. An elementary treatise on the differential and integral calculus. 185. Area between Two Curves.—If the area wereincluded between the two curves AB and ab, whose equa-tions are respectively y = f(x) andy = 0 (x), and two ordinates CD andEH, where OD = b and OH = a,we should find by a similar course ofreasoning, pa d A= / If (x) — ix)] d% Fig. 50 v b The determination of the area of a curve is called itsQuadrature. 186. The Circle.—The equation of the circle referred toits centre as origin, is y2 = a2 — x2; therefore the area ofa quadrant is represented by A = f\a2 — x2)? dx vx (a2 — x2)^ a2 .
. An elementary treatise on the differential and integral calculus. 185. Area between Two Curves.—If the area wereincluded between the two curves AB and ab, whose equa-tions are respectively y = f(x) andy = 0 (x), and two ordinates CD andEH, where OD = b and OH = a,we should find by a similar course ofreasoning, pa d A= / If (x) — ix)] d% Fig. 50 v b The determination of the area of a curve is called itsQuadrature. 186. The Circle.—The equation of the circle referred toits centre as origin, is y2 = a2 — x2; therefore the area ofa quadrant is represented by A = f\a2 — x2)? dx vx (a2 — x2)^ a2 . , xia /ri _. . „„„* = —i—5 *- + - sin1 - (See Ex. 4, Art, 151.) L Z a ajQ a2n ~ ~T;therefore the area of the circle = na2. Also, if OM = x, the area of OBDMbecomes A = f\a2 — x2f> dx [x (a2 — x2)2 a2 . , x~]x= [~2 + % Sm_1 Jo # (a2 — x2)* a2 . # 2 — + 2- Sm_1 a. Fig. 51, This result is also evident from geometric considerations16 0G2 QUADRATURE OF THE PARABOLA. X i for the area of the triangle OMD = \ {a2 — x2)?, and the area of the sector ODB = - sin-1 -• Z a Remark.—The student will perceive that in integrating betweenthe limits x = 0 and x — a, we take in every elementary slice PQRNin the quadrant ADBO; also integrating between the limits x = 0and x — x = OM, we take in every elementary slice between OBand MD.* 187. The Parabola.—From y2 — 2px, we have y = <\/%px. Hence, for the area of the part OPM,we have A = \/%p I x^clx — f V%p x% •; i. e., fxy. Fig. 52. Therefore the area of the segment POP, cut off by a chord perpendicular to the axis, is f of the rectangle PHHF. 188. The Cycloid.—From the equation x = r vers-1 - — */%ry — y2, we have dx ydy V%ry — y2 v»2r y2dy V%ry — y2 |7rr2 (See Ex. 6, Art. 151) = i the area of the cycloid. Sinceintegrating between the limits includes half the area ofthe figure.
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