EXAMPLE I. Required the area of a parabola, the abfcifs being 2, and the ordinate, perpendicular to the abfcifs, 6. Here the altitude is 2, and the base or double ordinate is 12; therefore X 2 X 12 = 16 is the area. EXAMPLE II. If the bafe of a parabolic fegment be 12, and its abfcifs 2 make an angle of 30 degrees with it; what will be the area? The fine of 30° being half radius, the altitude will be 2 x = 1; and hence × 12 area. 2 43 8 is the PRO Again, it will appear that, if in the two fegments AEB, BDC, be inferibed triangles AEB, BDC, of the fame bafes and altitudes with them, then thefe laft triangles will be of their circumfcribed fegments, and, confequently, of the triangle ABC. And if, in like manner, in the last made fegments be infcribed the greatest triangles, they will be of the triangles immedi ately preceding them, or of the first triangle. And fo on continually. Confequently all the infcribed triangles, taken together, will be expreffed by the feries b x (1+ + + + &c); 4 42 43 44 I 4.4 I I I I where b denotes the first triangle; and which, when the feries is infinitely continued, will denote the area of the parabolie fegment. I Corol. 6. Hence the infinite feries 1 + + + &c, 2 43 will be 1, as we also know from other principles; and confequently the fum of any finite number of terms of the series ++ &c, is less than¦; and the sum of the infinite I PROBLEM VI. To find the Area of a Fruftum, or Zene of a Parabola, included by two Parallel Right Lines and the Intercepted Curves of the Parabola. To one of the parallel ends, add the quotient arifing from the divifion of the fquare of the other by the fum of the faid ends; multiply the fum by the altitude of the fruftum, or diftance of the ends; and of the product will be the area. T Or divide the difference of the cubes of the diameters, by the difference of their fquares; and multiply the quotient by of the altitude. By the property of the parabola, DD-dd: a :: DD-dd add dd: DD- dd the altitudes of two complete fegments whose bases are the ends E, d of the fruftum; and confequently their difference, or the fruftum, is a × = 3a × DD + Dd + dd D+d Corollary. Hence a parabolic fruftum is equal to a parabola of the fame altitude, and whose base is equal to one end of the fruftum, increased by a 3d proportional to the fum of the ends and the other. Therefore, in the one end DC, produced, of a parabolic If the two parallel ends of the fruftum of a parabola be 10 and 6, and the part of the abfcifs, perpendicular to, and connecting the middles of, thofe ends, be 4; what will be the area? Here the abfcifs being perpendicular to the ends, will be the altitude of the figure, and therefore If the part of the abfcifs, connecting the middles of the two ends, make with them an angle of 48° 35, required the area; the other dimenfions being as in the first example. The fine of 48° 351 being 75 or hence 4 × 3 = the altitude X 3/4 B b = 3 very nearly, of that in the laft parabolic fruftum ABCD, take CE = the other end AB; raise EF perpendicular to EC and CD; draw DF, and perpendicular to it FG meeting CE produced in G: Then if upon the base CG be defcribed any parabola CHG touching the line ABH, it will be equal to the fruftum ABCD. EF2 For CGCE+EGAB+EG = AB+ =AB+ DE DC2 DC + AB laft example, and therefore the area here must be 3 of that above; that is, 32 X = 98 4 == 24 the area required. To find the Area included by the Focal Distance, the Line drawn from the Focus to the Curve, and the Contained Arc of the Parabola. RULE I. Upon the axe, or focal distance AB, produced if neceffary, having demitted the ordinate or perpendicular CD, cutting off the abfcifs AD; then 13 To the focal diftance AB add of the abfcifs AD; multiply the fum by the ordinate CD; and half the product will be the area of the part ACB. That is, (AB+ AD) × 1 DC = the area.* EXAMPLE. If the abfcifs be 2, and the ordinate 6; required the area of ACB. By cafe 1 prob. 2, we have 2 : 6 :: 6 : 18 the parameter; and hence 4 AB. 18 Then (4÷ + 3) × 3 = 5% × 3 = 151 the area required. * DEMONSTRATION. RULE Putting AD, DC=y, and AB = a; ACB = ACD ± BCD will be y ¿y ( = x ± a) = 3xy − {xy + may = (x y (x + a) = DC X (AB + AD). 2. E. D. Corollary. The area cut off by Ac is = AD × ¿Dc. = 1375 Subtract the focal diftance, or distance between the focus and the beginning of the arc, from the distance between the focus and the end of the arc, and multiply the remainder by the faid focal dif tance; then multiply the root of the product by the fum of the distance between the end of the arc and focus, and double the focal distance; and of this product will be the area. AB That is, (CB BA) AB X (2AB+ BC) is the area ACB.* EXAMPLE. Taking here the fame example as before; we have 4, and BC AD + AB = 2 + 4 = 6; and hence 42 × 2 × 9 + 61 = 15% = the area as before. I PROBLEM VIII. To find the Curve Surface of a Paraboloid. RULE I. To the fquare of the ordinate, or femi-diameter of the bafe, add 4 times that of the axe; and the fquare root of the fum will be the tangent to the curve at the bafe, and intercepted by the axe produced; let this tangent be called t, viz. t = √yy + 4xx; x being the axe, and y the ordinate. B b 2 Then, By the nature of the parabola, z=CB is AB + AD = a + x, or x = za, alfo y = √ ax = √(z − a) a; which written for them in the laft rule, give values of x and ولا √ (≈ − a) a × (2a + z) = the area ACB. 2. E, D. |