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Ex. 5. The equation to the evolute of the ellipse is

(a x) + (by)} = (a62)}; find the area enclosed by the curve.

Let ax = (a? 62) (cosp), by = (a? 62) (sin 0)3; then

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225.] It is sometimes convenient to refer the area, which is to be determined, and its bounding curve, to a system of oblique coordinate axes; say, to a system whose angle of ordination is w; then the area of the surface-element is dy dx sin w, and if a represents the required area, sin w dy dx ;


the integral being definite, and the limits being given by the circumstances of the problem.

Ex. 1. The sides of a triangle are x = 0, y=0,- + = 1, and the angle of ordination is w: find the area of the triangle.

Let (a —) = , then

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Ex. 2. The equation to an ellipse referred to a system of oblique axes, whose angle of ordination is w, is find the area of the ellipse.

Let y = 4 (0,2–22)}; then
the area = ["S*sin w dy dx

= n a, b, sin w. On comparing this with the value found in Ex. 3, Art. 220, it appears that a, b, sin w = ab.

Ex. 3. Find the area of a parallelogram whose sides are 2 a and 2b, and which are inclined to each other at an angle w;

the area = lo losin w dy dx = 4 ab sin w.

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SECTION 2.-Quadrature of Plane Surfaces.Polar Coordinates.

226.] When the bounding curve of the surface, whose area is required, is referred to polar coordinates, it is obviously convenient to refer also to the same system of coordinates every point in the surface, and to express in terms of them the surface-element which abuts at that point. Now this latter expression may be determined by the method of transformation of definite integrals, which has been explained in the preceding chapter ; and indeed we have therein proved, see (32), Art. 213, that if x=rcos , y=rsin 0, dx dy = r dr dd; and consequently if a denotes the required area, A = || rdr do ;

(8) this integral being of course definite, and the limits being given by the circumstances of the problem.

It is desirable however also to investigate the preceding expression on independent geometrical principles.

Now the problem is to find the area of the plane surface contained between a plane curve and two radii vectores separated by a finite angle; see fig. 26.

Let AP, PQP, be the curve whose equation referred to polar coordinates is r=f(0);

(9) and let it be required to determine the area of P, S Pn.

Let SP, = ro, SP, = rn, P, sa = 0,, P, SA = 0n; and let E be any point within the bounding lines ; draw through E the radius vector SEP, and also a consecutive one inclined to sp at an infinitesimal angle do; from s as centre and with se as radius draw the small circular arc Eg, and also another arc at an infinitesimal distance from it: then, if the polar coordinates to E are r and 0, EF = dr, Eg = r do, and the area-element = rdr do, the element being ultimately an infinitesimal rectangle; thus the integral of r dr do, with respect to r, between the limits 0 and f (), will give the area of the triangular slice SPQ, and the integral of all such triangular slices between 0, and 0, will give the required area. Consequently we have the following double integral; the area P., SP, = "/"rdrdo; (10)

Jeg vo which is the same as (8). If the r-integration is effected first, the superior limit being f(0) or the radius vector of the curve, and the inferior limit being 0,


the area P,SP= /* {f(0)do ; (11) and replacing f (0) by its value r given in (9), r referring to the curve,

1 or the area S Po Pin = 21 The geometrical meaning of which is that the r-integration gives the area of the sectorial slice sPQ, which is manifestly equal to are do; and the whole required area is equal to the integral of this quantity.

And let it not be supposed that any inaccuracy of result arises from the fact that the element of the area is not rectangular at the superior limit of the r-integration, that is, at the point P: for if two infinitesimal arcs RP, QT are described from s as a centre with radii sp and sq respectively, then, if sp = r = f(0), SQ=r+dr = f(@+do), the area s PQ is intermediate to spr and SQT; that is, is intermediate toa- and "

(r+dr)? do

-, the difference between which is an infinitesimal of the second order, and

pa do . must therefore be neglected. Hence is the correct expression for the infinitesimal sectorial area.

227.] Examples illustrative of the preceding. Ex. 1. To find the area of a sector of a circle. See fig. 27.

Let the radius of the circle = a, and the arc AB subtend at the centre an angle a; then

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Ex. 2. To find the area of a portion of a circle cut off by equal chords drawn through a point in its circumference. See fig. 28.

Let the radius of circle = a; and let BSA = B ́sa = a; the equation to the circle is r = 2 a cos 0. ... the area BSB' = 2 x the area BSA

2 a cose
= 217**erdr de

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Ex. 3. To find the area of a loop of the lemniscata whose equation is p2 = ao cos 20. The area of a loop = 2 /* /a 10m2 < r dr de

= a* / *cos 20 do = Ex. 4. To find the area of the loop of the folium of Descartes; see fig. 63, Art. 260, Ex. 10, Vol. I. As the equation referred to rectangular coordinates is

23 — 3 axy + y3 = 0;
. 3 a sin 0 cos 3a tan 0 sec .

(sin 0)3 + (cos 0)3 = 1 +(tan 6)3 ? let this value of r, which is the superior limit of the first integration, be represented by r; then the area = 2 /*/*r dr de

1"(tan 6)2 (sec 6)? do Jo {1+(tan 6)3}?

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Ex. 5. If the equation of the cardioid is r = a (1+cos 6), the

3 7 a whole area = 2:

Ex. 6. If the equation to a curve is r = a sin 30, the area of each loop is ma*; and the area of all the loops = ***

Ex. 7. If the latus rectum of a parabola = 4a, the area contained between two focal radii inclined at 0, and 0, to the least distance =

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228.] In all the preceding examples the r-integration has preceded the 6-integration; the effect of which order has been that the area is resolved into triangular elements with a common vertex at the pole s; and the sum of these is determined by the O-integration. Now the areas, which are ordinarily subjects of investigation, admit of this resolution : but if the 0-integration had been first effected, r being constant, it would have determined the area of a circular annulus, the radii to whose bounding circles would have been respectively r and r+ dr, and the subsequent r-integration would bave given the sum of all similar annuli; but the areas, which are commonly the subjects of these processes, do not conveniently admit of such a resolution, and the equations of the bounding curves do not commonly give convenient values of limits; and therefore, although theoretically the order of integration is indifferent, yet we choose that which is practically most convenient, and make the r-integration precede the 6-integration. The circle, I would observe, when the centre is the pole, is adapted to both orders with the same facility, because the limits of the two integrations are constant.

229.] We proceed to the investigation of areas whose limits are of a more complex character than those considered above.

Ex. 1. To find the area of a circular annulus, the radii of whose exterior and interior bounding circles are a and b.

2 pa The area of the annulus = Il rdr do

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(a? 6%) do = (a62). Ex. 2. To find the area contained between the conchoid of Nicomedes, its asymptote, and two given radii vectores ; fig. 29.

Let sa = a, AB = PQ = b; SP = r, BSP = 0; therefore the equation to the curve is r = a sec 0 +b; also sQ = a sec 8. Let 0, and 0, be the superior and inferior limits of 0; then

for fasec 8 +6 the area =

- r dr do

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