EXAMPLE. If an oblique parabola, whose base is inclined to its diameter in an angle of 30 degrees, be turned about its base; required the value of the solid generated by its revolution; the base of the parabola, or length of the spindle, being 80, and the diameter from the vertex to the base 32. Here AB = 80, DP = 32, and the angle DPC = 30°; then, as radius = 1:4 = sine LP or 30° :: PD = 32 : DC = 16 the radius of the greatest circle of the spindle. CD XAF X FB bz % = 20 X bb :: PC = V PD? – DC? = d: AP2 GH X PC FG = CD 4pc?z? = 4d x ; hence AG = AF + FG = -bb bb + 4b1-4d2 2 X ; and the fluxion p X GH? X AG of the solid bb 6b+ 4bd-8d2 will be pz X 400 Х x (b% - zz)2 = 76 x [63 x (6+4d) - 2627 x (b+82) + bz? x (6 + 20d) - 8d23]; the fluent of which, viz. 4pc?z3 bo [163 x (b + 4.d) - 1627 x (b +81) + bz? X (6 + 20d) - ] will be a general expression for the value of the segment; which, when z becomes = b, will be it pccb= 'snbcc for the content of the whole folid ADBEA. 2. E. D. х Therefore 32' x 80 x 35 x •785398 = •785398 X 43690} = 343 14'5693576, as in the example to prob. 15. PROBLEM XX. To find the Value of an Universal Paraboloid, or a Body Generated by the Revolution of a Parabola about Any Diameter. a С B II D Suppose ab to be the diameter about which the parabola cac revolves, cad a vertical section of the solid generated by ac, and dg agc that of the folid generated by AGC; also let gih be perpendicular, and Frif parallel to cc. Then, by the property of the parabola, CH = b : CH - FI = CH” (FE F EI)' = bb - (y = d)? :: GH = a : EB = 2y x (v F d) : * = x hence the fluxion of the solid 2cyyx will become 2 acy?j x 97,d; and the correct fluent gives (b + d)3 x (36 = d) - 33 x (3y = 4d) . bb ac X 6bb 666*[[b+d)(36=d)-(by*+d)(367°d] for the value of the solid required; in which the upper figns respect the folid dkfc generated by cf, and the under ones the folid DKFC generated by cf. Corol, I CE X BC Corol, 1. When E arrives at A, then y = 0, and 3b +d the expression becomes ac x (1 = d) x 6bb CB + 2CH с ХGH xсв3 x =, by substituting for Gh 6сн ch? X EA BC + 2CH its value сх св* XAB x for 6BC the solid CAD. And if cm be an ordinate to the diameter AB; we shall have BC : BC :: AB : AM ; BC + 2CH hence AB 'X 2 AB + AM will be = and 6вс consequently the above value of the solid can be 2AB + AM comes c X CB? X 6 AB X BC В с 6 64 – yt Corol. 2. When B coincides with h, d will be = 0, and then the rule becomes ac x =CXHI 2bb X (6 b + yy) for the frustum cffc generated about Hi, as in problem 12. Or cx ch’xLgh for the whole paraboloid, as in prob. 11. THE INVESTIGATION OTHERWISE. Supposing Fl to be an ordinate to the diameter AB, whose parameter is P; and putting m and n for the fine and cosine of the angle L, to the radius i; we shall have FL =VP X AL = V px, where x is now = Al; then fe = m Vpx, and LE = npx; hence AE = AL+LE = x+nvpx. Then the fluxion of the solid c x FE’ x Ae is cmopx x (x + Inxv); whose fluent gives cm°px x (x + invpx) = *cyy X (3x + 2n7px) = < X FE X (2 AE + AL) for the value of the folid AFK, generated by the revo. lution of af. And jc X CEP X (2AB + AM) for the solid CAD, as above. Scholium. Scholium. It is evident that the surface of this folid might be found by the method by which we determined that of an elliptic spindle ; and the surface of the parabolic spindle might easily be determined also, if there appeared any occasion for it; but not by the same method. DEFINITIONS. what hath been said of the hyperbola among the conic sections in general, may be added the following observations. To If avb touch the hyperbola in the vertex v, and be equal to the conjugate axe, viz. Av and ve each equal to half the conjugate 'axe; and if from the center center c, through the extremities A, B, right lines CA, CB, be drawn; those lines will be what are called asymptotes to the hyperbola ; that is, they are lines which continually approach to the curve. And, like as the conjugate axe is a tangent to the curve at the vertex of the transverse, and bounded by the asymptotes; so, if to the extremity any diameter dd, be drawn a tangent adb, and terminated by the asymptotes, then ab will be the conjugate to do. When the asymptotes form a right angle at the center, the hyperbola is said to be right-angled ; as also equilateral, because its axes are then equal to each other. D, of 1 PROBLEM I. To Construet an Hyperbola, having given the Transverse and Conjugate Axes AB and 2BC. The semi-conjugate BC being erected perpendicular to AB, and D being the middle of AB, or the center of the hy E f A D perbola; with the center D and radius Dc, describe an arc meeting AB, both ways produced, in f and f; which will be the two foci. Then, assuming several points E in the transverse produced, with the radii AE, BE, and centers f, F, describe arcs intersecting in the several points G; through all which points draw the hyperbolic curve. Dd PRO |