Let C represent the circumference of the larger section and c represent the circumference of the

a form of expression for the required area, but one that involves q, a line which is not a part of the frustum.

In order to eliminate it from the expression, it is necessary to find relations between it and lines of the frustum.

Substituting this value of (C – c) q in the last obtained expres

sion for F, we have,

F= cs+ Cs or F = (C+ c) s.

:. F = } (2 π R + 2 π r) s = π (r + R) s.

2. Show that (c+ C) equals the circumference of a middle section, and that the area of the frustum may be described as the circumference of the middle section by the slant height: F = 2π

(r + R)s.

3. Find an algebraic expression (formula) for the frustum of a cone in terms of the altitude EG.

D

MA
B

The expression last deduced shows that the area generated by revolving a segment of a straight line about another straight line in the same plane, will generate a surface which would equal the convex area of a cylinder having for its radius the perpendicular from the middle point of the generating line to the axis and for its altitude the projection of the given segment on the axis.

G

E

Q

FIG. 319.

4. Show that if a regular polygon of an even number of sides be revolved about a diameter of the circumscribed circle, which is also a diagonal of the polygon, as an axis, the area generated will equal the convex surface of a right cylinder having for the radius of its base the apothem of the polygon, and the diameter of the circumscribed circle for its altitude.

E

Q

M

FIG. 320.

Or expressed differently: The area generated by the rotation of a regular polygon, as above described, will equal the convex area of a right cylinder, having for the diameter of the base the diam

eter of the inscribed circle, and for its altitude the diameter of the circumscribed circle.

138. The surface of a cone or of a frustum of a cone may be considered as generated by a circle, the plane of which moves parallel to itself and the radius of which varies as its distance from a given point. For if we have a right circular cone or frustum, sections perpendicular to the axis will produce circles, the radii of which will be to each other as the distances of the sections from the vertex.

The surface of a sphere or of a zone (§ 116) may be generated by moving a circle parallel to itself from A

Q

toward X, remaining perpendicular to AX, and so changing the radius that its square shall equal the product of AQ and QX.

For if we have a sphere, the radius of any plane section will be a mean proportional between the segments into which the diameter, perpendicular to the section, is separated by it.

THEOREM. If a chord of a given circle be a tangent to a concentric circle, and the extremities of the chord be joined to the centre, and the figure thus determined be revolved about a non-intersecting diameter, the surface generated by the exterior arc will be greater than the surface generated by the chord, which will be greater than the surface generated by the interior intercepted arc.

By § 103, BD > BD > KE.

D X

P

E

If any radius CP be drawn intersecting the inner circle at J, and the chord at T, P will be at a greater distance from the axis than T, and T will be at a greater distance than J. So that the circumference generated by P will be greater than that generated by T; which in turn will be greater than the circumference generated by J.

Revolved about AX, BD will gen

M

C

K

B

FIG. 324.

erate a zone, BD the frustum of a cone, and KÈ a zone.

These may also be generated, as seen in the early part of this article, by circumferences, the planes of which are perpendicular to the axis.

In the cases now under consideration, the larger circumferences will be moved the greater distance and so will generate the larger area.

The zone formed by the revolution of BD will be greater than the frustum formed by the revolution of BD, and that in turn will be greater than the zone formed by the revolution of KE.

Hence the theorem is established.

139. PROBLEM. To find the expression for the surface of a sphere in terms of the square on its radius.

As a consequence of the last article, if a regular polygon of an even number of sides, together with its circumscribed and inscribed circumferences, be revolved about a diagonal passing through a pair of opposite vertices, the surface generated by the polygon will be less than the surface of the circumscribed sphere and greater than the surface of the inscribed sphere.

In Prob. 4, § 137, it is shown that the area generated by the polygon equals the convex surface of a right cylinder, having for the diameter of its base the diameter of the inscribed circle, and for its altitude the diameter of the circumscribed circle.

If we should retain the same circumscribing circle and should double the number of sides of the polygon, the diameter of the inscribed circle would be increased, and the area generated by the polygon of the increased. number of sides would be equivalent to a cylinder having the same altitude as before, but having a greater diameter of base.

Each time the number of sides of the polygon is doubled, the diameter of the cylinder having the equiva