Quadrature of the Parabola

A parabolic segment is the region bounded by a parabola and line. To find the area of a parabolic segment, Archimedes considers a certain inscribed triangle. The base of this triangle is the given chord of the parabola, and the third vertex is the point on the parabola such that the tangent to the parabola at that point is parallel to the chord. Proposition 1 of the work states that a line from the third vertex drawn parallel to the axis divides the chord into equal segments. The main theorem claims that the area of the parabolic segment is \tfrac43 that of the inscribed triangle. == Structure of the text ==

Conic sections such as the parabola were already well known in Archimedes' time thanks to Menaechmus a century earlier. However, before the advent of the differential and integral calculus, there were no easy means to find the area of a conic section. Archimedes provides the first attested solution to this problem by focusing specifically on the area bounded by a parabola and a chord. 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, with weighted segments of a parabola and a triangle suspended along the arms of a lever at specific distances from the fulcrum. When the center of gravity of the triangle is known, the equilibrium of the lever yields the area of the parabola in terms of the area of the triangle which has the same base and equal height. Archimedes here deviates from the procedure found in On the Equilibrium of Planes in that he has the centers of gravity at a level below that of the balance. The second and more famous proof uses pure geometry, particularly the sum of a geometric series. Of the twenty-four propositions, the first three are quoted without proof from Euclid's Elements of Conics (a lost work by Euclid on conic sections). Propositions 4 and 5 establish elementary properties of the parabola. Propositions 6–17 give the mechanical proof of the main theorem; propositions 18–24 present the geometric proof. == Geometric proof ==

Dissection of the parabolic segment The main idea of the proof is the dissection of the parabolic segment into infinitely many triangles, as shown in the figure to the right. Each of these triangles is inscribed in its own parabolic segment in the same way that the blue triangle is inscribed in the large segment. Areas of the triangles In propositions eighteen through twenty-one, Archimedes proves that the area of each green triangle is \tfrac18 the area of the blue triangle, so that both green triangles together sum to \tfrac14 the area of the blue triangle. From a modern point of view, this is because the green triangle has \tfrac12 the width and \tfrac14 the height of the blue triangle: Following the same argument, each of the 4 yellow triangles has \tfrac18 the area of a green triangle or \tfrac1{64} the area of the blue triangle, summing to \tfrac4{64} = \tfrac1{16} the area of the blue triangle; each of the 2^3 = 8 red triangles has \tfrac18 the area of a yellow triangle, summing to \tfrac{2^3}{8^3} = \tfrac1{64} the area of the blue triangle; etc. Using the method of exhaustion, it follows that the total area of the parabolic segment is given by :\text{Area}\;=\;T \,+\, \frac14T \,+\, \frac1{4^2}T \,+\, \frac1{4^3}T \,+\, \cdots. Here T represents the area of the large blue triangle, the second term represents the total area of the two green triangles, the third term represents the total area of the four yellow triangles, and so forth. This simplifies to give :\text{Area}\;=\;\left(1 \,+\, \frac{1}{4} \,+\, \frac{1}{16} \,+\, \frac{1}{64} \,+\, \cdots\right)T. Sum of the series To complete the proof, Archimedes shows that :1 \,+\, \frac{1}{4} \,+\, \frac{1}{16} \,+\, \frac{1}{64} \,+\, \cdots\;=\; \frac{4}{3}. The formula above is a geometric series—each successive term is one fourth of the previous term. In modern mathematics, that formula is a special case of the sum formula for a geometric series. Archimedes evaluates the sum using an entirely geometric method, illustrated in the adjacent picture. This picture shows a unit square which has been dissected into an infinity of smaller squares. Each successive purple square has one fourth the area of the previous square, with the total purple area being the sum :\frac{1}{4} \,+\, \frac{1}{16} \,+\, \frac{1}{64} \,+\, \cdots. However, the purple squares are congruent to either set of yellow squares, and so cover \tfrac13 of the area of the unit square. It follows that the series above sums to \tfrac43 (since == See also ==