mensions of the first exceed in some direction those of the second; and if it happens that the dimensions of one surface are in all directions less than the dimensions of another surface, it is evident that the first surface will be less than the second. Now, in whatever direction the plane BPD be made to pass, as it cuts the plane in BD, and the other surface in BPD, the straight line BD, will always be less than BPD; therefore the plane surface OABCD is less than the surrounding surface PABCD.

514. II. A convex surface OABCD (fig. 255) is less than any Fig. 255. other surface which encloses it by resting on the same perimeter ABCD.

Demonstration. We repeat here, that we understand by a convex surface a surface that cannot be met by a straight line in more than two points; still it is possible that a straight line may apply itself exactly to a convex surface in a certain direction; we have examples of this in the surfaces of the cone and cylinder. It should be observed moreover, that the denomination of convex surface is not confined to curved surfaces; it comprehends polyedral faces, or surfaces composed of several planes, also surfaces that are in part curved and in part polyedral.

This being premised, if the surface OABCD is not smaller than any of those which enclose it, let there be among these last PABCD the smallest surface which shall be at most equal to OABCD. Through any point O suppose a plane to pass touching the surface OABCD without cutting it; this plane will meet the surface PABCD, and the part which it separates from it will be greater than the plane terminated by the same surface; therefore by preserving the rest of the surface PABCD, we can substitute the plane for the part taken away, and we shall have a new surface, which encloses the surface OABCD, and which would be less than PABCD. But this last is the least of all, by hypothesis; consequently this hypothesis cannot be maintained; therefore the convex surface OABCD is less than any which encloses OABCD and which is terminated by the same perimeter ABCD.

515. Scholium. By a course of reasoning entirely similar it may be shown,

1. That if a convex surface terminated by two perimeters, ABC, DEF (fig. 256), is enclosed by any other surface termi- Fig 256. nated by the same perimeters, the enclosed surface will be less

Fig. 257. 2. That, if a convex surface AB (fig. 257) is enclosed on all sides by another surface MN, whether they have points, lines, or planes in common, or whether they have no point in common, the enclosed surface is always less than the enclosing surface.

For among these last there cannot be one which shall be the least of all, since in all cases we can draw the plane CD a tangent to the convex surface, which plane would be less than the surface CMD; and thus the surface CND would be smaller than MN, which is contrary to the hypothesis, that MN is the smallest of all. Therefore the convex surface AB is less than any which encloses it.

Fig. 258.

THEOREM.

516. The solidity of a cylinder is equal to the product of its base by its altitude.

Demonstration. Let CA (fig. 258) be the radius of the base of the given cylinder, H its altitude; and let surf. CA represent the surface of a circle whose radius is CA; we say that the solidity of the cylinder will be surf. CAx H. For, if surf. CA x H is not the measure of the given cylinder, this product will be the measure of a cylinder either greater or less. In the first place let us suppose that it is the measure of a less cylinder, of a cylinder, for example, of which CD is the radius of the base and H

the altitude.

Circumscribe about the circle, of which CD is the radius, a regular polygon GHIP, the sides of which shall not meet the circumference of which CA is the radius (285); then suppose a right prism having for its base the polygon GHIP, and for its altitude H; this prism will be circumscribed about the cylinder of which the radius of the base is CD. This being premised, the solidity of the prism is equal to the product of its base GHIP multiplied by the altitude H; and the base GHIP is less than the circle whose radius is CA; therefore the solidity of the prism is less than surf. CAx H. But surf. CAx H is, by hypothesis, the solidity of the cylinder inscribed in the prism; consequently the prism would be less than the cylinder; but the cylinder on the contrary is less than the prism, because it is contained in it; therefore it is impossible that surf. CA × H should be the measure of a cylinder of which the radius of the base is CD and the

altitude H; or in more general terms the product of the base of a cylinder by its altitude cannot be the measure of a less cylinder.

We say, in the second place, that this same product cannot be the measure of a greater cylinder; for, not to multiply figures, let CD be the radius of the base of the given cylinder; and, if it be possible, let surf. CD × H be the measure of a greater cylinder, of a cylinder, for example, of which CA is the radius of the base and H the altitude.

The same construction being supposed as in the first case, the prism circumscribed about the given cylinder will have for its measure GHIP × H; the area GHIP is greater than surf. CD; consequently, the solidity of the prism in question is greater than surf. CDx H; the prism then would be greater than the cylinder of the same altitude whose base is surf. CA. But the prism on the contrary is less than the cylinder, since it is contained in it; therefore it is impossible that the product of the base of a cylinder by its altitude should be the measure of a greater cylinder.

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

517. Corollary 1. Cylinders of the same altitude are to each other as their bases, and cylinders of the same base are to each other as their altitudes.

518. Corollary II. Similar cylinders are to each other as the cubes of their altitudes, or as the cubes of the diameters of the bases. For the bases are as the squares of their diameters; and, since the cylinders are similar, the diameters of the bases are as the altitudes (510); consequently the bases are as the squares of the altitudes; therefore the bases multiplied by the altitudes, or the cylinders themselves, are as the cubes of the altitudes.

519. Scholium. Let R be the radius of the base of a cylinder, H its altitude, the surface of the base will be л R2 (291), and the solidity of the cylinder will be л Ra × H, oг л Ra H,

LEMMA.

520. The convex surface of a right prism is equal to the perimeter of its base multiplied by its altitude.

Demonstration. This surface is equal to the sum of the rect

angles AFGB, BGHC, CHID, &c. (fig. 252), which compose it. Fig. 252.

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× 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 x 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. CDx 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.

1