EXAMPLES. 1. What is the solidity of the paraboloid ADB, whose height Dm is 84, and the diameter BA of its circular base 48? Here 482.7854 × 42 (= Dm) : 1809.5616 × 42-76001.5872-solidity required. 2. What is the solidity of a paraboloid, whose height is 60, and the diameter of its circular base 100? Ans. 235620. 3. Required the solidity of a paraboloid conoid, whose height is 30, and the diameter of its base 40? Ans. 18849.6. 4. Required the solidity of a paraboloid conoid, whose height is 50, and the diameter of its base 100? Ans. 196350. PROBLEM XX. To find the solidity of the frustrum of a paraboloid, when its ends are perpendicular to the axe of the solid. RULE.* Multiply the sum of the squares of the diameters of the two ends by the height of the frustrum, and the product again by .3927, and it will give the solidity. * Demon. The segment whose base is B, and altitude A, is AB, and that whose base is b and altitude a, is=ab, by the last problem: wherefore the frustrum, or the difference of the seg EXAMPLES. 1. Required the solidity of the parabolic frustrum ABCd, the diameter AB of the greater end being 58, that of the less end dc 30, and the height no 18. Here (582+302) × 18x.3927=(3364+900) × 18×.3927 =4264 × 18x.3927=76752 × .3927=30140.5104-solidity required. 2. What is the solidity of the frustrum of a parabolic conoid, the diameter of the greater end being 60, that of the less end 48, and the distance of the ends 18? PROBLEM XXI. Ans. 41733.0144. To find the solidity of an hyperboloid. RULE.* To the square of the radius of the base add the square of the middle diameter between the base and the vertex; and this sum multiplied by the altitude, and the product again by .5236, will give the solidity. ment is AB-ab. But B- -b : ▲—a (d) :: B: A= and B-- -b:d :: b bd B-b Bd B-d' : α= by the nature of the paraboloid; and these values of a and a being substituted for them will make AB- ab=· dB2-db2 2B--2b same as the rule. Q. E. D. =d× (B+b) which is the * Demon. Let t=transverse, and c=conjugate diameter of the generating hyperbola, p=3.1416, y, x, the or EXAMPLES. 1. In the hyperboloid ACB, the altitude Cr is 10, the radius Ar of the base 12, and the middle diameter nm 15.8745; what is the solidity? dinates, or semi-diameters of the ends of any frustrum of the hyperboloid, x=its altitude, and a distance of the less ordinate y from the vertex of the whole solid. the fluxion of the solid = pv2; =pc2; × = x2, we shall have At + A2+2ax + t2 But to convert this into the rules given in the text, let D, 8, d, be the greatest, middle, and least diameters, x=abscissa whose ordinate is d, and a=altitude. Then we shall have these three equations : t22=c2xt+xx x t2d2=c2×t+x—ža× x▬▬‡a From the sum of the two latter of which subtract the double of the former, and there will result × D2—252 +ď2 a2c2: and hence a2c2 2D2-462 + 2ď2 3 D2+48+ da substituted for it in the theorem above will give 6 xap for the content of the frustrum; which is the same as the following rule given in the text. And if d the least diameter be supposed to become in Here 15.87452+122 x 10 x .5236=251.99975+144× 10x.5236 = 395.99975 × 10 × .5236=3959.9975 × 5236 =2073.454691-solidity required. 2. In an hyperboloid the altitude is 50, the radius of the base 52, and the middle diameter 68; what is the solidity? Ans. 191847. PROBLEM XXII. To find the solidity of the frustrum of an hyperbolic conoid. RULE.* Add together the squares of the greatest and least semidiameters, and the square of the whole diameter in the middle, then this sum being multiplied by the altitude, and the product again by .5236, will give the solidity. finitely little, or nothing, the rule will become, ×ap=D2+452 ×a×.5236. Q. E. D. D2+452 6 * The demonstration of this rule is contained in that of the last problem. Or, if D=middle diameter, m=that at of the length, s=generating area of the hyperbola, L = length of the spindle, and p=3.1416. 4m2-3D2 3s Then will (D2+ 4m-3D L X —D)× PL = solidity of the spindle. And if the generating hyperbola be equilate ral, then will (38 X LD —L2)× дPL=solidity of the spin dle. EXAMPLES. 1. In the hyperbolic frustrum ADCB, the length rs is 20, the diameter AB of the greater end 32, that DC of the less end 24, and the middle diameter nm 28.1708; required the solidity. Here (162+122+28.17082) × 20 × .52359= (256+144 +793.5939) × 20 × .52359=1193.5939 × 20 × .52359= 23871.878.52359=12499.07660202=solidity required. 2. Required the solidity of the frustrum of an hyperbolic conoid, the height being 12, the greatest diameter 10, the least diameter 6, and the middle diameter 81. Ans. 667.59. And, if l-length of the frustrum, s=generating area, and the other letters as before; then will (2D2 +ď2+ 4m2-3D2-d2 3s 12 4m_3D_dxT+d—D)× pl=solidity of the middle frustrum of an hyperbolic spindle. But if the generating hyperbola be equilateral, the frus3s 12+D2-d2 D-d trum will be =(3d2—l2+ī X 1 6 pl. Note. The content of any spindle formed by the revolution of a conic section about its axis may be found by the following rule: Add together the squares of the greatest and least diameters, and the square of double the diameter in the middle between the two, and this sum multiplied by the length and the product again by .1309 will give the solidity. And the rule will never deviate much from the truth when the figure revolves about any other line which is not the axis. |