, and If this be taken from x = a to x = 0, we find the area of the quadrant to be ", and therefore that of the whole circle to be πα. If the equation to the circle be y = 2a x - ?, area A = Sex (a* = a*)} = . cire. area whose cosine is " + C. For the whole ellipse A =Ha* = mab. Qy' = 4a (2a – x). A = 2a {(2ax – a*)+ a vers + C. Taking this from x = 2a to x = 0, and doubling it on account of the symmetry on both sides of the axis of X, we find the whole area between the curve and its asymptote to be 47 a?. (7) The equation to the cissoid is y(2a – x) = 2013. Here A = – 20 (2ax – x2)} + 3. circ. area whose versine is - + C, and the whole area included between the asymptote and the two branches of the curve is 3T aʼ. (8) The equation to the cycloid is dy_(20 x – xP)? This being only a differential equation, it is necessary to use an artifice for the purpose of effecting the integration. If we integrate Sydx by parts we have Sydx = xy – sædy, = xy - sdw (2ax – w?). Taking this integral from x = 0 to x = 2a, and doubling it, we find the whole area of the cycloid to be 37a', or three times the area of the generating circle. (9) The differential equation to the tractrix being y (a– y')!' A = Sydæ = – Sdy(a’ – y°)}; and the whole area included between the curve and the positive axis is main (10) The equation to the catenary being y=fee +6), its area is clot-e *) = c(to – c*)}. (11) The equation to the evolute of the ellipse is ()*+63)-1. 3 Taß The whole area inclosed by the curve is This is best investigated by Dirichlet's method of evaluating definite integrals. See Chap. XI. (12) The equations to the companion to the cycloid are y = a0, I = 2(1 - cos ), Sydr = xy - fxdy = a* (sino - cos 8) + C. The whole area is 27 a?, or twice the area of the generating circle. When an area is referred to polar co-ordinates r and 0, its value is given by the double integral firdrdo taken be tween proper limits. Integrating with respect to r we have A = 1 [updo + C; and if we suppose C = 0, the integral A = ] [pde, in which there is substituted for r its value in terms of a given by the equation to the curve, is the value of the sectorial area swept out by the radius vector. In taking the integral between the limiting values of 0, the same precaution must be observed as in the case of rectilinear co-ordinates, that the interval shall not contain a value of which causes r to vanish or become infinite. If we suppose 0 to increase indefinitely, the same geometrical space will be repeatedly swept over by the radius vector at each revolution, so that, when the curve is not re-entering, the analytical area (if we may use the phrase) differs from the geometrical area : to obtain the latter we must subtract from the analytical area that portion which has been previously swept over. Thus if we wish to find the geometrical area included between the values 0 and 47 of 0, and if we put A, = }12*go do, A, = 1 | **yde, 72 = a' cos 28, A = }a'. This is the fourth part of the whole area of the curve, which is therefore equal to a’. In this case, if we had at once integrated from 0 = 0 to 0 = 1, or 0 = 27 we should have found the area to be zero. This anomaly would arise from our integrating through an r = a cos 0 + b, where a> b. If we wish to find the area included within ODCAHG, it is sufficient to integrate from 0 = 0 to that value of 0 which causes 7 to vanish, and then to double the result. Let a = cos(-9), then the area ODCAHG is equal to {{(a’ + 2bo) a +36(a? – 691} ; and the area OEBF is equal to }{(a? + 26°) (1 - a) – 36(a? – 62)}}. If b = a, the curve becomes the common cardioid, and its area is sta". area are (15) The equation to the conchoid of Nicomedes when referred to polar co-ordinates is r = a sec 0 + b, and its area is }{a’tan + 2ab log tan (6 + + *0} + C. (16) The curve whose equation is r = a sin 30 has six loops (see fig. 49), and it is sufficient to find the area inclosed by one of them. This is easily seen to be To and therefore the sum of the areas of the six loops is daa', or one half of the area of the circle which bounds them. (17) The equation to the spiral of Archimedes is r = a. Hence the area = "8* + C. 12 After n revolutions the analytical area swept out is a?" , m (97) "-; but to obtain the geometrical area we must sub 6 tract from it the area corresponding to (n − 1) revolutions, which gives us (312 – 31 + 1) **?" as the required geometrical area. In the same way we should obtain as the geometrical area corresponding to (n + 1) revolutions, the expression (3n2 + 3x + 1) (27) , and the difference between these or the space between the arcs after (n + 1) and after n revolutions is n (27) a’, which is n times the space between the arcs after the first and second revolutions. (18) In the hyperbolic spiral 70 = a. The area swept out by the radius vector from 0 to go is lar, which is equal to the triangle formed by the radius, the tangent and the sub-tangent. If the equation to the spiral be given by a relation between p and 1, we have r prdr 1 * J (ju? – pʻ) go? – p’ = a'. where c = a + 2b, a and 6 being the radii of the fixed and generating circles respectively. Hence Acil Code (pu2 – a') or rdr (pu? – a^)} |