April 20, 188S. 



KNOWLEDGE • 



24.1 



^ur i«atl)fmatiral Column. 



EASY LESSONS IX THE INTEGRAL CALCULUS. 

 ri'^nE study of the integral dealt with in our last, in its relation 

 J. to the area of circular sections (whether sector, se^ient, or 

 otherwise), leads naturally to the discussion of the areas of other 

 conic sections. 



The ellipse, however, requires no new application of the integral 

 calculus. Thus, if B Q A is a right elliptical quadrant, C the centre 

 of the ellipse, and the semi-axes C A and B C are a and 6 respec- 



tively, we know that putting C L = x and PL = !/, !/' = — {a'—x') 



and the area of the elementary rectangle Q L = \idx 



So that the area C B KX, in which CN=.ri, 



a J 



= - -* o ^ + ^^"' — 



a L 2 2 a J 



= -area CliX 

 a 



When we tarn to the hyperbola, however, we 6nd om-selves led 

 to a new integral, and a new application of the integral calculus. 

 (In passing, I wish the reader to notice that my cliief object in 

 going over thus the discussion of the simpler integrals, and those 

 more commonly employed, is to associate this part of the student's 

 : with the actual application of the calculus.) 



we know that y^ = x'—a'' (where C.\ = n), and taking an ordinate 

 K N, wo have, the area of an elementary rectangle P M = ydx, 



= v'i'^"' dx 

 so that the area A K X, in which C N = a', 



— I -J x' — d' dx 



to determine which we must integrate v x- — a-di, not as yet 

 among the quantities whose integral is known to us. Befoie trying 

 to do this, let us consider the conjugate hyperbola A'P'Q'. Here, 

 completing the construction shown in the figure, we have, if P'L = y, 



and the elementary rectangle FL=%''x'' + a- dx; so that the area 

 AK'X, in which CX = .r, 



= / Va' - x'dx 



involving also a new integral, but nearer in form (note the limits of 

 tlie integration) to what wo obtained in the case of the circle and 

 the ellipse. 



We still refrain from the work of integration, because we know 

 from a familiar property of the hyperbola, that there is yet another 

 way of dealing with the curve which will probably give us the a rea 

 more readily, and therefore show ns how to integral® vi^ — o' dx 

 and vx'-ra'fJ.r mthout onr being at the pains to deal directly with 

 these, after any of the tentative methods yet described. 



Take the rectangular hyperbola RAK with CZ and CZ', the 

 asymptotes, for the ases of rtference. Draw CAS, the axis, and 

 A G, P L, Q M (near to P L), K X, all perpendicular to C Z. 



Then CA = a and CG=\/2a = o, (for convenience). Also, let 



C L = ar, C 5I = a; + d.i*, CX = x., (to distinguish from the a-, of onr 



former inquiry). Then, if PL=!/, we know that, drawing PL' 



and AG' perpendicular to C Z', rectangle L'L = square G G' = a,'; 



- ir n' 



Hence ij = — ; elementary rect. Q L = — dx, and area 



I.e., x\i = - 

 G A K X. 



=yr^'£''"=-?["'°'''-'°^'"] 



=-9- '°s - 



- "l 



Xow dmw KX' perp. to C A S, and pnt C X' = Jj. Also join C K. 

 Then we note that AC X K= AC G A, so that, taking each in suc- 

 cession from the area C A K X, we have area C A K = area A G N K 

 (which we have just determined). 



Xow area A P KX'= AC X' K — area C AK ; therefore we now 

 have the area A PK X' in the toils. For 



ACX'K = iCX' . X'K = ixiv'i,»-a' 



a' CX 

 Area C A K = area G A K X = •;; log ttt^ 



But if we draw X Kj'.', X'Kn, as in the figure, we know that 



CN=X»' ?: 



V'2 



• We do not any longer write ex and afterwards change to dx, 

 bat the reader will, of coarse, understand that dx is not really a 

 If K PAR is part of a rectangular hyperbola, A the vertex, C \ finite increment, like L M, but is what this increment becomes as 

 the centre, and CZ,CZ',the asymptotes; and if C L = z!, P L=i/, ' Q il, in its approach to P L, is jnst about to merge into P L. 



