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 = ap, 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)πа, Sв = 2 ñ ñа, sc = 2 (n+1)πa; let PSA = Ꮎ ; therefore SP = {2(n−1)π+0}α, sp1 = {2n+0}a, which values it is convenient to represent by r and r,; the problem is to determine the area of APB, BP, C, C, which is expressed by the following definite integral :

therefore the area generated in the first revolution of the radius vector is 83a2; 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 de 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, and 0, being the limits of 0,

where r and

.'. the area =

rp dr

(r2 —p2)

1 fr rp dr

2

(13)

ro 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 r2-p2 = a2.

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

therefore the area contained between the pole and the curve

and as the area of the circular sector which is included in the above expression is ab, the area included between the circle and

a

(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 ex

pressed by

r2 do
2

of this small triangle, see fig. 36, = 11⁄2 sx × PQ =

pds

SY X

Hence

2

2

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 OPQв 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 Pп be the radius of curvature at P, and be denoted by p: then the area of the infinitesimal triangle Pпq is equal to

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

(15)

(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 = 4ax, its evolute, the radius of curvature at the vertex, and any other radius of curvature.

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

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

y
-1
x = a versin — (2ay—y2)};

a

SECTION 3.-The Quadrature of Surfaces of Revolution.

π

232.] In fig. 33, let APQ be a plane curve, and suppose it to generate a surface of revolution by revolving about a line or in its own plane, A'P'q' being its position, when half a revolution has been performed; and let the equation to AP be y = f(x); let OM y, Oм = x, MP = у, 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 2πу and 2 (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 2πy and 2 (y+dy); the area therefore of the convex surface of this band is intermediate to 2 yds and 2 (y+dy) ds; and neglecting the infinitesimal of the second order, the convex surface of the infinitesimal band is equal to 2 yds; 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

.. the surface = 2πff(x) {1 + (f'(x) )2} * dx ;

[ƒ(x) {1+

(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 = a2, so that

y ds

= a dx.

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

2= a["dx = 2na(x„—x.) ;

2πα

see Ex. 7, Art. 24, vol. I.

Ex. 2. To determine the surface of the paraboloid of revolution. Here let y2 = 4ax; so that

Ex. 3. To find the area of the surface described by the revo

lution of a cycloid about its base.

Here x = a (0—sin 0), y = a (1— cos 0);

so that

= 32 π a"" {(cos) -1}d.cos

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

The differential equation to the tractrix is

Ex. 5. The convex surface of a cone, whose generating line is ay-bx = 0, is πb (a2+b2)3.

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 ordi

nate.