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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 rap, 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)πα, Sв = 2n па, SC = 2 (n+1) π а; let PSA = 0; therefore SP = {2(n−1)π+0}a, SР1 = {2nπ+0}a, which values it is convenient to represent by r and r1; the problem is to determine the area of APB, BP, C, C, which is expressed by the following definite integral :

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therefore the area generated in the first revolution of the radius vector is 83 a2; 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,

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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 r2-p2 = a2.

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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

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therefore the area contained between the pole and the curve

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

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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

r2 do pressed by > 2

is the e-differential of the area.

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But the area

1

=

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r2de = pds; and if A is a sectorial area,

Α

A =

1Jpds;

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.

Р

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 PпQ is equal to

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

(15)

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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.

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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

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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

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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'r'q' being its position, when half a revolution has been performed; and let the equation to AP be y = f(x); let OM = ∞, 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 2y 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 27 (y+dy); the area therefore of the convex surface of this band is intermediate to 2 yds and 2π(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 =√
= √ 2 ny ds,

π

(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

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the surface = 2′′ ƒƒ(x) {1+ (ƒ′(x))2}1 dx ;

which is the form convenient in most cases; other

be explained in the sequel.

Ex. 1. To find the surface of a sphere.

(18)

processes will

The equation to the generating curve is 2+ y2 = a2, so that

y ds = a dx.

a

= Απα

dx = 4π a2.

.. the surface of the sphere Hence also a zone of a sphere contained between two planes perpendicular to the axis and at distances x, and a from the 0 centre is equal to

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Ex. 2. To determine the surface of the paraboloid of revolution.

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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—cos 0);

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so that

= 32 = a2 * {(cos)2 - 1} d. cos

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πα

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3 πa2.

0

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2

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

dy
dx

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.. the whole surface = = [2 my ds

- a dy;

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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 ordinate.

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