be of the trirectangular triangle. Now, the trirectangular triangle being of the surface of the sphere (577) is of 4πR2, or R. This multiplied by gives R, the same as above.

The proportion,

ABC surf. of hemisph. :: A + B + C 180° : 360°,

is readily put into a form which agrees with the enunciation as given in this scholium. Thus, surf. of hemisph. = 2πR3, whence

615. Theorem.-The volume of a sphere is equal to the area of its surface multiplied by of the radius, that is, πR', R being the radius.

DEM.-Let OLR be the radius of a sphere. Conceive a circumscribed cube, that is, a cube whose faces are tangent planes to the sphere. Draw lines from the vertices of each of the polyedral angles of the cube, to the centre of the sphere, as BO, CO, DO, AO, etc. These lines are the edges of six pyramids, having for their bases the faces of the cube, and for a common altitude the radius of the sphere (?). Hence the volume of the circumscribed cube is equal to its surface multiplied by R.

G

FIG. 352.

F

Again, conceive each of the polyedral angles of the cube truncated by planes tangent to the sphere. A new circumscribed solid will thus be formed, whose volume will be nearer that of the sphere than is that of the circumscribed cube. Let abc represent one of these tangent planes. Draw from the polyedral angles of this new solid, lines to the centre of the sphere, as ao, bo, and co, etc.; these lines will form the edges of a set of pyramids whose bases constitute the surface of the solid, and whose common altitude is the radius of the sphere (?). Hence the volume of this solid is equal to the product of its surface (the sum of the bases of the pyramids) into R.

Now, this process of truncating the angles by tangent planes may be conceived as continued indefinitely; and, to whatever extent it is carried, it will always be true that the volume of the solid is equal to its surface multiplied by R. Therefore, as the sphere is the limit of this circumscribed solid, we have the volume of the sphere equal to the surface of the sphere, which is 47R2, multiplied by R, i. e., to πR'. Q. È. D.

616. COR.-The surface of the sphere may be conceived as consisting of an infinite number of infinitely small plane faces, and the volume as composed of an infinite number of pyramids having these faces for their bases, and their vertices at the centre of the sphere, the common altitude of the pyramids being the radius of the sphere.

617. A Spherical Sector is a portion of a sphere generated by the revolution of a circular sector about the diameter around which the semicircle which generates the sphere is conceived to revolve. It has a zone for its base; and it may have as its other surfaces one, or two, conical surfaces, or one conical and one plane surface.

A a B

ILL-Thus let ab be the diameter around which the semicircle aCb revolves to generate the sphere. The solid generated by the circular sector AOa will be a spherical sector having a zone (AB) for its base; and for its other surface, the conical surface generated by AO. The spherical sector generated by COD, has the zone generated by CD for its base; and for its other surfaces, the concave conical surface generated by DO, and the convex conical surface generated by CO. The spherical sector generated by EOF, has the zone generated by EF for its base, the plane generated by EO for one surface, and the concave conical surface generated by FO for the other.

b
FIG. 358.

618. A Spherical Segment is a portion of the sphere included by two parallel planes, it being understood that one of the planes may become a tangent plane. In the latter case, the segment has but one base; in other cases, it has two. A spherical segment is bounded by a zone and one, or two, plane surfaces.

PROPOSITION XXXII.

619. Theorem.-The volume of a spherical sector is equal to the product of the zone which forms its base into one-third the radius of the sphere.

DEM.-A spherical sector, like the sphere itself, may be conceived as consisting of an infinite number of pyramids whose bases make up its surface, and whose common altitude is the radius of the sphere. Hence, the volume of the sector is equal to the sum of the bases of these pyramids, that is, the spherical surface of the sector, multiplied by one-third their common altitude, which is one-third the radius of the sphere. Q. E. D.

620. COR.-The volumes of spherical sectors of the same or equal spheres are to each other as the zones which form their bases; and, since these zones are to each other as their altitudes (604), the sectors are to each other as the altitudes of the zones which form their bases.

PROPOSITION XXXIII.

621. Theorem.—The volume of a spherical segment of

one base is 7А'(RA), A being the altitude of the segment, and R the radius of the sphere.

DEM.-Let CO = R, and CD = A; then is the volume of the spherical segment generated by the revolution of CAD about CO

For, the volume of the spherical sector generated by AOC is the zone generated by AC, multiplied by R, or 2πAR × R = πAR2. From this we must subtract the cone, the radius of whose base is AD, and whose altitude is DO. To obtain this, we have DO =R- A: whence, from the right angled triangle ADO, AD = √/R2 — (R − A)2 = √/2AR — A2. Now, the volume of this cone is

FIG. 354.

OD × πAD2, or π(R — A) (2AR — A2) = }π(2AR2 — 3A3R + A3). Subtracting this from the volume of the spherical sector, we have

πAR - π(2AR? 3A'R + A3) =

622. SCH.-The volume of a spherical segment with two bases is readily obtained by taking the difference between two segments of one base each. Thus, to obtain the volumes of the segment generated by the revolution of bCAc about aO, take the difference of the segments whose altitudes are ac and ab.

a

EXERCISES.

1. What is the circumference of a small circle of a sphere whose diameter is 10, the circle being at 3 from the centre?

Ans., 25.1328.

2. Construct on the spherical blackboard a spherical angle of 60°.

Of 45°. Of 90°. Of 120°. Of 250°.

SUG's. Let P be the point where the vertex of the required angle is to be situated. With a quadrant strike an arc from P, which shall represent one side of the required angle. From P as a pole, with a quadrant, strike an are from the side before drawn, which shall measure the required angle. On this last arc lay off from the first side the measure of the required angle,* as 60°, 45°, etc. Through the extremity of this arc and P pass a great circle (548). [The stu dent should not fail to give the reasons, as well as do the work.]

3. On the spherical blackboard construct a spherical triangle ABC, having AB 100°, AC = 80°, and A 58°. =

=

- 4. Construct as above a spherical triangle ABC, having AB = 75°, A = 110°, and B = 87°.

5. Construct as above, having AB = 150°, BC = 80°, and AC = 100°. Also having AB = 160°, AC = 50°, and BC = 85°.

6. Construct as above, having A= 52°, AC = 47°, and CB = 40°.

SUG'S.-Construct the angle A as before taught, and lay off AC from A equal to 47°, with the tape. This determines the vertex C. From C, as a pole, with an arc of 40°, describe an arc of a small circle; in this case this arc will cut the opposite side of the angle A in two places. Call these points B and B'. Pass circumferences of great circles through C, and B, and B'. There are two triangles, ACB and ACB'.

ure.

NOTE. The teacher can multiply examples like the three preceding at pleasThis exercise should be continued till the pupil can draw a spherical triangle as readily as a plane triangle.

7. What is the area of a spherical triangle on the surface of a sphere whose radius is 10, the angles of the triangle being 85°, 120°, and 150° ? Ans., 305.4 +.

8. What is the area of a spherical triangle on a sphere whose diameter is 12, the angles of the triangle being 82°, 98°, and 100°? 9. A sphere is cut by 5 parallel planes at 7 from each other. What are the relative areas of the zones?

10. Considering the earth as a sphere, its radius would be 3958 miles, and the altitudes of the zones, North torrid = 1578, North temperate = 2052, and North frigid=328 miles. What are the relative areas of the several zones?

SUG.-The student should be careful to discriminate between the width of a zone, and its altitude. The altitudes are found from their widths, as usually given in degrees, by means of trigonometry.

* For this purpose a tape equal in length to a semicircumference of a great circle of the #phere used, and marked off into 180 equal parts, will be convenient. A strip of paper may be used.

11. The earth being regarded as a sphere whose radius is 3958 miles, what is the area of a spherical triangle on its surface, the angles being 120°, 130°, and 150°? What is the area of a trirectangular triangle on the earth's surface?

12. Construct on the spherical blackboard a spherical triangle ABC, having A=59°, AC = 120°, and AB 88°. Then construct the triangle polar to ABC.

=

13. Construct triangles polar to each of those in Examples 3, 4, and 5.

14. In the spherical triangle ABC given A = 58°, B = 67°, and AC = 81°; what can you affirm of the polar triangle?

15. What is the volume of a globe which is 2 feet in diameter? What of a segment of the same globe included by two parallel planes, one at 3 and the other at 9 inches from the centre ?

16. Compare the convex surfaces of a sphere and its circumscribed right cylinder and cone, the vertical angle of the cone being 60°. 17. Compare the volumes of a sphere and its circumscribed cube, right cylinder, and cone, the vertical angle of the cone being 60°.

18. If a and b represent the distances from the centre of a sphere whose radius is r, to the bases of a spherical segment, show that the volume of the segment is [r (b − a) — 1(b3 — a')]. See (621, 622).