Now the altitudes AF, BG, CH, &c., of these rectangles are each equal to the altitude of the prism. Therefore the sum of these rectangles, or the convex surface of the prism, is equal to the perimeter of its base multiplied by its altitude.

521. Corollary. If two right prisms have the same altitude, the convex surfaces of these prisms will be to each other as the perimeters of the bases.

LEMMA.

522. The convex surface of a cylinder is greater than the convex surface of any inscribed prism, and less than the convex surface of any circumscribed prism.

Demonstration. The convex surface of the cylinder and that of Fig 252, the inscribed prism ABCDEF (fig. 252) may be considered as having the same length, since every section made in the one and the other parallel to AF is equal to AF; and if, in order to obtain the magnitude of these surfaces, we suppose them to be cut by planes parallel to the base, or perpendicular to the edge AF, the sections will be equal, the one to the circumference of the base, and the other to the perimeter of the polygon ABCDE less than this circumference; since therefore, the lengths being equal, the breadth of the cylindric surface is greater than that of the prismatic surface, it follows that the first surface is greater than the second.

By a course of reasoning entirely similar it may be shown that the convex surface of the cylinder is less than that of any Fig. 253. circumscribed prism BCDKLH (fig. 253).

Fig. 258.

THEOREM.

523. The convex surface of a cylinder is equal to the circumference of its base multiplied by its altitude.

Demonstration. Let CA (fig. 258) be the radius of the base of the given cylinder, H its altitude; and let circ. CA be the circumference of a circle whose radius is CA; we say that circ. CA x H will be the convex surface of the cylinder. For, if this proposition be denied, circ. CA × H must be the surface of a cylinder either greater or less; and, in the first place, let us suppose that it is the surface of a less cylinder, of a cylinder, for example, of which the radius of the base is CD and the altitude H.

Circumscribe about the circle, whose radius is CD, a regular polygon GHIP, the sides of which shall not meet the circumference whose radius is CA; then suppose a right prism, whose altitude is H, and whose base is the polygon GHIP. The convex surface of this prism will be equal to the perimeter of the polygon GHIP multiplied by its altitude H (520); this perimeter is less than the circumference of the circle whose radius is CA; consequently the convex surface of the prism is less than circ. CAx H. But circ. CA - H is, by hypothesis, the convex surface of a cylinder of which CD is the radius of the base, which cylinder is inscribed in the prism; whence the convex surface of the prism would be less than that of the inscribed cylinder. But on the contrary it is greater (522); accordingly the hypothesis with which we set out is absurd; therefore, 1. the circumference of the base of a cylinder multiplied by its altitude cannot be the measure of the convex surface of a less cylinder.

We say, in the second place, that this same product cannot be the measure of the surface of a greater cylinder. For, not to change the figure, let CD be the radius of the base of the given cylinder, and, if it be possible, let circ. CD × H be the convex surface of a cylinder, which with the same altitude has for its base a greater circle, the circle, for example, whose radius is CA. The same construction being supposed as in the first hypothesis, the convex surface of the prism will always be equal to the perimeter of the polygon GHIP, multiplied by the altitude H.

But this perimeter is greater than circ. CD; consequently the surface of the prism would be greater than circ. CD x H, which, by hypothesis, is the surface of a cylinder of the same altitude of which CA is the radius of the base. Whence the surface of the prism would be greater than that of the cylinder. But while the prism is inscribed in the cylinder, its surface will be less than that of the cylinder (522); for a still stronger reason is it less when the prism does not extend to the cylinder; consequently the second hypothesis cannot be maintained; therefore, 2. the circumference of the base of a cylinder multiplied by its altitude cannot be the measure of the surface of a greater cylinder.

We conclude then that the convex surface of a cylinder is equal to the circumference of the base multiplied by its altitude.

Fig. 259.

THEOREM.

524. The solidity of a cone is equal to the product of its base by a third part of its altitude.

Demonstration. Let SO (fig. 259) be the altitude of the given cone, AO the radius of the base; representing by surf. AO the surface of the base, we say that the solidity of the cone is equal to surf. AO SO.

1. Let surf. AO × SO be supposed to be the solidity of a greater cone, of a cone, for example, whose altitude is always SO, but of which BO, greater than AO, is the radius of the base.

About the circle, whose radius is AO, circumscribe a regular polygon MNPT, which shall not meet the circumference of which OB is the radius (285); suppose then a pyramid having this polygon for its base and the point S for its vertex. The solidity of this pyramid is equal to the area of the polygon MNPT multiplied by a third of the altitude SO (416). But the polygon is greater than the inscribed circle represented by surf. OA; consequently, the pyramid is greater than

surf. AO × SO,

which, by hypothesis, is the measure of the cone of which S is the vertex, and OB the radius of the base. But on the contrary the pyramid is less than the cone, since it is contained in it; therefore it is impossible that the base of the cone multiplied by a third of its altitude should be the measure of a greater cone.

2. We say, moreover, that this same product cannot be the measure of a smaller cone. For, not to change the figure, let OB be the radius of the base of the given cone, and, if it be possible, let surf. OB × ¦ SO be the solidity of a cone which has for its altitude SO, and for its base the circle of which AO is the radius. The same construction being supposed as above, the pyramid SMNPT will have for its measure the area MNPT multiplied by SO. But the area MNPT is less than surf. OB; consequently the pyramid will have a measure less than

surf. OB SO,

and accordingly it would be less than the cone, of which AO is the radius of the base and SO the altitude. But on the contrary the pyramid is greater than the cone, since it contains it; therefore it is impossible that the base of a cone multiplied by a third of its altitude should be 'the measure of a less cone.

We conclude then, that the solidity of a cone is equal to the product of its base by a third of its altitude.

525. Corollary. A cone is a third of a cylinder of the same base and same altitude; whence it follows,

1. That cones of equal altitudes are to each other as their bases;

2. That cones of equal bases are to each other as their altitudes;

3. That similar cones are as the cubes of the diameters of their bases, or as the cubes of their altitudes.

526. Scholium. Let R be the radius of the base of a cone, H its altitude; the solidity of the cone will be R2 × ¦ H, or R3 H.

THEOREM.

527. The frustum of a cone ADEB (fig. 260) of which OA, DP, Fig. 260. are the radii of the bases, and PO the altitude, has for its measure

× OP × (AO + DP + AO × DP).

Demonstration. Let TFGH be a triangular pyramid of the same altitude as the cone SAB, and of which the base FGH is equivalent to the base of the cone. The two bases may be supposed to be placed upon the same plane; then the vertices S, T, will be at equal distances from the plane of the bases; and the plane EPD produced will be in the pyramid the section IKL. We say now, that this section IKL is equivalent to the base DE; for the bases AB, DE, are to each other as the squares of the radii AO, DP (287), or as the squares of the altitudes SO, SP; the triangles FGH, IKL, are to each other as the squares of these same altitudes (407); consequently the circles AB, DE, are to each other as the triangles FGH, IKL. But, by hypothesis, the triangle FGH, is equivalent to the circle AB; therefore the triangle IKL is equivalent to the circle DE.

Now the base AB multiplied by SO is the solidity of the cone SAB, and the base FGH multipled by SO is that of the pyramid TFGH; the bases therefore being equivalent, the solidity of the pyramid is equal to that of the cone. For a similar reason the pyramid TIKL is equivalent to the cone SDE; therefore the frustum of the cone ADEB is equivalent to the frustum of the pyramid FGHIKL. But the base FGH, equivalent to the circle of which the radius is AO, has for its measure

Fig. 259.

× AO; likewise the base IKLл × DP, and the mean pro

portional between × AU and л × DP is л AО × DP; there-
x AO
fore the solidity of the frustum of a pyramid or that of the frus-
tum of a cone has for its measure

or

--2

OP × ( ̃ × AV + л × DP + л × AO × DP) (422),

528. The convex surface of a cone is equal to the circumference of its base multiplied by half its side.

Demonstration. Let AO (fig. 259), be the radius of the base of the given cone, S its vertex, and SA its side; we say that the surface will be circ. AO × SA. For, if it be possible, let circ. AO × 1 SA be the surface of a cone which has S for its vertex, and for its base the circle described with a radius OB greater than AO.

Circumscribe about the small circle a regular polygon MNPT, the sides of which shall not meet the circumference of which OB is the radius; and let SMNPT be a regular pyramid, which has for its base the polygon, and for its vertex the point S. The triangle SMN, one of those which compose the convex surface of the pyramid, has for its measure the base MN multiplied by half of the altitude SA, which is at the same time the side of the given cone; this altitude being equal in all the triangles SNP, SPQ, &c., it follows that the convex surface of the pyramid is equal to the perimeter MNPTM multiplied by SA. But the perimeter MNPTM is greater than circ. AO; therefore the convex surface of the pyramid is greater than circ. AO SA, and consequently greater than the convex surface of the cone, which, with the same vertex S, has for its base the circle described with the radius OB. But on the contrary the convex surface of the cone is greater than that of the pyramid; for, if we apply the base of the pyramid to the base of an equal pyramid, and the base of the cone to that of an equal cone; the surface of the two cones will enclose on all sides the surface of the two pyramids; consequently the first surface will be greater than the second (514), and therefore the surface of the cone is greater than that of the pyramid, which is comprehended within