1 . A ACB = x A 6Cc. sin. A But for the same prism, the section I axis 6Cc is evidently constant. This proof which only applies to triangular prisms, may be extended to those of whose ends are any polygons whatever, (and :: to cylinders), by dividing them into their component triangular prisms. 158. Let the radius of the circle = 1. Then, by the question y = vers. = 1 cos I, the equation to the curve. :: Sydx = 1 - sin. x, the area required. Again, dy = sin. 2 dx? (2n 2 2 or when y = 1, which values of the co-ordinates give the positions of the several points of contrary flexure. 159. By Problem 132, Vol. II., the area (T) of a plane S A expressed in terms of its sides a, b, c and semi-perimeter is 2 T= » S S (c + a - b). (c-a-) -a)(9-0).(-o) (-0 THE CUBATURE OF SOLIDS. 0000 160. Every Triangular Prism is decomposable into three equal Pyramids, of bases and altitudes, equal to those of the prism, and the prism = its base x its altitude, :, any triangular its base x its altitude pyramid = 3 If the base of the pyramid be a polygon, it may be decomposed into triangular pyramids of the same altitude, the sum of the bases being = that of the whole pyramid. Hence the volume of any pyramid whatever = (base x altitude). 3 ) 161. Let a, 6 be the axes of the ellipse, then 6% al y? = (262 ) 62 are equations to the ellipse, in the former of which x is measured along a from its origin in the curve ; in the latter x is measured along 6 from its extremity. Hence the volumes of the solids (A, B, generated about 2a, 2b, are 62 A = S nyodx = x3a ? (30 - x) a2 t 62 3 respectively. Let x = 2a in the former, and 2b in the latter, then the whole volumes are 4ab2x A= and B = Now when b = a, the spheroid will be a sphere S= 403 3 x 3 162*622 = B2 9 3 :. A:B :: B : S. Q. E. D. 162. Let the tangent parallel to the diameter of the semicircle be the line abscissæ, and let the abscissæ x originate at the point of contact, then by property of the o, we have (r - y) = m - x2 • s ro — x2 - x2 :: the volume V, generated between the convex part of the O, and the tangent, is expressed by Let x =r; then the whole exterior volume 2V = (10 — 37), And subtracting this from the cylinder described by the diameter about the tangent which = 772 X 2r = 2r37, we get the volume V', generated by the 0, expressed by Otherwise, circumference of the circle described in the revolution by its cen 3п tre of gravity. But the radius of this circle = 3 163. Letr be the radius of the circle, then by prob. 161, 4713 the volume of the spheroid V, = and that described by the A V' will evidently consist of two cones of the same base are and r Hence 3 :: V : V :: 4 : 2 :: 2 : 1. y = x 2rx dy? = 477x d.ro 1 Now the surface of a right cone = ? circumference of its base X slant side (as is known) :: S' = ar XrV2 + xr xr ana 2 164. Let Aa = 2a' (Fig. 68,) be the given diameter, and Bb = 26' its conjugate, then PM being any ordinate, if through it a plane I plane ABC be made to pass, cutting from the spheroid the segment Ap PP', it is required to find the volume of that segment. |