If h be the altitude of a cylinder the, radius of whose base is r, for its convex surface we shall have wrh

for its solidity wrth

(4)

(5)

and for the same reason we shall have wrth' for the solidity of another cylinder, whose altidude is h', and the radius of whose base is r', hence the solidities of the two cylinders are to each other as the altitudes multiplied by the squares of the radii of the bases. If their altitudes and radii of their bases are proportional, in which case the cylinders will be similar, we shall have

(6,)

or multiplying byr and dividing by h, r': r that is the solidities are as the cubes of the radii of their bases,

as before shown in geometry.

Also if the altitude of a cone is h, and r the radius of the base, its convex surface may be expressed by

+

2ra

2

= r2 × h2

(7)

Or let k = the slant height of the cone, then will its lateral surface be

ra × k = rka

(8)

which result will be also obtained if we take or the area of

the base, and increase it in the ratio ofr: k, viz.:

or if we multiply the convex surface of the cone by one-third

of its distance from the centre of the base, (Prop. IX. B. III.

El. S. Geom.) we shall obtain the same result.

The distance from the centre of the base to the surface may

If h' = the altitude of a frustum of the cone, then may the

h'k

the slant height

hsh

required, which call k'; let the radius of the smaller base of the frustum ber', then will the lateral surface be

= rk'a+r'k'

2rx+2r'

2

-xk' (11) = the greater base, and wr2 = the smaller base (Prop. X, B. III. El. S. Geom.) the solidity=+ h'(πr2 + ar2 + ar2/2)-(12)

wrh'

πrh

And since and each expresses the solidity of a

3

3

cone on one or the other of the bases, and whose altitude is equal to that of the frustum, hence if one of those expressions is taken from that of the frustum, the remainder will express a conesected frustum.

wrth

Thush (πr2 + ar2 + √ προγ2) =(πr2 + √2) - (13)

3

which is a conesected frustum, having a conical cavity on its larger base and h (wr2 + y2)

(14)

expresses a conesected frustum the cavity of which is formed on the smaller base.

Or if we multiply the convex surface by its distance from the centre of either base, we shall have the solidity of a conesected frustum, whose cavity is taken from the opposite base (Prop. XI, Cor. B. III, El. Šol. Geom.)

Thus rka + r'k', formula (11) the expression of the lateral

rh

surface, multiplied formula (10) the distance of the surface

k

from the centre of the larger base gives

rha + rr'hk'

k

(15)

When the two bases of the frustum are equal, the coinsected frustum becomes a conesected cylinder, and r and r', k and h' becomes identical. Hence the expression becomes

where k represents the altitude.

(16)

9. Applying the same notation to express the sphere, we have for the surface of any sphere whose radius is r, 4wr; and 4arrar, will be its solidity, (Prop. XXI. B. III. El. Sol. Geom.)

(1)

If the surface of a spherical zone is required, it may be expressed by the product of the altitude of the zone multiplied by the circumference of the sphere; let h=the altitude, and we have 2trh for the spherical surface of the zone, (2.) The solidity of the sector of which this is the spherical base,

(3.) Now, since the sector CBAD (see the diagram to the Problem on the 187 page,) may be considered to be made up of a cone CBD and a segment BDA; in order to express these portions, it will be necessary to find the area of the circular section BD; for this purpose, since CP=CA-AP =r-h, and CB=r, we have in the right angled-triangle BPC, BP=√CB-PC2=√r-r2+2rh-h2=√2rh-h2,

(4.)

The radius of the circular section BPD, hence the area of

Which may be resolved into wh2 (r - h,)

(8.)

2wrth+3wrhwh

=xrh-h,

3

Hence, the solidity of the segment is equal to the product of a circle, whose radius is the altitude of the segment multiplied by the radius of the sphere, minus a third of this altitude.

10. Let r' be the radius of a sphere inscribed in a vertical polyedroid, and r the radius of the circumscribed sphere. Then (Prop. XVIII. B. III, El. Sol. Geom.) the surface of the polyedroid will be 2r'«×2r=4rra,

- (1.)

And since the surface of the circumscribed sphere is=4r, formula (1, Art. 9,) it follows that the surface of the polyedroid and that of its circumscribed sphere, are to each other as r': r, since those are the only variable quantities which enter into their expressions.

If h be the height of any zone of the sphere, its surface, formula (2, Art. 9,) will be 2rxh=2rht,

(2)

The surface of the corresponding zone of the polyedroid, will be 2r'×h=2r/hr, (P. XVIII. Cor. B. III. El. S. G.) (3.) And hence the surface of any zone or segments of the polyedroid and sphere made by the same perpendicular to their common axis, will be in the ratio of their whole surface, viz.

r'

: r, or

r

(4.)

The base of any segment of the polyedroid made by a plane passing through a circle common to the polyedroid and sphere, may be found from formula (5, Art. 9.)

Let 2wrh - Th2, be the base of acommon segment of the sphere and polyedroid, and h the common altitude of the polyedroidal and spherical sector.

The solidity of the polyedroidal sector will be

And since the corresponding spherical sector is rha, for

mula (3, Art. 9) it follows that the solidities of the corresponding sector of these solids are as the ratio ofr: r', as has been shown with regard to the convex surface. The cone whose base is the base of the segment, and whose vertice is the centre of the polyedroid or sphere, formula (6, Art. 9,) may be expressed 2arth - 3wrh2+wh

3

hence the segment will be

=h(r+r+rh-h2)

If the polyedroidal segments consists only of a vertical cone, its solidity will be 2wrh - wh2)×h=wrh-wh, subtract this from the spherical segment on the same base, formula (7, Art. 9,) rh-h, and we have arh2. (7.) which is the value of that portion of the segment of the sphere not included in that of the inscribed polyedroid, which is such a portion of the sphere as would be generated by the revolutions of a circular segment as BD, about the axis DC, passing through the centre of curvature, and perpendicular to the arc of the segment at the point of contact D.

B

D

A

C

PROBLEM.

It is required to find when the spherical segment and the cone composing a spherical sector are equal to each other.

Let ABED represent a sphere generated by the revolution of a semicircle ABE about its diameter AE. The sector ABC, by this revolution, generates a spherical sector, which is composed A of a spherical segment generated by the revolution of a semisegment ABP, and

B

of a cone generated by the revolution of the right-angled triangle BPC.

D

E

The solidity of the sector, formula (3, Art. 9.) will be arh.

The solidity of the cone, arh - rh+ , formula (6.)

3'

Now, in order that the cone may be equal to the segment, the sector, which is the sum of both, must be double the cone:

hence, far'h=ţ«r2h=2«rh2+1⁄2«h',

dividing by 2, transposing, &c.,

Of these two solutions it is evident that only h=r-r can satisfy the conditions of the question, sincer+√ is more than 2r, or more than the diameter of the sphere.

EXAMPLES FOR EXERCISE.

1. What is the solidity of the spherical segments of which the frigid zones are the convex surfaces, the altitude of each segment being 327 miles, and the radius of the base 1575,28 miles? Ans. 1282921583 solid miles nearly.

2. What is the solidity of the spherical segments of which the temperate zones are the convex surface, the radius of the superior base being 1575,28 miles, that of the inferior 3628,86 miles, and the altitude 2053,7 miles?

Ans. 55021192817 solid miles nearly.

3. What is the solidity of the spherical segment of which the torrid zone is the convex surface, the radii of the bases being 3628,86 miles, and the altitude 3150,6?

Ans. 146715018499 solid miles nearly.

4. Having two vats or two tubs in the form of conical frusta, whose dimensions are as follows, viz.: the first has a base whose diameter is 3 feet, its altitude is 34 feet, and the slant height of its side is 4 feet: the diameter of the base of the second is 3 feet, its altitude is 5 feet, and the curve surface is 60 square feet, what must be the dimensions of one capable of containing as much as the other two, if the diameter of the bottom and top, and the altitude are in the proportion of 2 21 and 3.

5. What is the difference in surface of a vertical hexedroid circumscribing a sphere whose diameter is 10, and the whole surface of a conesected frustum of a cone inscribed in the same sphere, and whose wanting base is 6, and perfect base 4?