increased or diminished, the reasoning would be in no wise changed!

ter,

Hence the theorem; if the line drawn through the cen

etc.

SCHOLIUM.-The volume generated by any portion of the semi-polygon, as that composed of the two isosceles A's BOC, COD, is measured by

Sur. perimeter BCD x 10m.

THEOREM XXIX.

The volume of a sphere is measured by its surface multiplied by one third of its radius.

Let a sphere be generated by the revolution of the semicircle ACE, about its diameter, AE, as an axis; then will the volume of the sphere be measured by

sur. semi-circ. OA × 10A.

m

0

For, inscribe in the semi-circle any regular semi-polygon, as ABCDE, and let it, together with the semi-circle, revolve about the axis AE. The semi-polygon will generate a volume which has, for its

measure,

Sur. perimeter ABCDE × 30m, (Th. 28), in which Om is the apothem of the polygon.

E.

Now, however great the number of sides of the inscribed regular semi-polygon, this measure for the volume generated by it, will hold true; but when we reach the limit, by making the number of sides indefinitely great, the perimeter and apothem become, respectively, the semi-circumference and its radius, and the volume generated by the semi-polygon becomes that generated by the semi-circle, that is, the sphere. Therefore,

Vol. sphere sur. semi-circ. OA × 10A.

THEOREM XXXI.

The volume generated by the revolution of the segment of a circle about a diameter of the circle exterior to the segment, is measured by one sixth of times the square of the chord of the segment, multiplied by the part of the axis included between the perpendiculars let fall upon it from the extremities of the chord.

E

10

Let BCD be a segment of the circle, whose center is 0, and AH a part of a diameter exterior to the segment. Draw the chord BD, and from its extremities let fall the perpendiculars, BF, DE on AH; also draw Om perpendicular to BD. The spherical sector generated by the revolution of the circular sector BCDO about AH, is measured by zone BD × †B0, (Scholium 1, Th. 29), = 2«.BO × EF × ‡B0 = }«B0 × EF; and the volume generated by the isosceles triangle BOD is measured by

Om x EF, (Cor. 1, Th. 27).

H

The difference between these two volumes is that generated by the circular segment BCD, which has, therefore, for its measure,

2

«EF (BO' — Om') = 3*EF × Bm', (Th. 39, B. I).

Vol. segment BCD = f«EF × †BD' = f«BD2 × EF.

Hence the theorem.

THEOREM XXXII.

The volume of a segment of a sphere has, for its measure, the half sum of the bases of the segment multiplied by its altiude, plus the volume of a sphere which has this altitude for diameter.

Let BCD be the arc of a circle, and BF and DE perpendiculars let fall from its extremities upon a diameter, of which AH is a part; then, if the area BCDEF be revolved about AH D as an axis, a spherical segment will be generated, for the volume of which it is proposed to find a measure.

The circular segment will generate a volume measured by BD x EF, (Th. 31); and the frustum of the cone generated by the trapezoid BDEF will have, for its measure,

}«BF2 × EF+‡«DE2 × EF +‡«BF × DE × EF, (Th. 22), · }«EF(BF2 + DE2 + BF × DE).

=

But the sum of these two volumes is the volume of the spherical segment, which has, therefore, for its

measure,

¿«EF (BD2 + 2BF2 + 2DE2 + 2BF × DE)

From B let fall the perpendicular Bn on DE; then will

Bn2 + Dn2 = EF2 + Dn2;

and since BD2 = Bn2 + Dn2

we have BD2 = EF2 + DE2 + BF2 — 2DE × BF.

By substituting this value for BD2, in the above measure for the volume of the segment, we find

¿xEF(EF2+DE2+BF2—2DEX BF÷2BF2+2DE2 + 2BF×DE) = \«EF (EFa +3DE'+3BF') = ¿xEF® + EF ( ̃DE2+xBF")

etc.

2

Which last expression conforms to the enunciation. Hence the theorem; the volume of a segment of a sphere,

Cor. When the segment has but one base, BF becomes zero, and EF becomes EA; and the final expression

which we found for the volume of the segment reduces

to

DE

«EA3 + EA ×

2

Hence, A spherical segment having but one base, is equiva lent to a sphere whose diameter is the altitude of the segment, plus one half of a cylinder having for base and altitude the base and altitude of the segment.

SCHOLIUM. When the spherical segment has a single base, we may put the expression, tЕÃ3 + ЕA × under a form to indicate a

3

DE2
"
2

convenient practical rule for computing the volume of the segment. Thus, since the triangle DEO is right-angled, and OE=0A-EA, we have

2

DE2 = DO' — OE' = OA - OA2 + 20A X EA-EA

By substituting this value for DE2 in the expression for the volume of the segment, we find

Hence, the volume of a spherical segment, having a single base, is measured by one third of times the square of the altitude of the segment, multiplied by the difference between three times the radius of the sphere and this altitude.

RECAPITULATION

Of some of the principles demonstrated in this and the preceding Books.

Let R denote the radius, and D the diameter of any circle or sphere, and H the altitude of a cone, or of a segment of a sphere; then,

Volume or solidity of a sphere = {«R3, or f«D3.

Volume of a spherical sector

Volume of a cone, of which

=

R2 × H.

PRACTICAL PROBLEMS.

1. The diameter of a sphere is 12 inches; how many cubic inches does it contain?

Ans. 904.78 cu. in.

2. What is the solidity of the segment of a single base that is cut from a sphere 12 inches in diameter, the altitude of the segment being 3 inches? Ans. 141.372 cu. in.

3. The surface of a sphere is 68 square feet; what is its diameter ?

Ans. D

=

4.652 feet.

4. If from a sphere, whose surface is 68 square feet, a segment be cut, having a depth of two feet and a single base, what is the convex surface of the segment?

Ans. 29.229+ sq. ft.

5. What is the solidity of the sphere mentioned in the two preceding examples, and what is the solidity of the segment, having a depth of two feet, and but one base?