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Adding the equations

area ABCDE = (AG + GO + OH + HE) X 2 od

area ABCDE 2 R X 2 ad = 4 Rud. Now let the number of sides of the inscribed polygon be increased indefinitely. Then the area ABCDE approaches S, and d approaches R as a limit. Hence S=47R.

(213)

Q.E.D.

759. Cor. 1. The areas of the surfaces of two spheres are to each other as the squares of their radii.

760. Cor. 2. The area of a zone is equal to its altitude multiplied by the circumference of a great circle.

The zone generated by the revolution of arc BC is equal to GO X 2 R. (See diagram for Prop. XXVII.)

761. Cor. 3. Zones on the same or on equal spheres are to each other as their altitudes.

762. SCHOLIUM. Prop. XXVII, in connection with Props. XXIV, XXV, and 746, enables us to find the absolute size of spherical surfaces.

Ex. 1222. Find the surface of a sphere whose radius is 10.

Ex. 1223. Find the surface of a zone whose altitude is 5, if the radius of the sphere is 20.

Ex. 1224. Find the number of square feet in the surface of a lune whose angle is 40°, if the radius of the sphere is 20 ft.

Ex. 1225. Find the number of square feet in the surface of a spherical triangle, if its angles are respectively equal to 80°, 90°, and 100°, if the radius of the sphere is 20 ft.

Ex. 1226. Find the number of square feet in a hemispherical dome whose diameter is 40 ft.

Ex. 1227. The surface of a sphere is 100 square meters. Find the radius.

Ex. 1228. Find the angle of an equilateral spherical triangle on a sphere whose radius is 2, if the area of the triangle is equal to a.

SPHERICAL VOLUMES

763. DEF. A spherical sector is the solid generated by any plane sector of the semicircle when the semicircle revolves about its diameter.

764. The base of the spherical sector is the zone generated by the revolution of the arc.

A

B В

D

Thus, the solid generated by the revolution of AOB about OD is a spherical sector.

765. DEF. A spherical pyramid is a portion of a sphere bounded by a spherical polygon and the faces of the corresponding polyedral angle.

The spherical polygon is called the base of the pyramid.

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Thus, OABCD is a spherical pyramid having base ABCD.

766. DEF. A spherical segment is a portion of a sphere bounded by two parallel planes.

The sections (circles) made by the parallel planes are the bases, and the distance between the planes is the altitude of the segment.

If one of the bounding planes is tangent to the sphere, the segment is called a segment of one base.

767. DEF. A spherical wedge (or ungula) is the portion of a sphere bounded by a lune and the planes of its sides.

PROPOSITION XXVIII. THEOREM

To prove

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

Hyp. V is the volume of a sphere whose radius is R and whose surface is S.

V=SXR Proof. Conceive any polyedron circumscribed about the sphere, and let each vertex of the polyedron be joined to the center of the sphere.

The polyedron is then divided into pyramids, each having one face as a base and Ras the common altitude.

Therefore, the volume of the polyedron is equivalent to the sum of the bases multiplied by one-third of R, or it is equivalent to the surface multiplied by one-third of the radius.

But by increasing the number of faces of the polyedron indefinitely, its volume will approach the volume of the sphere and its surface, the surface of the sphere as a limit. Hence V=SxjR.

(213)

Q.E.D.

769. Cor. 1. If V denotes the volume, R the radius, and D the diameter of the sphere, V= Rs or V

6

DPT

770. COR. 2. The volumes of two spheres are to each other as the cubes of their radii.

771. Cor. 3. The volume of a spherical pyramid is equal to the area of its base multiplied by one-third the radius of the sphere.

772. Cor. 4. The volume of a spherical sector is equivalent to the area of its base multiplied by one-third the radius of the sphere.

The proof is analogous to the proof of Prop. XXVIII.

773. Cor. 5. If h denotes the altitude of a spherical sector, V its volume, and Z the surface of the base,

R
V= 2 x

3

· But

Z=2R Xh.

(760)

.. V=R’h.

774. Cor. 6. The volume of a sphere is to the volume of the circumscribed cylinder as two to three.

For the volume of the circumscribed cylinder is 2 #R?.

Ex. 1229. Find the volume of a sphere whose radius is 10.

Ex. 1230. Find the volume of a spherical sector whose base is 20, if the radius of the sphere is 10.

Ex. 1231. Find the radius of a sphere whose volume is 100 cu. ft.

Ex. 1232. Find the volume of a sphere whose surface is equal to 10 square meters.

Ex. 1233. Find the surface of a sphere whose volume is equal to M.

PROPOSITION XXIX. - PROBLEM

775. To find the volume of a spherical segment.

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Given. A spherical segment generated by the revolution of ABCD about EF.

Required. To find the volume V of the spherical segment, if AB=1, CD = 11, and CA = h.

Join OD and OB.

Then the volume generated by ABCD is equal to the spheri. cal sector generated by DBO, plus the cone generated by ODC, minus the cone generated by BAO. Hence

V= { #Roh + } #CD.CO - **AB. AO,

= * [2 Roh +(R? COʻ) CO (RA0°) AO]

= { + [2 R’h + R*(CO A0) (CO® A03]. But CO AO=h, and COAO can be factored, hence: V= { 7 [2 Roh + Rh n(COʻ + CO A0+ A0%]

пh V: [3 R? (CO' + CO X A0+ A0%]. .. (1) But

k' =(CO AO) = COʻ– 2 CO X A0+ AO", and COʻ+ CO X AO + AO

2 OC. AO ={COʻ+ } AO

AO-(CO ;40+40)

h2 = }(COʻ+ A03 20)

2

h2
= { ( 22 — p2 + RP 7,2).
}–

2

h = 3 R2 3(1879)

2
Hence by substituting this value in (1)

ha
V=
3

2
пh

ach3 (702 + ri?) +

sphe

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