8 The simplest example in modern language is the area of the parabola. A modern approach would be to find this area by calculating the integral
Archimedes used the method of exhaustion to compute the area inside a circle. Archimedes used the method of exhaustion as a way to compute the area inside a circle by filling the circle with a sequence of polygons with an increasing number of sides and a corresponding increase in area.
The lever and its properties were already well known before the time of Archimedes, and he was not the first to provide an analysis of the principle involved. [5] The earlier Mechanical Problems, once attributed to Aristotle but most likely written by one of his successors, contains a loose proof of the law of the lever without employing the concept of centre of gravity.
Archimedes used the method of exhaustion to calculate the area under a parabola in his work Quadrature of the Parabola. Laying the foundations for integral calculus and foreshadowing the concept of the limit, ancient Greek mathematician Eudoxus of Cnidus ( c. 390–337 BC ) developed the method of exhaustion to prove the formulas for cone and ...
Archimedes in his The Quadrature of the Parabola used the sum of a geometric series to compute the area enclosed by a parabola and a straight line. Archimedes' theorem states that the total area under the parabola is 4/3 of the area of the blue triangle. His method was to dissect the area into infinite triangles as shown in the adjacent figure ...
The area of the surface of a sphere is equal to four times the area of the circle formed by a great circle of this sphere. The area of a segment of a parabola determined by a straight line cutting it is 4/3 the area of a triangle inscribed in this segment. For the proofs of these results, Archimedes used the method of exhaustion attributed to ...
The ratio of the volume of a sphere to the volume of its circumscribed cylinder is 2:3, as was determined by Archimedes. The principal formulae derived in On the Sphere and Cylinder are those mentioned above: the surface area of the sphere, the volume of the contained ball, and surface area and volume of the cylinder.