vex surfaces, or of their solid contents, shall be less than any given difference.

Let V be the vertex of a right cone, A B D its base, and C the centre of the base. And, first, let P be the given difference of the surfaces; and let Q be the convex surface of some circumscribed pyramid. Then, because (as in III. 31.) a regular polygon may be inscribed in the circle ABD, the apothem CE of which approaches to the radius CD within any given difference, a polygon may be inscribed, such that CD2 CE may be to CE in a ratio less than any assigned; less, therefore, than that of P to Q. Let A B F G H be such a polygon; and let a similar polygon KLMNO be circumscribed

about the circle, so that the side KL may touch the circle in D (III. 27. Cor. 2.). Join VA, VB, &c., VK, V L, &c., and through E draw EU parallel to DV, to meet CV in U, and join UA, UB, &c.* Then, it may easily be shown, that the convex surfaces of the two pyramids, which have the points U, V, for their vertices, and the inscribed and circumscribed polygons for their bases, are made up of similar triangles, which are to one another in the same ratio, each to each, viz. that of CE2 to CD2 (II. 42. Cor.). Therefore, the convex surfaces of the pyramids are to one another in the same ratio (II. 23. Cor. 1.); and the difference of their convex surfaces is to that of the lesser pyramid as CD2-CE2 to CE2 (II. 20.), that is, in a less ratio than that of P to Q. But the convex surface of the lesser pyramid is less than the surface of the pyramid VABFGH

The lines UA, UB, &c. are omitted in the

figure.

(I. 26. Cor. and I. 12. Cor. 2.), and therefore by much more less than Q. Much more, therefore, is the difference of the convex surfaces less than P (II. 18.Cor.); and more still is the difference of the surfaces of the inscribed and circumscribed pyramids, which have the common vertex V, less than P.

Next, let S be the given difference of the solid contents; and let T be the solid content of some circumscribed pyramid. As before, let the regular polygon A BFGH be inscribed in the circle ABD, such that CD2 - CE2 may be to CE2 in a less ratio than that of S to T; let a similar polygon be circumscribed, and the inscribed and circumscribed pyramids completed. Then, because these pyramids have the same altitude, their solid contents are to one another as their bases, that is, as C E2, CD2 (IV.32.): therefore, the difference of the contents is to that of the inscribed pyramid as CD2-CE2 to CE2 (II. 20.) that is, in a less ratio than that of S to T. Therefore, because the content of the inscribed prism is less than T, much more is the difference of the contents less than S (II. 18. Cor.).

Therefore, &c.

Cor. 1. Any right cone being given, a regular pyramid may be inscribed (or circumscribed), which shall differ from the cone in convex surface, or in solid content, by less than any given difference; for the difference between the cone and either of the pyramids, whether in surface or in content, is less than the difference of the two pyramids (6.).

Cor. 2. Any two similar right cones being given, similar regular pyramids may be inscribed (or circumscribed), which shall differ from the cones in convex surface, or in solid content, by less than any the same given difference.

PROP. 8.

The convex surface of a right cone is equal to half the product of its slant side by the circumference of its base.

For, if half this product be not equal to the convex surface of the pyramid, it must either be greater or less than that surface. If greater, it must also be greater than the convex surface of some circumscribed pyramid (7. Cor. 1.)greater, that is, than half the product of the slant side of the cone by the perimeter of the circumscribed polygon (I. 26. Cor.), which is the base of the pyramid; for the triangles which form the convex surface of the pyramid have for their bases the sides of the circum

their convex surfaces respectively, and (9. Cor. 1.) RA, r2 a their solid contents. But, because the cones are similar (def. 6.) R: A::r: a; and be cause the slant sides S, s are the hypotenuses of right-angled triangles, which have the sides A, R and a, r about the right angles proportional (II. 32.) S: A .sa: therefore (II. 37. Cor. 3.), RS : A2::rs: a2, alternando (II. 19.) RS :rs :: A2; a2, and (p. 47, Rule ii.) T RS: rs: A2; a2, that is, the convex surfaces of the cones are as the squares of their axes.

Again, because R: A; r: a, R2 : A2 r2: a2 (II. 37. Cor. 4.), and (p. 47, Rule ii.) RA; A3: r2 a; a3: therefore alternando (II. 19.) RA: r2 a :: A3

a3, and (p. 47, Rule ii.) R2 A : r2 a :: A3 a3, that is, the solid contents of the cones are to one another as the cubes of their axes. Therefore, &c.

V. Draw any slant side V a A; from A draw any straight line AE perpendicular to VA (Ï. 44.); suppose AE to be taken equal to the circumference A B D, and join VE, and through a draw a e parallel to AE (I. 48.), to meet V E in e. Then, because the circumferences A B D, abd, are as their radii (III. 33.), that is, as VA, V a (II. 31.), that is, again, as A E, ae (II. 31.), and that AE is equal to ABD, ae is equal to abd (II. 18.). Now the triangle VAE is equal to the convex surface of the cone VABD (6.), because (I. 26. Cor.) it is equal to half the product of V A the slant side, and A E, which is equal to the circumference A BD; and, for the like reason, the triangle Vae is equal to the convex surface of the cone Vabd; therefore, the convex surface of the frustum is

pyramid, having its base L M N equal to the base ABD, and in the same plane with it, and its vertex K in the sanie parallel to the base with the vertex of the cone. Then, because the cone and pyramid have equal bases, and the same altitude, Let the plane abd be produced to cut they are equal to one another (9. Cor. 3.). the pyramid in the triangle Imn: then, because I mn is similar to L M N (IV. 12. and IV. 15.), they are to one that is, (II. 37. Cor. 4.) as K12 and another as I m2 and L M2 (III. 42. Cor.), K L2, or (IV. 14.) as V a2 and V A2: but the bases abd and ABD are to one another in the same ratio (III. 33. and II. 31.), and LMN is equal to ABD: therefore, also, Imn is equal to abd (II. 12. and II. 18.), and (9. Cor. 3.) the cone Vabd is equal to the pyramid Klm n. Therefore

(I. ax. 3.) the frustum of the cone is equal to the frustum of the pyramid, But the latter (IV. 33.) is equal to the sum of three pyramids, having the same altitude with the frustum, and for their bases the bases of the frustum, and a mean proportional between them; and (9. Cor. 3.) each of these pyramids is equal to a cone having the same altitude and an equal base. Therefore, also, the frustum of the cone is equal to the sum of three cones, having the same altitude with it, and for their bases the bases of the frustum, and a mean proportional between them.

Therefore, &c.

Cor. If a straight line A a be made to revolve about any axis, VC, in the same plane with it, the surface generated by such straight line shall be equal to the product of the straight line and the circumference generated by its middle point F,

For the generated surface is that of a cylinder if the line be parallel to the axis; and, in every other case, that of a frustum of a right cone. In the former case, the reason is sufficiently manifest (3.). In the latter, it may be shown, that if FG be drawn parallel to AE (in the first figure of the proposition) to meet V E in G, FG will be equal to the circumference generated by the point F: also, because VF is equal to half the sum of VA and Va, FG is equal to half the sum of AE and a e (II. 30. Cor. 2.); therefore the circumference generated by the point F is equal to half the sum of the circumferences A B D, abd; and hence by the proposition, the convex surface of the frustum, that is, the convex surface generated by the line A a is equal to the product of A a and the circumference generated by its middle point F.

Scholium.

Although the propositions of this section have, for greater brevity and simplicity, been stated and demonstrated only with regard to the right cylinder and right cone, it will be found that Props. 2. and 7. apply equally to the oblique cylinder and oblique cone, to which the demonstrations may be without difficulty adapted, and hence it may be demonstrated, almost in the words of Props. 4. 9. 11. 5. and 10. that the solid content of an oblique cylinder is equal to the product of its base and altitude; the solid content of an oblique cone to onethird of the product of its base and altitude; the solid content of a truncated oblique cone to the sum of the solid contents of three cones, having the same altitude with it, and for their bases its two bases, and a mean proportional between them; and, lastly, that the surfaces of similar oblique cylinders and cones are to one another as the squares of their axes, and their solid contents as the cubes of the axes. With regard to the convex surface of the oblique cylinder, it may likewise be shown in a simi. lar manner (see IV. 29. Scholium) to be equal to the product of its side (or axis) by the perimeter of a plane section perpendicular to it.

It remains to observe, with regard to Props. 3. and 8., that the remarkable property by which the convex surfaces of the cylinder and cone have been defined, viz. that of containing, the first, straight lines parallel to a given straight line, and the other straight lines diverging

from a given point, leads to another property of those surfaces, from which the measures assigned in Props. 2. and 6. may be very readily inferred. This property is, that they are developable, that is, they may be conceived to be unfolded and spread out upon a plane. Now, it is easy to perceive, that if the surface of a right cylinder be so developed, it will form a rectangle, which has for its base the circumference of the circle, which is the base of the cylinder, and for its altitude the altitude of the cylinder; whence it follows, that the convex surface of a right cylinder is equal to the product of its altitude by the circumference of its base. In like manner the developed surface of a right cone will form a circular sector, the arc of which is equal to the circumference of the base of the cone, and its radius to the slant side; whence it follows, that the convex surface of a right cone is equal to half the product of its slant side by the circumference of its base.

SECTION 2.-Surface and content of the sphere.

PROP. 12.

If an isosceles triangle ABC be made to revolve about an axis which lies in the same plane with it and passes through the vertex A, and if a perpendicular AD be drawn from the vertex to the base, and E F be that portion of the axis which is intercepted by perpendiculars drawn to it from the extremities of the base; the convex surface generated by the base shall be equal to the product of EF by the circumference of a circle having the radius AĎ; and the solid generated by the triangle shall be equal to one-third of the product of this surface by the perpendicular AD.

First, of the surface generated by the base B C. This is evidently the convex surface of a truncated cone having the axis E F, and EB, FC for the radii of its bases; and is therefore (11. Cor.) equal to the product of BC, and the circumference generated by its middle point D, that is, (if D G be drawn parallel to BE to meet EF in G) the circumference which has the radius DG. Now, if E L be drawn

D

B

A F

parallel to BC, E L will be equal to BC, (I. 22.), and the triangles LE F, ADG will be similar (I. 18.), because the sides of the one are perpendicular to the sides of the other, each to each; therefore EL or BC is to EF as AD to DG (II. 31.), that is, as the circumference of a circle which has the radius AD to the circumference of a circle which has the radius DG (III. 33. and II. 12.); and therefore (II. 28. Schol. Rule I.) the product of BC and the latter circumference is equal to the product of EF and the former. But the convex surface in question is equal to the product of BC and the circumference of which has the radius D G. Therefore (I. ax. 1.) that surface is likewise equal to the product of E F, and the circumference which has the radius A D.

Next, of the solid generated by the triangle ABC. Let CB and FE be produced to meet one another in V. Then the solid in question is the difference of those generated by the triangles A CV and A B V. Now, the solid generated by the triangle ACV is equal to the sum or difference of two cones, having the altitudes AF, VF respectively, and for their common base the circle of which C F is radius-equal, that is (9.), to one-third of the product of A V, which is the sum or difference of the altitudes, by half the radius C F, and the circumference which has the radius CF (III. 32.); or to one-third of the product of AD, half VC, and that circumference (for ADxVC is equal to AVX C F), or lastly, to one-third of the product of AD, and the surface generated by VC (8.). And in the same manner it may be shewn, that the solid generated by the triangle ABV is equal to one-third of the product of A D, and the surface generated by VB. Therefore the difference of these solids, that is, the solid in question, is equal to one-third of the product of A D, and the surface generated by B C. It has been supposed in the above demonstrations, that EF and B C are not parallel. If BC be parallel to E F, the surface generated by BC will be that of a right cylinder having the axis EF, whence the first part of the proposition is manifest; and the solid generated by the triangle ABC will be equal to two-thirds of the cylinder (9.), whence the second part of the proposition.

Therefore, &c.

Cor. The proof of the second part of

the proposition is equally applicable, whether A B C be isosceles, or otherwise. Therefore, if any triangle A B C be made to revolve about an axis which lies in the same plane with it and passes through its vertex A; and if AD be drawn perpendicular to the base, the solid generated by the triangle shall be equal to one-third of the product of AD and the surface generated by the base BC.

PROP. 13.

G

F

E A

L

M

C

N

If the half AFGHKB of any regular polygon of an even number of sides revolve about the diagonal AB; the whole surface of the solid generated by its revolution shall be equal to the product of AB by the circumference of a circle whose radius is the apothem CE of the polygon; and its solid content shall be equal to one-third of the product of this surface by the apothem CE. From the points F, G, H, K, draw FL, GM, ÍN, KO perpendicular to AB, (I.45.) and join CF, CG, CH, CK. Then, because C is the centre of the polygon, the triangles CAF, CFG, &c. 1 are isosceles triangles, having the common vertex C, and the perpendiculars drawn from C to their respective bases equal each of them to the apothem CE. And, because these triangles revolve about the axis AB passing through C, and that A L, LM, &c. are the parts of the axis intercepted by perpendiculars drawn from the extremities of the base of each; the portions of the whole surface in question, generated by AF, FG, &c., are equal, respectively, to the products of AL, LM, &c., by the circumference of the circle which has the radius C E (12.). Therefore the whole surface is equal to the sum of these products, that is, to the product of A B by the circumference of the same circle.

K

B

Again, because the portions of the whole solid which are generated by the triangles CA F, CFG, &c. are the third parts, respectively, of the products of the portions of surface generated by A F, FG, &c. by the apothem CE (12.); the whole solid is the third part of the sum of these products, that is, the third part of the product of the whole surface by the apothem C E.

Therefore, &c.