But for the same prism, the section axis bCc is evidently

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.

T

or when y

1, which values of the co-ordinates give the posi

tions of the several points of contrary flexure.

159. By Problem 132, Vol. II., the area (T) of a plane A expressed in terms of its sides a, b, c and semi-perimeter, is

But when the right-angled (c being the hypothenuse),"

-b)

( − a). ( — − b . ) = (c + a − b) . (c — a − b)

— c2 —(a−b)° = c2 —a2 + b2 + 2ab

S

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 × altitude).

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 b from its extremity.

Hence the volumes of the solids (A, B,) generated about 2a, 2b, are

A = ƒ xy'dx =

x2 (3a - x)

= Sxy'dx =

b2 3

respectively. Let x = 2a in the former, and 26 in the latter,

Now when ba, the spheroid will be a sphere S =

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

.. the volume V, generated between the convex part of the

O, and the tangent, is expressed by

Let xr; then the whole exterior volume 2V =

(10 -3%),

And subtracting this from the cylinder described by the diameter about the tangent which ar2 × 2r = 2r3, we get the volume V', generated by the, expressed by

By Guldin's Theorem, the volume V' area of the O X

circumference of the circle described in the revolution by its cen

tre of gravity. But the radius of this circle =

163.

Let r be the radius of the circle, then by prob. 161,

▲ V' will evidently consist of two cones of the same base πr2 and

altitude r, each of which is known to equal r3 × —-- Hence

.. V: V' :: 4 : 2 :: 2 : 1.

Again, to compare their surfaces S, S', we have

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 plane ABC be made to pass, cutting from the spheroid the segment Ap PP', it is required to find the volume of that segment.