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PROPOSITION XXXV. THEOREM.

784. The area of a zone is equal to the product of its altitude by the circumference of a great circle.

A

B

E

D

Let ABCDE be the circumference of a great circle, BC any arc of this circumference, and OA the radius of the sphere. And, let PR be the altitude of the zone generated by arc BC.

We are to prove

=

zone BC PR X 2 п 0 А.

If the semicircle A B C D revolve about the diameter A D as an axis, the semi-circumference A B C D will generate the surface of a sphere; the arc B C, a zone,

and the chord B C, a surface whose area is PR × 2 π OI. §759

Now if we bisect the arc B C, and continue this process indefinitely, the surface generated by the chords of these arcs will approach the zone as its limit;

the OI will approach the radius of the sphere as its limit; while PR will remain constant.

.. at the limit, zone BC= PRX 2 π OA.

Q. E. D.

785. COROLLARY 1. Zones on the same sphere, or equal spheres, have the same ratio as their altitudes.

786. COR. 2. A zone is to the surface of the sphere as the altitude of the zone is to the diameter of the sphere.

=

П

A P

787. COR. 3. Let arc A B generate a zone of a single base. Then, zone A B = AP X 2 π O A. Hence, zone A B XAD π A B2. (§ 307.) That is, a zone of one base is equivalent to a circle whose radius is the chord of the generating arc.

=

ON THE VOLUME OF THE SPHERE.

PROPOSITION XXXVI. THEOREM.

788. The volume of a sphere is equal to the area of its surface multiplied by one-third of its radius.

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Let R be the radius of a sphere whose centre is 0,

surface, and V its volume.

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Conceive a cube to be circumscribed about the sphere.

S its

From O, the centre of the sphere, conceive lines to be drawn to the vertices of each of the polyhedral A, B, C, D, etc. These lines are the edges of six quadrangular pyramids, whose bases are the faces of the cube, and whose common altitude is the radius of the sphere.

The volume of each pyramid is equal to the product of its base by its altitude.

§ 574

.. the volume of the six pyramids, that is, the volume of the circumscribed cube, is equal to the surface of the cube multiplied by

R.

Now conceive planes drawn tangent to the sphere, cutting each of the polyhedral of the cube.

We shall then have a circumscribed solid whose volume will be nearer that of the sphere than is the volume of the circumscribed cube.

From O conceive lines to be drawn to each of the polyhedral of the solid thus formed, a, b, c, etc.

These lines will form the edges of a series of pyramids, whose bases are the surface of the solid, and whose common altitude is the radius of the sphere;

and the volume of each pyramid thus formed is equal to the product of its base by its altitude.

.. the sum of the volumes of these pyramids, that is, the volume of this new solid, is equal to the surface of the solid multiplied by R.

Now, this process of cutting the polyhedral by tangent planes may be considered as continued indefinitely,

and, however far this process is carried, it will always be true that the volume of the solid is equal to its surface multiplied by R.

+

But the sphere is the limit of this circumscribed solid.

...V=SX R.

П

$199

Q. E. D.

789. COROLLARY 1. Since S4 π R2 (§ 762), V=4π R2 X If we denote the diameter of the sphere by

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790. COR. 2. Denote the radius of another sphere by R' and

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That is, spheres are to each other as the cubes of their radii.

R3

R/8

791. COR. 3. The volume of a spherical sector is equal to the product of the area of the zone which forms its base by one-third the radius of the sphere.

Let R denote the radius of a sphere, C the circumference of a great circle, H the altitude of the zone, Z the surface of the zone, and the volume of the corresponding sector.

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792. COR. 4. The volumes of spherical sectors of the same sphere, or equal spheres, are to each other as the zones which form their bases, or as the altitudes of these zones.

For, let V and V denote the volumes of two spherical sectors, Z and Z' the zones which form their bases, H and H' the altitudes of these zones, and R the radius of the sphere.

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793. COR. 5. The volume of a spherical segment of one base, less than a hemisphere, generated by the revolution of a circular sector AO B about the diameter A D, may be found by subtracting the volume of the cone of revolution generated by OBC from that of the spherical sector A O B.

In like manner, the volume of a spherical segment of one base, greater than a hemisphere, generated by the revolution of

A B'C' may be found by adding the volume of the cone of revolution generated by O B'C' to that of the spherical sector generated by A O B'.

794. COR. 6. The volume of a spherical segment of two bases, generated by the revolution of C B B'C' about the diameter A D, may be found by subtracting the volume of the segment of one base generated by ABC from that of the segment of one base generated by A B' C.

EXERCISES.

1. Given a sphere whose diameter is 20 inches; find the circumference of a small circle whose plane cuts the diameter 4 inches from the centre.

2. Construct, on the spherical blackboard, spherical angles of 30°, 45°, 90°, 120°, 150° and 135°.

3. Construct, on the spherical blackboard, a spherical triangle, whose sides are 100°, 80° and 70° respectively. What is true of its polar triangle?

4. Find the surface and volume of a sphere whose radius is 10 inches; also find the area of a spherical triangle on this sphere, the angles of the triangle being 80°, 85° and 100° respectively.

5. If 7 equidistant planes cut a sphere, each perpendicular to the same diameter, what are the relative areas of the zones ?

6. Given, two mutually equiangular triangles on spheres whose radii are 10 inches and 40 inches respectively; what are their relative areas?

7. Let V denote the volume of a spherical pyramid, S its base, E the spherical excess of its base, and R the radius of the sphere; show that S

=

R2 E, and V

=

π

TR3 E.

8. Given, the volume of a sphere 1728 inches; find its radius.

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