greatest attraction: viz. x x (*+y)== ̄ ̄ (2+y°*+f(x9= + Cf (x)• = 0. Prop. 36. To solve Prop. 32, the force being supposed inversely as the mth power of the distance, and the generating polygon being composed of triangles having such a law of density as that in the scholium to lemma 3. By using the value found in that scholium, and proceeding, in other respects, as in the similar propositions already given, we find, for the equation of the curve touching the sides of the polygon, x x (2*+9')** — (2*+ (1+r) ;)= + Crya = 0. (x2+ Prop. 37. Let Prop. 32 be yet once more resolved, on the supposition that the force is inversely as the mth power of the distance; and the density, in the triangles forming the generating polygon, either uniform, or as any function of x and T. If we make use of the first value of A in lemma 3, we get, for the equation of the curve touching the sides of the polygon, x 22 +(2-m) (4-m) 3(x2+32) (x2+y2) (x2+ (1+r2) y2)”—1 — { 1 —(2—m) rtyt 3.5 (x2+y2)2 &c.} + C = 0. When ro, or the polygon becomes a circle, this equation is reduced to + Co, as was found in another man x (x2+y2)′′±1 ner, in Cor. 1, Prop. 35. If r is finite, the above expression will terminate when m is a whole positive even number; and consequently the guiding curve will then be algebraic. But, if m be amongst the numbers 5, 7, 9, 11, &c., we must use the other expression found in the lemma, and there arises, for the guiding curve, the transcendent equation xq (x, m, y, ry) + Cry ≈ 0. If m = 1, the equation is x √x2+ y2 when m = 3, Ꮖ ry x2+3a × x2+ (1+r2) y2+(x2+) arc (tang. = (2 ry √x2 + y2 × arc (tang. = + Cry = 0; and, finally, ry + Cryo. √x2 + y2 In like manner, might be solved Prop. 31 and 34, the force and density being as in Lemma 3, but this I leave to the reader. Prop. 38. The force being inversely as the mth power of the distance (where m is any whole positive number), and the density either uniform or any function of x and T,* the base of the infinitely long cylinder of greatest attraction has, for its equation, * • = + C = 0; (x2+y2) 2 for it will appear from lemma 3, and its corollaries, that, whether m be odd or even (that is to say when it is any number in the series 1, 2, 3, 4, 5, &c. ), the attraction of an infinite cylinder will be of the form * What this means with respect to a cylinder, is shewn at the end of the scholium to Prop. 33; and with respect to a solid of revolution in Prop. 33. A = Dƒƒ xf (x, T) 'Î*, D being a function of m; 112 (x2+T2) hence the truth of the proposition is manifest. And because the equation of the curve generating the solid of revolution of greatest attraction (on the same hypotheses of force and density) has been shewn to be following remarkable x + Co, we have the (x2+ya+1 Theorem. m being any whole positive number, and the density either uniform or as any function of x and T, the same curve which, by revolving, generates the solid of revolution of greatest attraction, when the force is inversely as the mth power, shall be the base of the infinitely long cylinder of greatest attraction, when the force is inversely as the (m + 1th) power. Numberless other interesting questions might be proposed, relating to solids of greatest attraction; for instance, we may inquire what must be the curve bounding the base of a cylinder of given mass and length so that it shall exercise the greatest action in a direction parallel to its axis. But as this kind of inquiry proceeds exactly in the same way as the other (only we must use the attraction B, instead of A, in Prop. 1), it is unnecessary to lengthen a paper which has already been extended too far. APPENDIX TO §. III. Of the Attraction of an infinitely long Prism, whose Base is any right lined Figure whatever. Prop. A. Let the rectangle bb' c'c, fig. 20, be the section or base of a prism, infinitely extended on both sides of it, and let the line psu bisect the opposite sides bb', cc' of the rectangle. It is required to find the attraction of the infinitely long solid, on the point p, in the direction psu. Let C be the centre of the rectangle, put k = sC, a= น = PC; draw rm perpendicular to sCu, and put x= Cm. Now it appears, from Cor. 2, Prop. 1 of the paper (putting A for the required attraction) that A = 4/pm x arc (tang. = 4x arc (tang. = pm =4x arc (tang. ) — 4x arc (tang. = rm a u+x1 a 1+x) u + x 1. the last = axx (u+z)2 + a2; put u + x = z, x = ż, x = z which is 44L. (a° + ≈3) š term of which is 4 (az-au) ż 4u arc (tang.) so that a u x • "+*) + 4aL. a a (a2 + (u + x)o)§, or A = 4 (x + u) arc (tang. = 4.) 2π +4аL. (a2 + (u + x)2)1⁄2, which fluent being taken from x = — - k to xk gives A = 4 (u + k) arc (tang. = a „1—1) + 4aL. (a2 + (u + k2 a 144) — 4 (u — k) arc (tang. :) u + k I )1⁄2 — qaL. (aˆ† (u — ko)ž. If we choose to express this by the lines and angles of the figure (20), it is A = 4 × pux arc, cpu 4 Prop. B. Let the section of the prism be an isosceles triangle; the attracted point p being in the line psm (fig. 21), which passes through the vertex s to the middle of the base r'p'. Draw rm parallel to the base, and put r tang. rsm; call ps, u; sin, x; then rm = rx; and we have for the attraction of the infinite solid A = 4x arc (tang. (i 1 + 4. S+ u2 1+ u + x rx ruxx (u + x)2 + r2x2 ruxx 24 u + x x + x2 = 4 x ps x arc, bps + 4 x bs x L pc pb' the last term is Tx u+x 4 1+12 4 arc (tang. = − 4 ruxx 4S ruxx Fi <=%, x=Z-a, xż, and it becomes which fluent is rxx (u+x) — rr2¿ 4 ruxx น 1+r3 √ (x+a)2 + 122 if we put a = 17. Make, moreover x +α 2 4 1 + 14,{ruL.√z+ræ — u arc (tang. = 1)}, we have then at length rx u + x) A = 4x arc (tang. = 77 arc (tang. x+)} + cor. a γα Cor. If the position of the attracting solid be reversed, as in fig. 22, call ps, u, and the attraction will be given by the same formula; only the fluent (if it begin at the point) must now 4 ‚‡, { ruL √(x + x)' + r*a* — u |