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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,
= sin. 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
(c + a - b). (c-a-)
THE CUBATURE OF SOLIDS.
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).
Let a, 6 be the axes of the ellipse, then
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 =
? (30 - x)
62 3 respectively. Let x = 2a in the former, and 2b in the latter, then the whole volumes are
and B =
Now when b = a, the spheroid will be a sphere S=
= 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
circumference of the circle described in the revolution by its cen
tre of gravity. But the radius of this circle =
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
:: V : V :: 4 : 2 :: 2 : 1.
y = x 2rx
1 Now the surface of a right cone = ? circumference of its base X slant side (as is known) :: S' = ar XrV2 + xr xr
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.