the area of the polygon = n area of triangle oAc = N.AB.OB

Now when the number of the sides of the polygon is infinitely increased, the perimeter and the area coincide respectively with

π

the perimeter and the area of a circle; in which case becomes

n

an infinitesimal angle, and its sine must be replaced by the arc to the radius unity, and its cosine by unity. Hence we have

Ex. 6. To determine the convex surface and content of a right circular cone.

Let A, fig. 5, be the vertex of the cone, and c be the centre of its circular base; let AC = a, сBb, and let pq be a side of a regular polygon of n sides circumscribing the circular base, so that PCQ =

2π n

; whence it is plain that

Therefore the area of the triangle APQ AB X BP

therefore the whole convex area of the circumscribed pyramid of n sides, each of which is similar to APQ, = nb {a2+b2} + tan

n

But when n is infinitely increased, the surface of the pyramid coincides with the convex surface of the cone, and replacing

that is, the convex surface of a cone is equal to the product of the slant side by the semi-circumference of the base.

This proposition may also be arrived at by the following process. If the convex surface is developed into a plane surface, and it is clearly capable of such developement, it becomes a sector of a circle, of which the centre is the vertex of the cone,

the length of the arc is the circumference of the base of the cone, and the radius is the slant side of the cone. And as the sector may be resolved into infinitesimal triangles with a common vertex, and all of the same altitude; so the area of it is equal to half the rectangle of which two adjacent sides are the radius of the circle and the length of the containing arc: and thus as the containing arc = 2b, and the radius = (a2 + b2), the convex surface of the cone = πb (a2 + b2).

As the area of the base of the cone =

= πb2+πb {a2+b2}}.

πb2, the whole surface

Again, by Euclid XII, 7, the content of the pyramid APQC is one-third of that of the prism of the same altitude and the same base; and therefore

Let the number of the sides of the polygon circumscribing the base be infinitely increased, in which case the pyramid as

π

Now it is plain that ab2 is the content of a cylinder of altitude a described on the circular base whose area is b2; hence it follows, that the content of a cone is one-third of that of its circumscribed cylinder.

Ex. 7. To determine the surface and content of a sphere of given radius.

==

a,

Suppose the circle ACBD, fig. 6, to be a plane central section of the sphere, or to be that by the revolution of which, about its diameter BOA, the sphere is generated. Let the radius and be divided into n equal parts, each of which is therefore and suppose MN to be one of these parts, then

equal to

a

n

a

MN=

n

Draw the ordinates MP, NQ, including between them the arc PQ of the circle. Then, if PQ is infinitesimal, and which therefore must be considered straight, the zone of the sphere generated by the revolution of PQ about Ao is equal to a rectangle, of which one side is PQ and the other is the circular path described by P, as it revolves; that is,

the surface of the zone = PQ × 2 π × MP.

And since the surface of the whole sphere is equal to 2n times the surface of such a zone, we have

the surface of the sphere 4πа2,

=

and is equal therefore to four times the area of a great circle of the sphere.

Now suppose a cylinder to be described about the sphere, and to be of the same altitude as the sphere, and to be generated by the revolution of the rectangle AEFB about the diameter BA; and suppose two planes drawn through м and N, perpendicular to BOA, so as to intercept a zone of the cylinder whose height is P'Q' or MN: then, since a is the radius of the cylinder, the surface of the intercepted zone=2πax MN; but from above,

a

corresponding surface of zone of sphere =2πα =2παχ Μ Ν.

n

Hence, if a cylinder be described about a given sphere, and if the surface of the cylinder be divided into zones by planes perpendicular to its axis, the surfaces of the intercepted zones of the cylinder and the sphere are equal; and therefore,

The whole surface of the sphere is equal to the convex surface of its circumscribed cylinder.

Again, to determine the content of a sphere.

Let a polyhedron be described about the sphere; let s = the area of a face, and a = the radius of the sphere; then the content of the pyramid, whose vertex is at the centre of the sphere and

as

3

base is s, is : and a similar expression is true of each face; let the number of faces be infinitely increased, and let each face become infinitesimal; then the polyhedron ultimately coincides

with the sphere, and the sum of the s's becomes the surface of the sphere, and is therefore equal to 4a2; therefore the conΑπα == 3

tent of the sphere

Ex. 8. To determine the area of a parabola.

Let OPQBA, fig. 7, be an area bounded by the parabola OPB, the axis oa, and the ordinate AB which is parallel to the tangent at the vertex o.

Let AB be divided into n equal parts, and through the several points of division let lines be drawn parallel to oa; let K and L be respectively the rth and the (r+1)th points of division, and

b

therefore KL = ; then, if n is taken very large, the area of

PQLK = KLX KP.

But

n

and as this expression is true of every slice similar to PQKL, the whole area is the sum of all such, which are obtained by giving tor successively the values 1, 2, 3, ... (n − 1), n; therefore

PRICE, VOL. I.

H

Hence the area of such a parabola is equal to two-thirds of that of the circumscribed rectangle.

This last example, however, more properly belongs to the Integral Calculus.

The student will find no difficulty in applying the infinitesimal method to the proof of the following problems:

(1) The tangent of an ellipse makes equal angles with the focal distances.

(2) The tangent of a parabola makes equal angles with the focal distance and a line through the point of contact and parallel to the axis.

(3) The curve which cuts a series of confocal ellipses at right angles is a confocal hyperbola.

(4) The area of an ellipse is to that of a circle whose radius is the semi-axis major of the ellipse as the axis minor is to the axis major.

Many other problems will be introduced hereafter.