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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 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 area of the surface of a sphere is equal to quadruple the area of a great circle of this sphere. The area of a segment of the parabola cut from it by a straight line is 4/3 the area of the triangle inscribed in this segment. For the proof of the results Archimedes used the Method of exhaustion of Eudoxus.
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 ...
In Quadrature of the Parabola, Archimedes proved that the area enclosed by a parabola and a straight line is 4 / 3 times the area of a corresponding inscribed triangle as shown in the figure at right. He expressed the solution to the problem as an infinite geometric series with the common ratio 1 / 4 :
This method was further developed and employed by Archimedes in the 3rd century BC and used to calculate the area of a circle, the surface area and volume of a sphere, area of an ellipse, the area under a parabola, the volume of a segment of a paraboloid of revolution, the volume of a segment of a hyperboloid of revolution, and the area of a ...
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 ...
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 {\displaystyle x} , although higher powers become complicated without algebra.