Sidebilder
PDF
ePub

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 Pp' 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 PP'p are I plane Abab, PM' is I lines Pp, Dd, and Dd is evidently the diameter of the circle. Hence

PM = DM' x M'd,
But by a well known property of the ellipse

DM' X M'd: PM' x M'p :: CB : CB',
or y' : 2a, x, – x;" :: 6 : 6'?.
putting CB', the semi-conjugate axis = b.
PM' = y, , Pp = 2a, and PM' = x,

6% :: y;

(2a, X, – 3,7) (1)

[ocr errors]

:: the section Ppp is an ellipse whose axes are

a, anda, and

aré :, in the constant ratio of 1:6', 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 CNI Pp, since we may conceive the volume BbAB to be generated by the parallel motion of the ellipse Bb, varying in magnitude but not in form, we shall have

V=s area of the ellipse x d. CN. But since the area of an ellipse = " X (the product of its diameters,) and that of BC6 = * X bb', (as we learn from equation

PM: 1,) and :, the area of the similar ellipse Pp = 766 x

7

[ocr errors]
[ocr errors]

(a” – 2),

abb abb' s (a" – x') dCN =

sin. C S (a" – x) dx or V = ob' sin. C * (dx )

(2)

V=

Let x = a. Then, since ab = a'l' sin. C,
27bb'a' sin. C 2bʻa

(3)
3

3

which is known from other principles to be = spheroid.

Hence V

mbb' sin. C V

X (2a's + x3 3a%2x) = PpA

3a? the volume required.

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 kpown to =

1 its base

3

x altitude. Hence the volume required = 1 B x h.

=

3

Otherwise. 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

'?B.x2

B.x3
dx =
ho

3ha
Bh3 Вh
:. V
3h

3

V' =

[ocr errors]

166. The bases of the whole cone and part cut off, are

76?, 7c ; :. if h, h' be their altitudes, the frustum F will be expressed by

7 (3* h con)
3

6
But hatli = h'.

F =

[merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][ocr errors]

167. Let the equation to the generating parabola be

y = mx. Then the frustum F will be expressed by

(2* — *'?) 2

F =

(4 + 2) (4. – X') generally.

2 Let *= 0, y = 5, y = 0 Then F =

πα ama ba + ca

(6 + c). 2

2

m

[merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][ocr errors][merged small]

- 9 = 1. 2

"
:: y = 2
the equation to the parabolic curve.
Hence V = wydx = the paraboloid.

2r +1 Now the cylinder (V) of same base and altitude = wyor = *+.

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

Tahti

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. Wy + very - ya the equation to the cycloid. Hence the volume V generated by APm is expressed by

= *
rydy y(r - y) dy

2ry - y
= *Sydy 2ry - y
=*rdy 2ry - ya — *f (r- y) dy v 2ry — yo
Ξ X area AMQ

V = «Sy'da

+

[ocr errors]

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

[merged small][ocr errors][merged small]

Again, the volume of the cylinder đescribed by Vt is

V' = (2r): x TV = 447% X 26"

= 87*p3. :: the volume of the solid described by the cycloidal area VAT is expressed by

V - V = 7°

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 I circular ends, and d is 1 elliptical ones, and they meet in the centre; :. if r be the radius of the circular ends, and 8 = inclination of B to them, we shall have, by projections,

B =

[ocr errors]

But it is also evident that d = a.cos. O,

:: V = ara,

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

[ocr errors][ocr errors][merged small][merged small][ocr errors][merged small][merged small]

abo
- 6+

a-X
we have
V= f wydt = C - box – ab’r l. (a — x)

1
= abr.l.

6*mx.

172. Letr be the radius of the sphere, then the area of a great circle is. ar?, and the volume of a cone of that base and alti

27
2r 376

4r3.

1 tude = 2r is op? X

3
3

2 3 (See 161).

1

the sphere.

173. The octagon is composed of 8 equilateral-triangular faces.

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

3A 2 right L = 3A and eight of those A cover the sphere which is = 4 great circles,

:. 8 (3A - *) = 47

:. A = in Now a being a side of the spherical A, we have by the fundamental theorem of spherics

cos.'a
Cos. A =
sin. a

2

2

COS. a

= COS.

1. cos. a = 0, or a =

« ForrigeFortsett »