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Ex. 3. To determine the area contained between two successive convolutions of the spiral of Archimedes ; fig. 30.

Let the general form of the equation to the spiral be r = ad, o being the whole angle through which the radius vector has revolved; and let sa, sb, sc, severally be the values of the radius vector after n-1, n, and n+1 complete revolutions, so that SA = 2(n-1) , SB = 2n 7a, sc= 2(n+1)Q; let psa = 0; therefore SP = {2(n-1)+0}a, sp, = {2n7 +0}a, which values it is convenient to represent by r and rų; the problem is to determine the area of APB, BP, C, C, which is expressed by the following definite integral :

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1)3–2n3 + (n-1)} 873 = 8n73 as; therefore the area generated in the first revolution of the radius vector is 873 a”; and hence that generated in the nth revolution is n times that generated in the first.

230.] If the equation to the curve is given in terms of r and p, instead of finding the equivalent expression in terms of r and 0, and then integrating as in the preceding articles, it is more convenient to pursue the following course :

Let r dr do be integrated first in respect of r; and supposing the limits of r to be the radius vector of the curve and 0, we have, 0 and 0, being the limits of 0,

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(13)

But by (24), Art. 271, Vol. I, nado =

(p2-p2)+'

1 p rp dr .. the area = 2 Joo (op2p2)}" where r and r, are the radii vectores of the bounding curve corresponding to 0 and 0, respectively ; in which expression p must be replaced by its value in terms of r, and the r-integration then effected.

Ex. 1. To find the area contained between the involute of the circle and two limiting radii vectores ; see fig. 31.

The equation to the curve is ge - = a.

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Ex. 2. To find the area contained between an epicycloid and its base-circle during one revolution of the generating circle; see fig. 42, Vol. I. By (9), Art. 268, Vol. I, the equation to the curve is

(a +26)2

p= ani
therefore the area contained between the pole and the curve

pa+20 pr dr
do (po2 - p2)
a +26 a+20 7 (702 — a?) dr

a la {(a +26)2 — 7o2}}
a+2b pa+28 r(r2 – a2 + dr

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do

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= ** (a+b)(a +2b); and as the area of the circular sector which is included in the above expression is tab, the area included between the circle and the epicycloid = * * (3a+2b).

In reference to the expression for the area which is given in (12) it is to be observed, that the triangle osp, which is expressed by 90, is the 6-differential of the area. But the area of this small triangle, see fig. 36, = sx * PQ = pas. Hence me do = p ds; and if A is a sectorial area, A = [pds ;

(14) which is another expression for the area.

231.] The method of the present section is also applicable to the following problem :

To find the area contained between a curve, its evolute, and any two limiting radii of curvature.

In fig. 32 let OPQB be the plane curve on which p and Q are two consecutive points, p being (x, y), and PQ being an infinitesimal arc, and therefore equal to ds; let pn be the radius of curvature at P, and be denoted by p: then the area of the infinitesimal triangle png is equal to

(15) and as the required area is the sum of all these, we have

(16) in which p and ds must be expressed in terms of a single variable, the limits of integration being assigned by the conditions of the problem.

Ex. 1. To determine the area contained between the parabola, whose equation is y2 = 4 ax, its evolute, the radius of curvature at the vertex, and any other radius of curvature.

the area = pds,

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and therefore the area contained between the curve, the evolute, and the radii of curvature at the vertex and at the extremity of

56. the latus rectum is equal to , a.

Ex. 2. To find the area contained between the cycloid, its evolute, and two given radii of curvature. In fig. 32, let o, the starting point, be the origin; then

x = a versin-1% – (2ay-y2);
:: p= 2 (2ay)*, ds = (224,)'dy ;

po y dy .. the area beginning at o = 2a |

), (2 ay-y2):

= 2a{-(2 ay – yo)}+a versin-1}; and therefore the area o B’B = 21 a?.

SECTION 3.—The Quadrature of Surfaces of Revolution. 232.] In fig. 33, let APQ be a plane curve, and suppose it to generate a surface of revolution by revolving about a line ox in its own plane, a'r'e' being its position, when half a revolution has been performed; and let the equation to AP be y=f(x); let on = X, MP = y, PQ = ds; P and q will, in a complete revolution, describe circles whose radii are respectively y and y+dy, and therefore the paths traversed severally by P and Q are 2 and 27 (y+dy): supposing the curve to be continuous and to have no points of inflexion between P and Q, the element PQ will describe a circular band whose breadth is ds, and the circumferences of whose bounding circles are 2 ty and 27 (y+dy); the area therefore of the convex surface of this band is intermediate to 27 yds and 27 (y +dy) ds; and neglecting the infinitesimal of the second order, the convex surface of the infinitesimal band is equal to 2ny ds; and therefore, as it is an infinitesimal bandelement of the surface, the surface = /2ny ds,

(17) the integral being of course definite, and the limits being given by the conditions of the problem.

As y = f(x) is the equation to the generating plane curve, dy = f'(x) dx; and

ds = (dx2 + dy2)#

= {1+($'(x))2}* dx ; .:. the surface = 27 | f(x) {1+(f'(x))a}dx ; (18) which is the form convenient in most cases; other processes will be explained in the sequel. Ex. 1. To find the surface of a sphere.

The equation to the generating curve is x2 + y2 = a*, so that y ds = a dx.

.. the surface of the sphere = 470 / dx = 47 ao. Hence also a zone of a sphere contained between two planes perpendicular to the axis and at distances X, and X, from the centre is equal to

20a **dx = 210(—%); see Ex. 7, Art. 24, vol. I.

Ex. 2. To determine the surface of the paraboloid of revolution. Here let ya = 4ax; so that

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Ex. 3. To find the area of the surface described by the revolution of a cycloid about its base.

Here x = a (0— sin 0), y = a (1-сos 6); so that

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Ex. 4. To determine the area of the surface described by the revolution of the tractrix about the axis of x.

The differential equation to the tractrix is

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= 2 . Ex. 5. The convex surface of a cone, whose generating line is ay -bx = 0, is a b (a? +62).

Ex. 6. Find the equation to the plane curve which by its revolution about the x-axis generates a surface, the area of which is proportional to (1) the extreme abscissa ; (2) the extreme ordinate.

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