If h = R, the fraction becomes; if h = 2 R, it becomes ; if h = R, it becomes.

Note. In the above investigation, no allowance has been made for the refraction of light, which must of necessity modify the extent of surface visible.

PROPOSITION XI.

THEOREM. If a triangle and a rectangle, having the same base and the same altitude, turn simultaneously about the common base, the solid described by the revolution of the triangle will be a third of the cylinder described by the revolution of the rectangle.

Let ABC be the triangle, and CD the rectangle.

C

A

D

F

B

To the axis, draw the perpendicular AD; the cone described by the triangle ABD is the third part of the cylinder described by the rectangle AFBD, (B. VIII, Prop. v, Cor. ;) also the cone described by the triangle ADC is the third part of the cylinder described by the rectangle ADCE: hence the sum of the two cones, or the solid described by ABC, is the third part of the two cylinders taken together, or of the cylinder described by the rectangle BCEF.

If the perpendicular AD falls without the triangle, the solid described by ABC will be the difference of the two cones described by ABD and ACD; but, at the same time, the cylinder

E

D

C

B

described by BCEF will be the difference of the two cylinders described by AFBD and AECD. Hence the solid described by the revolution of the triangle will still be a third part of the cylinder described by the revolution of the rectangle having the same base and altitude.

Scholium. The circle of which AD is radius has for its measure AD2: hence × AD3× BC measures the cylinder described by BCEF, and ¦ ☛ × AD3 × BC measures the solid described by the triangle ABC.

PROPOSITION XII.

THEOREM. If a triangle be revolved about a line drawn at pleasure through its vertex, the solid described by the triangle will have for its measure the area of the triangle multiplied by two-thirds of the circumference traced by the middle point of the base.

Let CAB be the triangle, and CD the line about which it revolves.

Produce the side AB till it meets the axis CD in D; from the points A and B, draw AM, BN perpendicular to the axis.

The solid described by the

B

NK M

D

P

triangle CAD is measured (B. VIII, Prop. xI, Schol.) by T×AM'× CD; the solid described by the triangle CBD is measured by

BN'x CD: hence the dif

ference of those solids, or the solid described by ABC,

will have for its measure

(AM2 — BN3) × CD.

To this expression another form may be given. From I, the middle point of AB, draw IK perpendicular to CD; and through B, draw BO parallel to CD: we shall have AMBN = 2 IK, and AM-BN = AO; hence (AM+BN) × (AM-BN,) or AM2-BN'=2 IK × AO. Hence the measure of the solid in question is expressed by TXIK× AOx CD. But if CP is drawn perpendicular to AB, the triangles ABO, DCP will be similar, and give the proportion

AO : CP :: AB : CD;

hence AOx CD=CP × AB, which CP x AB is double the area of the triangle ABC; hence we have AO x CD =2 ABC; hence the solid described by the triangle ABC × IK, or, which is

ABC is also measured by the same thing, by ABC × & circ. IK, circ. IK being equal to 2 × IK. Hence, the solid described by the revolution of the triangle ABC has for its measure the area of this triangle multiplied by two-thirds of the circumference traced by I the middle point of the base.

× AB×IK× CI. But the triangles ABO, CIK are similar, and give the proportion

AB BO or MN: : CI: IK;

hence ABX IK=MNx CI: hence the solid described

by the isosceles triangle ABC will have for its measure × MNX CI2.

Cor. 2. The general solution appears to include the supposition that AB produced will meet the axis; but the . results would be equally true, though AB were parallel to the axis.

Thus the cylinder described by AMNB is equal to . AM2. MN; the cone described by ACM is equal to AM'. CM, and the cone described by BCN to T. AM2. CN. Add the first two

B

N

M

C

solids, and take away the third: we shall have the solid described by ABC equal to . AM2. (MN+ ¦ CM − 1 CN;) and since CN-CM=MN, this expression is reducible to. AM. MN, or CP2. MN, which agrees with the conclusion above drawn.

lygonal sector AOD lying all on one side of the diameter FG be supposed to perform a revolution about this diameter, the solid so described will have for its measure

. OI. MQ, MQ being that portion of the axis which is included by the extreme perpendiculars AM, DQ.

For, since the polygon is regular, all the triangles, AOB, BOC, &c., are equal and isosceles. Now, by the last corollary, the solid produced by the isosceles triangle

AOB has for its measure T OI. MN; the solid described by the triangle BOC has for its measure

. OI. NP, and the solid described by the triangle COD has for its measure. OI2. PQ : hence the sum of those solids, or the whole solid described by the polygonal sector AOD will have for its measure . OI. (MN+NP+IQ,) or . OI. MQ.

(112.) This proposition is a particular case of a more general theorem, called Guldin's theorem, or the centrobaryc method, which is as follows:

The volume generated by the movement of a plane surface, so moved as that no two consecutive positions shall intersect, is measured by multiplying the area of the moving plane into the distance passed through by its centre of gravity.

In the above proposition, we know that the centre of gravity of the triangle CAB is in the line CI, at its distance measured from C, (Art. 81.) Hence, the circumference described by this centre of gravity is of the circumference traced by the middle point I. Consequently the volume thus generated by the revolution of the triangle ABC has for its measure the area of this triangle multiplied by of the circumference traced by I, the middle point of the base.

same thing, AB'× AS. That is, the solidity of a right cone may be found by multiplying its base by of its height.