"It is to Archimedes... that we owe the nearest approach to actual integration to be found among the Greeks. His first noteworthy advance... was concerned with his proof that the area of a parabolic segment is four thirds of the triangle with the same base and vertex, or two thirds of the circumscribed parallelogram. This was shown by continually inscribing in each segment between the parabola and the inscribed figure a triangle with the same base and... height as the segment. If A is the area of the original inscribed triangle, the process... leads to the summation of the seriesA + \frac{1}{4}A + (\frac{1}{4})^2A + (\frac{1}{4})^3A+...or...A[1 + \frac{1}{4} + (\frac{1}{4})^2 + (\frac{1}{4})^3+...]so that he really finds the area by integration and recognizes, but does not assert, that(\frac{1}{4})^n \to 0~\text{as}~n \to \infty,this being the earliest example that has come down to us of the summation of an infinite series. ... In his treatment of solids bounded by curved surfaces he arrives at conclusions which we should now describe by the following formulas: Surface of a sphere,4\pi a^2 \cdot \frac{1}{2} \int\limits_{0}^{\pi} \sin\theta d\theta = 4\pi a^2.Surface of a spherical segment,\pi a^2 \int\limits_{0}^{a} 2\sin\theta d\theta = 2\pi a^2 (1-\cos\alpha).Volume of a segment of a hyperboloid of revolution,\int\limits_{0}^{b} (ax + x^2) dx =b^2(\frac{1}{2}a + \frac{1}{3}b).Volume of a segment of a spheroid,\int\limits_{0}^{b} x^2 dx = \frac{1}{3}b^3.Area of a spiral, \frac{\pi}{a} \int\limits_{0}^{a} x^2 dx = \frac{1}{3} \pi a^2.Area of a parabolic segment, \frac{1}{A^2} \int\limits_{0}^{A} \bigtriangleup^2 d\bigtriangleup = \frac{1}{3} A."