By way of lemma, let us first prove that all sections of a spheroid, made by planes mutually parallel, are similar ellipses, which may be effected either by the formulæ in page 11, or more simply thus ;

Let P'p' be the line of intersection of the planes of PP'p with a circle Dd described by the revolution of any point D in the generating ellipse; then since DP'd and PPp are plane AbaB, lines Pp, Dd, and Dd is evidently the diameter of the Hence

PM' is circle.

P'M' DM' x M'd,

But by a well known property of the ellipse

DM' x M'd: PM' x Mp :: CB" : CB', or y: 2a, x, x,2 :: b2 : b′2.

putting CB', the semi-conjugate axis b.
y,, Pp =2a, and PM' = x,

PM'

.. the section PpP' is an ellipse whose axes are

b

are.. in the constant ratio of b: b', and it is evident that the axes of all sections parallel to PpP' are in the same ratio. Hence all sections parallel to one another are similar. (See Prob. 68).

Now, making CM = x, PM = y, and drawing CNPp, since we may conceive the volume BыAB to be generated by the parallel motion of the ellipse Bb, varying in magnitude but not in form, we shall have

Varea of the ellipse x d. CN.

But since the area of an ellipsex (the product of its diameters,) and that of BCbx bb', (as we learn from equation 1,) and.. the area of the similar ellipse Pp bb' ×

PM2

sin. C ƒ (a' — x2) dx

sin. C × (a3x − =2)

(2)

Let xa. Then, since ab a'b' sin. C,

which is known from other principles to be =—-- spheroid.

By this method may be found the volumes of all solids whose parallel sections are similar figures.

165. The base B of the solid may be decomposed into triangles, and consequently the whole volume into as many pyramids of the same altitude h, each of which being known to =

3

x altitude. Hence the volume required = Bxh,

Otherwise.

its base

As in the preceding problem it may be easily shewn that all sections parallel to the base B are similar to it, and also that (h, x being dist. of B, and B' from the given point) the volume corresponding to B' is expressed by

by

166.

The bases of the whole cone and part cut off, are πb2, πc2;

.. if h, h' be their altitudes, the frustum F will be expressed

Now the cylinder (V) of same base and altitude = xy'x =

.. V : V' :: 1 : 2r + 1; m ; 2n - m.

169.

Let PM (Fig. 59,) = Am = x and Pm = y, (Amt being the tangent parallel to the base TV) and AB the axis of the cycloid 2r &c. &c.

Then x = AQ + MQ = vers. y + √2ry → y3 the equation to the cycloid.

Hence the volume V generated by APm is expressed by

V = =ƒy'dx = = ƒ{

= Sydy 2ry - y3

=*rfdy√2ry — ya — « ƒ (r − y) dy √ 2ry — ya

-T

2x

(r−

3

2= • (2ry — y3)3.

Let x 2r. Then the whole volume described by 2ATt is

Again, the volume of the cylinder described by Vt is

V' = π (2r)2 × TV = 4′′r2 × 2r%

..the volume of the solid described by the cycloidal area VAT is expressed by

170.

Let a be the part of the axis of the cylinder intercepted by the parallel bases; these bases being equal, if d be the distance of their planes, it is evident that the volume comprised between them, is expressed by

V d x Base = d x B.

Now a is

=

circular ends, and d is elliptical ones, and they meet in the centre; .. if r be the radius of the circular ends, and = inclination of B to them, we shall have, by projections,

which being independent of the magnitude and inclination of the ends, gives the proof required.

=

Vfxy'dx = C — «b3x — ab3x l. (a —

x)

172.

Let r be the radius of the sphere, then the area of a

great circle is wr2, and the volume of a cone of that base and alti

Let A be any one of the equal angles of corresponding spherical equilateral A, of the circumscribing sphere whose radius is 1, then the surface of any one of them is expressed by

ЗА 2 right = 3A -T

and eight of those A cover the sphere which is 4 great circles, .. 8 × (3A — π) = 4%

Now a being a side of the spherical ▲, we have by the fundamental theorem of spherics