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equal to the surface of a cylinder having the altitude E G, and a base with the lesser radius CA. In the arc E F inscribe a portion of a regular_polygon EMNOPF, whose sides shall not reach the arc A B, and draw CI perpendicular to one of the sides of the polygon. By (231) the surface generated by the revolution of the polygon will be equal to that of a cylinder whose altitude is EG, and the radius of whose base is CI. Hence the surface produced by this polygon must be greater than the surface of a cylinder whose altitude is E G, and whose radius is AC. But this last is, by hypothesis, equal to the surface produced by the revolution of the arc EF. From whence it follows, that the spherical surface produced by the revolution of the arc E F is less than the surface produced by the polygon inscribed in it; but the former surface includes the latter entirely within it, and therefore cannot be less than it. Hence it follows, that the surface of the spherical segment EF is not less than the surface of the cylinder whose altitude is E G, and whose radius is E C.

Next let the spherical surface be greater than that of the cylinder. Let the proposed spherical surface be that which would be produced by the revolution of the arc A B round AC. We are to prove that this surface is not greater than that of a cylinder whose altitude is AD, and whose radius is AC. If possible, let it be greater than this cylindrical surface. But if the surface generated by the arc A B be greater than that of the cylinder whose altitude is A D and radius A C, for the same reason the surface produced by the arc BH must be greater than that of the cylinder whose altitude is DH and whose radius is A C; and hence the whole surface of the sphere would be greater than that of the circumscribing cylinder, which is contrary to what was proved in (233). Hence the surface generated by the arc AB is not greater than a cylindrical surface whose altitude is AD and whose radius is AC.

Hence it appears, that any plane drawn intersecting a sphere and its circumscribing cylinder parallel to the bases of the cylinder, divides the cylindrical and spherical surfaces into parts which are equal each to each. Hence, if two such planes be drawn, the cylindrical and spherical surfaces which they include between them will be the differences between the equal cylindrical and spherical surfaces which they cut off towards either base of the cylinder, and therefore those differences are equal. (237) Cor. 1.-The spherical surface included between two parallel planes intersecting the sphere is therefore equal to the circumference of a great circle, multiplied by the perpendicular distance between the planes. (238) Cor. 2.-Let the distance between the planes be a, and let r be the radius of the sphere. The circumference of a great circle

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is 2 7 r, and the surface between the planes is 27 ra. If r' be a mean proportional between 2r and a, we have 2ura = are Hence the surface is equal to the area of a circle whose radius is a mean proportional between the distance between the planes and the diameter of the sphere. (239) Cor. 3.—If one of the planes be a tangent plane, the surface becomes that of a spherical segment, and the mean proportional between the diameter and perpendicular is the chord E F of the generating arc. Therefore the surface of a spherical segment is equal to a circle whose radius is the chord of half the arc formed by a section of the segment by a plane through its axis.

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PROPOSITION XII. (240) If the triangle BAC, and the rectangle BCEF,

having the same base and the same altitude, revolve together round their common base BC, the solid described by the revolution of the triangle will be one third of the cylinder

described by the revolution of the rectangle. Draw the perpendicular A D. The cone described by BAD is one third of the cylinder described by BEAD, and in like manner the cone described by CAD is one third of the cylinder described by CEAD (210). When the perpendicular A D falls within the base, the solids in question are the sums of these cones and cylinders, and when it falls without the base A D they are their differences. In either case the truth of the proposition is therefore apparent.

If the perpendicular fall on the extremity of the base, the solids in question will be simply a cylinder and cone having the same base and altitude, and the proposition is reduced to (210). (241) COR.-Since the cylinder is equal to 7. ADX BC, the solid described by the triangle will be į . AD' x B C.

B

B

D

PROPOSITION XIII.

(242) The triangle CAB being supposed to revolve

round any line C D passing through the vertex C, to determine the volume of the solid produced by its revolution.

MKN

X CD.

Produce the side A B until it meet the axis in D, and draw the right lines A M and B N perpendicular to the axis CD.

The volume of the solid described by the triangle CAD is . AMR x CD (240). The solid described by the triangle C B D is in like manner 17 X BN x CD. Hence the difference of these solids, or the solid described by A B C is fm . (AM - B N')

Let I be the middle point of A B and draw I K perpendicular to CD. Hence AM, I K, and B N are in arithmetical progression, and therefore 2IK = AM + BN. Also, if B O be drawn parallel to CD, AO=AM – BN. Therefore 2 1 K x AO = (AM + BN) (AM-BN)=AM'- B N. Hence the solid described by the triangle CAB is equal to fax I KX AOX CD. If C P be drawn perpendicular to AB, the triangles DCP and A BO are similar, and we have therefore

AO: AB=CP: CD,

::: AO X CD= A B x CP. But A B x CP is twice the area CAB. Hence the solid described by CA B is equal to 17 XIK x area CAB; or, which is the same, it is equal to CAB multiplied by the circumference whose radius is I K. Hence " the volume of the solid described by the revolution of the triangle CAB is equal to two thirds of the area of the triangle CAB multiplied by the circumference traced by I, the middle point of the base."

We have, in the preceding proof, supposed that A'B produced will meet the axis CD. The same result will, however, be obtained if A B be parallel to the axis. In this case the volume of the cylinder described by A MNB is equal to 1. A M. MN; the cone described by ACM is equal to 17. A MR.CM, and the cone described by BCN is equal to 17. AM. C N. Add the first two volumes, and subtract from their sum the third, and we shall have the volume described by ABC equal to 7. AM (MN + 4CM – ICN); and since CN-CM = M N, the volume is equal to 7. AM .MN = $.CP.MN; which is equivalent to the result already found. (243) Cor. 1.-If the triangle A CB be isosceles the point P will coincide with I, and the area CAB will be equal to AB X ICI, and the volume of the solid will be a X A BC XIK, or XABXIKXCI. But the triangles ABO and CIK are similar, and therefore

P

M

N

M K N

AB : BO=CI : IK, or AB : MN=CI : IK,

::: AB X I K= MN XCI. Hence the solid described by the isosceles triangle A B C is equal to ja X MNX CI'.

PROPOSITION XIV.

(244) Let A B, BC, CD, .... be several sides

of a regular polygon, O its centre, and OI the radius of the inscribed circle; if the polygonal sector AOD, lying all on the same side of the diameter FG, be supposed to revolve round this diameter, the volume of the solid described by it will be equal to f*. O I'.MQ; MQ being that part of the axis included between perpendiculars AM, DQ from the extremes of the

polygonal sector. Since the polyon is regular, the several triangles A OB, BOC, ...... are equal and isosceles. By (243) the volume described by the triangle A OB is equal to št. 01. MN. In the same manner the volume described by BOC is equal to .01'.NO, and so on. Hence the whole solid described by the polygonal sector being the sum of these is equal to 17.0I'.(M N + NO + .... ), or 7.01 MQ

B

PROPOSITION XV.

(245) The volume of a spherical sector is equal to its

spherical surface multiplied by a third of the radius, and the volume of the whole sphere is equal to the surface of the sphere multiplied by a third of the radius.

M.

I

E

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М.

A

P

N

B

P

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Let A B C be the circular sector which by its revolution round AC describes the spherical sector. The surface of the spherical segment described by the arc A B is equal to 2. AC . AD (236). Now the volume of the spherical sector ACB is equal to

7. AC'. AD, that is, the surface of the spherical segment multiplied by a third of the radius.

First, suppose that T. AC. AD is

UK greater than the sector AC B and equal to the sector ECF. In the arc E F inscribe a portion of a regular polygon, so that its sides shall not meet the arc A B. Let the polygonal sector ECF be supposed to revolve at the same time with the arc E F round EC. Let C I be the radius of the circle inscribed in the polygon, and let FG be drawn perpendicular to EC. The volume described by the polygonal sector will be equal to $,CI'. E G. (244); but C l is greater than AC, and E G is greater than AD; for if AB and EF be drawn, the similar triangles E F G and A B D give

EG:AD= FG:BD=CF:CB, and therefore EG is greater than AD. Hence it follows, that 1.7.CI. EG is greater than in .CA . AD. The former is equal to the volume described by the polygonal sector, and the latter is, by hypothesis, that of the spherical sector ECF. Hence the volume described by the polygonal sector is greater than that of the spherical sector ECF, of which it is a part, which is absurd. Hence the spherical surface multiplied by a third of the radius is not greater than the volume of the spherical sector.

Secondly, this product is not less than the volume of the spherical sector. Let EC F be the circular sector which by its revolution describes the spherical sector, and suppose, if possible, that f.CE. E G is equal to a smaller sector ACB. The former construction being retained, the volume of the polygonal sector will be fa . CI'. E G. But CI is less than CÈ, and therefore the volume of the polygonal sector is less than T.CE'. EG, which, by hypothesis, is equal to the volume of the spherical sector

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