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Cor. 3.-The lateral surface of the frustum of a cone is equal to half the sum of the circumferences of the ends multiplied by the slant height of the frustum.




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The surface of the frustum TBA is equal to the difference between the surfaces of the two cones SAB and SVT.

Let R and r denote the radii OB and QT, and let BT = h.

To find ST, the slant height of the smaller cone, let ST = x;

.. Rrh+x:x;

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But R denotes half the circumference of one end, and * denotes half the circumference of other end.

the rule.


If to this we add R2 + r2, the areas of the ends, we have

total surface = {R2 + (R+ 1) + r2}.

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If in this formula r=0, we have, as before, the expression for the total surface of the complete cone.

If r = R, we have the expression for the total surface of the cylinder, as might have been expected.

PROP. VII. Lemma. - The surface generated by a straight line revolving about an axis in its plane, is equal to the projection of the line on the axis multiplied by the circumference of the circle whose radius is the perpendicular erected at the middle of the line, and terminated by ́ the axis.

Def.-The projection of a point A upon line XY is the foot a of the perpendicular let fall from the point upon the line.

The projection of a finite straight line AB upon the line XY is the distance ab between the projections of the extremities of AB.

Let AB be a straight line, revolving about the axis XY, ab its projection on the axis, HO the perpendicular to it at its middle point H, terminating in the axis.

To prove that

surface generated by AB = ab
x circumference, having OH
as radius.

Draw AM parallel and HK perpendicular to the axis.

The surface generated by AB is evidently that of a frustum of a


Hence (Prop. VI.),

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Now, the triangles ABM and OHK are similar, their sides being mutually perpendicular to each other.




AB: AM :: HO: HK,

AB: ab:: HO : HK,

AB: ab: 27HO: 2′′HK.

.. AB : ab :: circumference having HO as radius : circumference having HK as radius;

.. circumference having HK as radius=

ference having HO as radius.


x circum


Substituting in (1) this value for circumference having HK as radius;

.. surface generated by AB=ab x circumference having HO as radius.

Cor.-If AB is parallel to the axis, then the surface generated is that of a cylinder, and ab = AB, HO = radius of base.


PROP. VIII. The surface of a spherical zone is equal to the product of its altitude by the circumference of the sphere.

Def.-A sphere is a solid bounded by a surface, all points of which are equally distant from a point within it, called the centre.

A sphere may be generated by the revolution of a semicircle about its diameter as an axis, for a surface thus generated will evidently have all its points equally distant from the centre.

It can be shewn that every section of a sphere made by a plane is a circle, and as the greatest possible section is one made by a plane passing through the centre, such a section is called a great circle, any other a small circle.

The poles of a circle of the sphere are the extremities of the diameter of the sphere, which is perpendicular to the plane of the circle, and the diameter is called the axis of the circle.

A spherical zone is a portion of the sphere intercepted between two parallel planes. If one of the planes touch the sphere, the zone is said to be of a single base.

Let the sphere be generated by the revolution of the semicircle EBF about the axis EF, and let the arc AD generate the zone whose area is required.

Let the arc AD be divided into any number of equal parts, AB, BC, CD, &c. The chords AB, BC, CD, &c. form part of the perimeter of a polygon, which differs from a regular polygon only in this, that the arc subtended by one of its sides, as AB, is not necessarily an aliquot part of the whole circumference. The sides being equidistant from






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the centre, the perpendiculars OH, OK, &c. are equal. Therefore we have, by the preceding Lemma,

Surface generated by AB = ab × circumference of which OH is radius.

Surface generated by BC = bc × circumference of which OH is radius.

Surface generated by CD = cd × circumference of which OH is radius.

Taking the sum, we have

surface generated by AB+ BC + CD = (ab + bc + cd) x circumference of which OH is radius.

This being true, whatever be the number of chords AB, BC, &c. let that number be increased indefinitely.

Then the surface generated by the chords approaches indefinitely to the area of the zone generated by the arc AD, and OH approaches nearer and nearer to the radius of the sphere, and ultimately coincides with it; hence at the limit we have

surface of zone = ad × circumference of sphere.

Let S denote the surface of the zone whose altitude is h, the radius of the sphere being r.

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Cor. 1.-Zones of the same sphere are evidently to cach other as their altitudes.

Cor. 2.-Let the arc AD generate a zone of a single

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That is, the surface of a zone of one base is equal to that of a circle whose radius is the chord of the generating arc of the zone.

PROP. IX. The surface of a sphere is equal to the product of its diameter by the circumference of the sphere.

This follows directly from the preceding proposition, since the surface of the whole sphere may be regarded as a zone, whose altitude is the diameter of the sphere. Cor. 1.-Let S = the surface of sphere whose radius is r.

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or the surface of the sphere is equivalent to that of four great circles.

Cor. 2. The total surface of the cylinder circumscribing the sphere whose radius is r, is evidently

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Cor. 3.-Let S and S' denote the surfaces of two spheres whose radii are r and r.

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