Page:Encyclopædia Britannica, Ninth Edition, v. 16.djvu/30

Rh 20 MENSURATION W_ = _ ^. , and therefore (fig. 23) dx a-y /&quot; ( a- - e-x- ) i a 2 - arc of quadrant AB = / &amp;lt; ^T^-T ( ax &amp;gt; where c-=-- -5 This integral may be shown to be equal to the series a rapidly converging series when c is a small fraction. By taking more and more terms of the above series we can B_ approximate as nearly as we please to the circumference of an ellipse. For example, we have quadrant AB to a first approximation ; hence whole circumference Fig. 23 51. Area of an Ellipse. We have at once / j /-a _ area = 4/ ydxi = / Va a - a; -c&e . Jo aJo / _ . But / Va?-x 2 dx is the area of the quadrant of a circle of radius a. Thus area of ellipse = 4 -- ( 44) a 4 The following proof is worth the reader s attention. By a well- known theorem in conic sections the orthogonal projection of a circle on a given plane is an ellipse. Now, if A denote the area of any plane figure, A the area of the projected figure, and 8 the angle between the planes it can easily be shown, by dividing the two areas by planes indefinitely near to each other and perpendicular to the common section of the planes, that A cos 9 = A. In the case of the circle and ellipse ira 2 and cos0 = ; a hence area of ellipse = ira 2 x = -nab. Co 52. Area of an Ellipse in terms of a Pair of Conjugate Diame ters. Let a and b denote the semiconjugate diameters, and a the angle between them, then by an elementary property of the ellipse ab = a b sina; hence area of ellipse = ira b sin a. D. The Hyperbola. 53. Area of a Segment of an Hyperbola. The equation of an y? 7/ 2 hyperbola being - - = 1, we have Of 0&quot;* y = Vx 2 - 2 ; hence (fig. 24) I fx _ area of the segment PAP = 2 - / V# 2 - a 2 dx 54. Area of a Sector of an Hyperbola. The sector PAP C is equal to triangle POP - segment PAP 55. Area of a Zone of an Hyperbola. In fig. 24 the zone PP Q Q = segment QAQ - segment PAP - av.y a - Sift - ab log/ aij * + fi ], where ay 1 + bx 1 J x v y l and x z, y z are the coordinates of P and Q respectively. If the axes of coordinates be inclined at an angle a, we multiply the above results by sin a to obtain the correct areas. Fig. 24. Fig. 25. 56. Area bounded by an Hyperbola and its Asymptotes. The equation of an hyperbola referred to its asymptotes is of the form xy=c 2. Let CM (fig. 25) = ^, CM = x. 2, Q M = ?/i.Q M = j/ 2) then, if o be the angle between the asymptotes, /* X area of QMM Q =/ l ysinadsK fx l dx . , . , / a-j . . / 2/2 = C 2 sma/ = c 2 smologJ =c-snialog, ^^ Jx-i x x-2/  y C , V XT = and x. 2 = . 2/i 2/2 and 2ab Now c 2 = - (a) a Again, let MM = a 1 -a 2 =| , then c 2 =^i2/i = ^ . . . , sin = rr, , and therefore cp + tr , therefore 2/2 - l/i Again, since ^X 1 y l sin o = Jc 2 sin o = x. 2 y. 2 sin a , we have triangle QCM = Q CM , and hence the sector QCQ = QMM Q . The corresponding results for a rectangular hyperbola are ob tained by substituting in the above formulae a? for c 2 and 1 foi SECTION III. PLANE IRKEGTILAR, KECTILINEAL, AND CURVI- LINEAL FIGURES. A. Irregular Rectilineal Figures. 57. The area of any irregular polygon can be found by dividing it into triangles, trapeziums, &c. , in the most convenient manner, and adding together all the areas. For example, ABCDEF (fig. 26) FC G Fig. 26. Fig. 27. It may sometimes happen that some of the component figures have to be subtracted instead of added ; for example, ABODE (fig. 27) = AFHE + BCG-AFGB-EDH. 58. Again, the irregular rectilineal figure PjP.,. . . P ? P 9 (fig. 28) can be broken up into a series of triangles and trapeziums as shown in the figure, and hence its area can be found. 59. A figure made up of straight lines may be measured by cutting it up into triangles by lines drawn from some one vertex to the others. For example (fig. 29), ABCDEF = ABC 4- ACD + ADE + AEF.
 * /l e V 3 ^ l - B - 5 t - &c.),
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