In the same manner it may be proved, that the surfaces produced by the revolution of the lines BD and AP about the axis DC, are equal to

NDXcirc GC,

and COXcirc GC.

The surface of the whole solid, therefore, (Euc. 1.2.) is equal to CDXcirc GC.

But a

The demonstration is applicable to a solid produced by the revolution of a polygon of any number of sides. polygon may be supposed which shall differ less than by any given quantity from the circle in which it is inscribed; (Sup. Euc. 4. 1.) and in which the perpendicular GC shall differ less than by any given quantity from the radius of the circle. Therefore, the surface of a hemisphere is equal to the product of its radius into the circumference of its base; and the surface of a sphere is equal to the product of its diameter into its circumference.

Cor. 1. From this demonstration it follows, that the surface of any segment or zone of a sphere is equal to the product of the height of the segment or zone into the circumference of the sphere. The surface of the zone produced by the revolution of the arc AB about ON, is equal to ON×circ CP. And the surface of the segment produced by the revolution of BD about DN is equal to DNxcirc CP.

Cor. 2. The surface of a sphere is equal to four times the area of a circle of the same diameter; and therefore, the convex surface of a hemisphere is equal to twice the area of its base. For the area of a circle is equal to the product of half the diameter into half the circumference; (Art. 30.) that is, to the product of the diameter and circumference.

Cor. 3. The surface of a sphere, or the convex surface of any spherical segment or zone, is equal to that of the circumscribing cylinder. A hemisphere described by the revolution of the arc DBP, is circumscribed by a cylinder produced by the revolution of the parallelogram DdCP. The convex surface of the cylinder is equal to its height multiplied by its circumference. (Art. 62.) And this is also the surface of the hemisphere.

So the surface produced by the revolution of AB is equal to that produced by the revolution of ab. And the surface produced by BD is equal to that produced by bd.

Ex. 1. Considering the earth as a sphere 7930 miles in diameter, how many square miles are there on its surface? Ans. 197,558,500.

2. If the circumference of the sun be 2,800,000 miles, what is his surface ? Ans. 2,495,547,600,000 sq. miles.

3. How many square feet of lead will it require, to cover a hemispherical dome whose base is 13 feet across?

Ans. 265.

PROBLEM VIII.

To find the SOLIDITY of a SPHERE.

70. 1. MULTIPLY THE CUBE OF THE DIAMETER BY .5236.

Or,

2. MULTIPLY THE SQUARE OF THE DIAMETER BY OF THE CIRCUMFERENCE.

Or,

3. MULTIPLY THE SURFACE BY OF THE DIAMETER.

1. A sphere is two thirds of its circumscribing cylinder. (Sup. Euc. 21. 3.) The height and diameter of the cylinder are each equal to the diameter of the sphere. The solidity of the cylinder is equal to its height multiplied into the area of its base, (Art. 64.) that is putting D for the diameter,

DXD X.7854 or D3x.7854.
DXD2x.7854

And the solidity of the sphere, being of this, is

D3X5236.

2. The base of the circumscribing cylinder is equal to half the circumference multiplied into half the diameter; (Art.30.) that is, if C be put for the circumference,

CXD; and the solidity is C×D2.

Therefore, the solidity of the sphere is
of CxD2=D3×↓C.

3. In the last expression, which is the same as C×D×1D, we may substitute S, the surface, for CxD. (Art. 69.) We then have the solidity of the sphere equal to

Or, the sphere may be supposed to be filled with small pyramids, standing on the surface of the sphere, and having their common vertex in the center. The number of these may be such, that the difference between their sum and the sphere shall be less than any given quantity. The solidity of each pyramid is equal to the product of its base into of its height. (Art. 48.) The solidity of the whole, therefore, is equal to the product of the surface of the sphere into of its radius, or of its diameter.

71. The numbers 3.14159, .7854, .5236, should be made perfectly familiar. The first expresses the ratio of the circumference of a circle to the diameter; (Art. 23.) the second, the ratio of the area of a circle to the square of the diameter (Art. 30.); and the third, the ratio of the solidity of a sphere to the cube of the diameter. The second is of the first, and 17 the third is of the first.

As these numbers are frequently occurring in mathematical investigations, it is common to represent the first of them by the Greek letter. According to this notation,

T=3.14159,

*=.7854, | x=

=.5236.

If D-the diameter, and R=the radius of any circle or sphere;

Then, D=2R D2-4R D3=8R3.

=

Ex. 1. What is the solidity of the earth, if it be a sphere 7930 miles in diameter?

Ans. 261,107,000,000 cubic miles.

2. How many wine gallons will fill a hollow sphere 4 feet in diameter ?

Ans. The capacity is 33.5104 feet=250 gallons.

3. If the diameter of the moon be 2180 miles, what is its solidity? Ans. 5,424,600,000 miles.

72. If the solidity of a sphere be given, the diameter may be found by reversing the first rule in the preceding article; that is, dividing by .5236 and extracting the cube root of the quotient.

Ex. 1. What is the diameter of a sphere whose solidity is 65.45-cubic feet? Ans. 5 feet.

2. What must be the diameter of a globe to contain 16755 pounds of water? Ans. 8 feet.

PROBLEM IX.

To find the CONVEX SURFACE of a SEGMENT or ZONE of a

sphere.

73. MULTIPLY THE HEIGHT OF THE SEGMENT OR ZONE INTO THE CIRCUMFERENCE OF THE SPHERE.

For the demonstration of this rule, see art. 69.

Ex. 1. If the earth be considered a perfect sphere 7930 miles in diameter, and if the polar circle be 23° 28′ from the pole, how many square miles are there in one of the frigid zones?

If PQOE (Fig. 15.) be a meridian on the earth, ADB one of the polar circles, and P the pole; then the frigid zone is a spherical segment described by the revolution of the arc APB about PD. The angle ACD subtended by the arc AP is 239 28', And in the right angled triangle ACD,

Then, CP-CD-3965-3637-328-PD the height of the segment.

And 328x7930x3.14159-8171400 the surface.

2. If the diameter of the earth be 7930 miles, what is the surface of the torrid zone, extending 23° 28′ on each side of the equator?

If EQ (Fig. 15.) be the equator, and GH one of the tropics, then the angle ECG is 230 28'. And in the right angled triangle GCM,

R: CG sin ECG: GM-CN-1578.9 the height of half the zone.

The surface of the whole zone is 78669700.

3. What is the surface of each of the temperate zones? The height DN-CP-CN-PD=2058.1

And the surface of the zone is 51273000.

The spherical sector, (Fig. 24.) produced by the revolution of ACBD about CD, may be supposed to be filled with small pyramids, standing on the spherical surface ADB, and terminating in the point C. Their number may be so great, that the height of each shall differ less than by any given length from the radius CD, and the sum of their bases shall differ less than by any given quantity from the surface ABD. The solidity of each is equal to the product of its base into of the radius CD. (Art. 48.) Therefore, the solidity of all of them, that is, of the sector ADBC, is equal to the product of the spherical surface into of the radius.

Ex. Supposing the earth to be a sphere 7930 miles in diameter, and the polar circle ADB (Fig. 15.) to be 23° 28′ from the pole; what is the solidity of the spherical sector ACBP? Ans. 10,799,867,000 miles.