other: so that their arcs Bb, and Ee, may be confounded with the tangent to the radius at the meeting of it with the surface of the sphere, in the same manner as this has been done in determining, in Plane Geometry, the circumference of the circle from its diameter, by the approximating poligon; then the radius CB, being drawn, CBb, will be a right angle, drawing the Bn, perpendicular upon pb, the PBn, will also be a right angle, and the triangle Bbn, will be similar, the triangle CBP, for the angles bBn = CBP, and Bbn = BCP; this furnishes the proportion:

thence Bb

PB : CB = Bn : Bb;

СВ × Вп

PB

Under the same supposition the part of the circumference bB, being allowed to be considered as a straight line by the principle of the calculation of the side surface of a truncated cone, ascertained in the last proposition, the side surface of this truncated cone, of which Bb, is the slent altitude, will be:

s=Bb x (PB + pb);

Upon the same ground of the near contiguity of the planes through EB, and eb, the PB, and pb, that is the two radii of the parallels can be considered as equal; by which this expression becomes:

s = 2π Bb × PB,

placing in this expression the value of Bb, just found by the proportion, it becomes:

and by the compensation of PB, in numerator and denominator:

s = 2.4.r. Bn.

The Bn Pp, is any minute part into which the radius may

be divided: so that

=

Bn

Pp

will represent this radius.

expressed in the same minute units, or subdivisions, which have been used above.

This proposition is general, and applies equally to any part of the circumference of the sphere between two parallels, thence also to their multiples expressed by the same units;

thence to the whole sphere, which would have

2r

Bn

for

its expression in units of Bn; therefore, the surface of the whole sphere will be equal to the product of these two quantities, that is:

As the surface of the circle of the radius r, has been found in Plane Geometry, (§ 83,) to be equal to 2, the comparison of these results shews, that the surface of the sphere is four times that of the great circle of the same sphere; and this result is evidently independent of all arbitrary supposition, being grounded entirely upon the determination of the circumference of a circle from its diameter; which, as has been shewn in its place, is obtained by an extreme approximation; the Pp, or distance of the parallel planes forming the subdivisions of the radius, or any part of the same, is exactly corresponding to the subdivision of the circle by the regular polygon, in which the circumference is subdivided by the approximation.

If it is desired to have this expression reduced to the diameter is to be substituted for r,

of the sphere, the value of r = −

D

that is, the circumference of the great circle of the sphere multiplied by its diameter, is equal to the surface of the sphere.

Scholium 1. Every part of the surface of the sphere will equally be obtained, by the substitution of the value of the part of the radius to which it corresponds, in the units of the subdivision employed; which evidently will always disappear by the compensation that takes place in the multiplication: as for in

stance, there will be, according to the above principles:

§ 45. PROBLEM. To determine the solid content of a sphere.

APPLIC. Let the figure (122,) of the foregoing Problem, represent the sphere of which the solid content is to be determined.

Determination 1. Having by the preceding Problem determined the surface of the sphere, it is evident that drawing radii from any three or more points of it, taken equally as minute subdivisions, as have always been done for this object, they will form the bases of pyramids, having their vertices in the centre of the sphere; and that therefore, the solid content of the sphere will be again equal to the sum of the solid contents of all these pyramids. Every pyramid being the third part of the product of its base into its altitude, the expression for the surface found above, being multiplied by the third part of the radius, will give the solid content of the sphere, that is:

2. If in the general expression, the diameter is desired to

be introduced instead of the radius the value of r =

D

or,

2

Or the sixth part of the number, given in Plane Geometry, would form a constant number, which multiplied into the cube of the diameter of a sphere, will give its solid content.

Scholium. Any sector of the sphere will of course be expressed by the part of the circumference to which it corresponds, so for instance for the sector from the centre C, resting upon the part

of the circumference cut off by EB, as above used, there will be obtained:

(Remark.) It may be observed by the preceding Problem, and by this: that in all cases where a surface is obtained, there will always be two lineal dimensions resulting, to be multiplied: as here, the r2, or D2; when a solid is determined, three lineal dimensions will be multiplied, as here, r3, or D3. Thus always denoting the nature of the result, in accordance with the calculation, and the nature of the subject, under consideration, or obtained in result.

§ 46. PROBLEM. To compare the solid content of the cylinder, the hemisphere, and the cone, of equal base and altitude.

=

APPLIC. Let BCAH, be a square BCA, the fourth part of a circle, and BCH, a triangle, (fig. 123,) where AC, the radius of the circular part BH, the base of the triangle, and the sides CA, and CB, are all equal; it is required to determine the ratio that the solids described by their revolution around the BC, as axis, will have to one another.

Determination. The triangle CBH, by revolving upon BC, as axis, will describe a cone, the radius of the circular base of which will be BH, and the altitude BC, which will be equal to one another, thence produce an upright rectangular cone. The quadrant BAC, will describe a hemisphere, which will have the same radius as the base of the cone. The square ACBH, will describe a cylinder, having the same CA, for the radius of its base, and its equal CB, for its altitude. Any section like DEFG, being made through the three bodies perpendicular to their common axis, and another plane passing through defg, being parallel to this, the distance Dd, between them, being such a minute subdivision of the radius as has been used in the preceding cases; there will be such parts of the three solids generated as having the same altitude Dd, Ee, or Ff, and which on account of this supposed subdivision, will form cylinders of equal altitudes: therefore will be to one another as their

circular bases; that is: as the squares of their radii. The triangles CDE, and CDF, are right angled in D; they give, therefore, by 36, of Plane Geometry, the following comparison between the squares of their sides, viz:

4

CF2 = DC2 + DF2.

The triangles CBH, and CDE, are similar; thence CD = DB; the radii BH, BC, CF, and CA, are equal; DG, is parallel to CA, or BH, and equal to them, and to CF; the substitution of these equal quantities in the above, will give

DG2 = DE2 + DF2.

Such is, therefore, the ratio of the similar surfaces described by these lines, therefore of the circles that are described with them as radii; and if multiplied all through by the same quantity, their ratio will be the same.

=

By the revolution of the triangle, the quadrant, and the square, these lines describing such circles, if the surfaces, the ratio of which is represented by these squares, be multiplied by a common factor, as Dd = Ee = Ff Gg, which has been. taken as a subdivision of the radius CB, such as to consider the bodies described as cylinders of the same altitude, these cylinders will have the same ratio as their bases; there will therefore be

DG2. Ff = DE2. Ff + DF2. Ff, And multiplying them as in the foregoing section, by the altitude divided by these Ff, considered as units, in which the radius CB, is expressed; this CB, being the common altitude of the three solids, and called = h, the result will be BH2. h. Ff_DG2. h. Ff. DE2 h. Ff DF2. h. Ff

Ff

or,

=

BH2. h DG2. h

=

The similarity of the triangles

=

Ff

+

Ff.

DE2. h + DF2. h.

CDE, and CBH, gives the ratio of the whole the same as that of the parts. The BH2, represents the base of the cylinder; the DE2, that of the cone; and the DF2, that of the hemisphere, described by the revolution of the square, triangle and quadrant, in any of their corresponding parts; thence the cylinder will be equal to the sum of the cone and hemisphere together.