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Fig.150.

Fig.147.

Fig.151.

Fig.149.
Fig.152.

ing triangle DEF of the other ; and we say—Two sym-
metrical spherical triangles are equal in area.
299. The two opposite triangles CMK, LFI, (fig. 147)
being symmetrical, are equal in area ; if we add to each of
these equal areas the triangle CMI, we shall have CMK
-- CMI = CMI + LFI; but the triangles CMK, CMI,
constitute the lunary surface whose angle is CIM ; the
sum of two triangles CMI, LFI, in the nearer hemisphere,
is therefore equivalent to this lunary surface. We have
therefore this general rule:—If two semi-circumferences
cut each other in a hemisphere, the sum of two opposite
triangles will be equal to a lunary surface whose angle
is equal to the angle made by the semi-circumferences.
300. Let ABC (fig. 151) be a spherical triangle, and
DEFGHI the circumference of a great circle which does
not cut this triangle. Produce the sides of the triangle both
ways till they meet this circumference; and we shall
have IBD + GBF = Lum.B, HAG + DAE = Lun.A.,
FCE + HCI = Lun. C. It is manifest that the sum of
these sia triangles is equal to the hemisphere plus twice
the triangle ABC, that is, equal to 2 ABC + Lun. 2D;
we shall therefore have 2 ABC + Lun. 2 D = Lun.
(A + B + C), or ABC = Lun. (A + B + C – 2 D).
We say therefore—The area of a spherical triangle is
equivalent to half the lunary surface whose angle is equal
to the sum of the three angles of the triangle minus two
right-angles.
Suppose a, b, c, to designate the arcs which measure
the angles A, B, C, of any spherical triangle; and let C
denote the circumference of a great circle, and R the
radius. We shall have, for the area of the spherical tri-
angle, R × (a + b + c # C). * .. -
301. To ascertain the volume of the body generated
by the revolution of the polygonal sector a b c d'O (fig.
149), it will be necessary to examine the body generated
by the isosceles triangle ABO (fig. 152) revolving about
the straight line OD as an axis. Produce AB till it
meet the axis in D; and from the points A, B, draw
AM and BN, perpendicular to OD.
The volume of the body generated by the triangle
AOD, will have for its measure ; it X (AM)” X OD,
(174). The body generated by the triangle OBD, has
for its measure ; it (BN)* × OD; therefore the differ-
ence of these bodies, or the body generated by ABO,

will have for its measure ; it. [(AM*) — (BN*)] × OD. Fig.152. This expression may take another form. From I, the middle of AB, draw IK perpendicular to OD, and through B draw BP parallel to OD; we shall have AM -- BN = 2 IK, and AM – BN = AP ; consequently (AM + BN) × (AM – BN) = (AM)” — (BN)*, (176), - 2 IK X AP. The volume of the body generated by ABO, may therefore be expressed by # or X IK X AP × OD. But if we draw OI, as OAB is an isosceles triangle, OI will be perpendicular to AB; and the two triangles APB, OID, are similar, and give

. OI OD the proportion AP of AB hence AP × OD = OIX

AB; and substituting OIX AB for its value in the above expression we have ; X IK X AB. × OI. But the similar triangles ABP, OIK, give the proportion

A : - #. wherefore AB X IK = OIX BP = OI X MN; and substituting this value of AB × IK in the expression for the volume of the body in question, we have # or X OI2 × MN. The area of a circle whose radius is OI is expressed by it X (OI)”. We therefore say— The volume of a body generated by the revolution of an s isosceles triangle about a straight line which meets it only in its summit, and is in the same plane, has for its measure fu'o thirds of the area of a circle whose radius is the height of the triangle, multiplied by the part of the avis intercepted by perpendiculars drawn to it from the vertices at the base of the triangle.

302. [If the axis of revolution is parallel to the base of the triangle, MN will equal AB, the surface generated by AB will be a cylindrical surface, and the volume of the body generated by the triangle will be two thirds of the volume of a cylinder the radius of whose base is OI, the height of the triangle, and whose height is AB, the base of the triangle. It follows, therefore, that the volume of a body generated by the revolution of the same triangle about its base as an axis, will be equivalent to one third of the cylinder of the above dimensions.]

303. Returning to figure 149, it will be readily seen that the portion of the polygon a b c d, inscribed in the quadrant ad is composed of isosceles triangles whose common height is the radius of the inscribed circle. The volume of the body generated by a Ob, will have for its

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Fig.149.

Fig.148.

measure # It X (Oh)" × a e ; that generated by b Oc
will have # It X (Oi)° X ef; and that generated by
dOc will have 3 or X (O !)* >< f(). As Oh, Oi and Ol
are equal, the sum of these bodies, or the body generat-
ed by the portion of the inscribed polygon a b c d, will
have for the meaaure of its volume, # 1 × (Oh)” X (a e --
e f-- f O), or ; at X (O h)” X a O ; and if the portion of
the inscribed polygon occupied the entire semicircle, the
body generated by it, in a revolution about the diameter
a p, will have for its measure # It X (O h)* X a p.
If we apply this result to the body generated by the
circumscribed polygon, we shall have for the measure of
its volume, 3 or X (a O)2 × AP.
By increasing the number of sides of the polygons,
the inscribed polygon increases continually, and the limit
of this increase is the semicircle ; but with this increase
of the number of sides, the factor Oh increases, and
the limit of this increase is O a , and as the other factors

remain the same, the limit for the product is ; it x

(O a)2 × a p.
But by increasing the number of sides of the exterior
polygon, it becomes continually less, and the limit of
this diminution is the semicircle; but with this increase
in the number of sides, the factor AP is continually di-
minished, and the limit of this diminution is a p ; and
as the other factors remain the same, the limit of the
product expressing the volume of the body generated by
the portion of the polygon circumscribed about the semi-
circle, is # It X (Oa)* X a p. It appears therefore that
# It X (O a)* X a p, cannot be the measure of a volume
greater or less than that of the body generated by the
semicircle whose radius is a O, revolving about the di-
ameter ap ; but this body is the sphere whose radius
is a O.
lf we denote a O by R and a p by 2 R, the expres-
sion becomes ; at X R* X 2 R, or # 1 × R°; and may
take the form 4 at . R* X R. But 4 Ti R2 is the area
of the sphere whose radius is R ; we therefore say—The
sphere has for the measure of its volume, the area of its
surface, multiplied by one third of its radius; or the area
of a great circle multiplied by two thirds of the diameter.
304. If we have a cone and a cylinder whose common

height is equal to the diameter of the sphere (fig. 148), and whose common base is equal to a great circle of the Fig.148. sphere; we may express these convex surfaces as sollows:

that of the cone (5): 7t. R2 ; sphere 4 it. R2 ; cylinder = 4 at . R2. The volumes of these bodies have the following expressions:

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Remark. The sphere is therefore two thirds of the circumscribed cylinder. The cone is one third of the circumscribed cylinder. The area of the sphere is equivalent to that of the convex surface of the circumscribed cylinder.

305. The bodies of revolution which have now been discussed, are said to be similar when the figures which generate them are similar. Two cones generated by similar triangles revolving about homologous sides, have their heights in proportion to the radii of their bases.

Let A represent the area of a come whose height is H and the radius of whose base is R, and whose side is S; we shall have the equation d = 2 it. B. × 4 S, (272). For another similar cone we shall have a = 2 it. r X , s. If we compare these expressions, cancelling the common

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. A may write - = - We therefore say— (s.

7.2 T s 2 T /;2 The areas of similar comes are to each other as the seeond powers of their corresponding dimensions. 306. Two similar cylinders are generated by similar rectangles (fig. 144). Their areas will have for their ex- Fig.144. pressions, A = 2 it. It X H, a = 2 it. r X. h. If we compare these neglecting the common factors in the terms of

H R2 - 2 the second ratio, we shall have #= R × H_ _II*.

r x sh T r2 T h” that is—The areas of similar cylinders are to each other as the second powers of their homologous measures. 307. All spheres are similar, because every sphere has for its generatrix a semicircle, and all semicircles

are similar. The areas of two spheres have for their

expressions, A = 4 it. R*, a = 4 it. r2 ; comparing these

and cancelling the common factors in the second ratio, 2

... . A R g will give a = z: ; that is—The areas of two spheres

are to each other as the second powers of their radii. 308. If we compare the volumes of two similar cones, which are expressed as follows; V = } it. R* x H, and V R2 × H R 3 V = } t. ro X / ; we shall have — = sco × 11 io — * Q) r2 × h 7.3

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= 7; as, r = 7 : that is—The volumes of similar cones are to each other as the third powers of their homologous lines. 309. The volumes of two similar cylinders have for their expressions, V' = r. R2 × H, v' = r. ro X. h. If

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volumes of similar cylinders are to each other as the third powers of their homologous measures. 310. Two spheres, being always similar bodies, and having for the expressions of their volumes, V" = # It. R*, and !!! - 3 or”, oy be readily compared; 3 : and will give v. ~ R* D". That is—The volQ) 3 als umes of two spheres are to each other as the third powers of their radii, or as the third powers of their diameters.

311. It was observed (227) that a regular polyedron has all its laces equal regular polygons, all its diedral angles equal, and all its polyedral angles equal. Let us inquire into the number of regular polyedrons which may be constructed. in the first place, it is manifest that the faces of these regular bodies must be equilateral triangles, squares or regular pentagons; for the angles of a regular hexagon are equal to four right angles; no polyedral angle therefore, can be formed with regular hexagons, or with regular polygons of a greater number of sides. Of those bodies contained by equilateral triangles, the

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