MENSURATION. 



Mensura- 

 tion of 



r\ R . its, 



17. 



*50 



The base =. 7854 X 18.5-=268. 803 sq. inches, Part 

 1. Prob. 9. 



Therefore the solid content=2G8.S03 x 8=2150.121 

 cubic inches. 



PROBLEM V. 



To find the solid content of any pyramid or cone. 



HULK. Multiply the area of the base by the height, 

 and one third of the product is the solid content. 



This rule has been proved in GEOMETRY, Part. II. 

 Sect. 2. Prop. 17. and Sect. 3. Prop. 3. 



Ex. Each side of the base of a triangular pyramid is 

 3 feet, and its height is 10 feet : What is its solid con- 

 tent? 



By Prob. 2. of Sect. 1. the area of the base= 3.897 11 

 square feet. 



Hence f^lL^JL ^ 33.97 11 cub. feet the solid 



Fig. 18. 



The greater end 

 The lesser 

 The mean = ./(41.57 X 23.38) 



= 41.57 

 = 23.38 

 = 31.18 



3)96.13 



The mean area 32.04. 

 32.04 X 9 =288.36 cubic feet the solid content. 



PROBLEM VII. 



To find the surface of a sphere, or of any segment or 

 zone of it. 



RULE. Multiply the circumference of the sphere by 

 the height of the part required, and the product will 

 be the curve surface, whether it be a segment, a zone, 

 or the whole sphere. 



This rule has been investigated in FLUXIONS ( 163.) 

 We shall here give a different investigation. 



Let HIKL be a square described about a circle, and Fi 2- 1!) - 

 AB a diameter joining two opposite points of contact. 

 Take D d an indefinitely small arc, and draw DE, d e 

 perpendiculars to AB, and produce them to meet the 

 side of the square in F andf. Suppose now the circle 



content. 



Ex. 2. The diameter of the base of a cone is 8 inches, 

 and its height is a foot . What is its content ? 



The area of the base=.7854 x 8=6.2832 sq. inches. 



6.2832x12 f and square to revolve about AB as an axis; the cir- 



And the solid content = =25.1328 cub. cum ference will generate a spherical surface, and the 



side of the square will generate a cylindric surface. 

 Let s denote the spherical zone generated by the arc 



PROBLEM VI. d D, and c the corresponding cylindric surface gene- 



rated by the straight line F/; also, put n for 3.1416. 

 Then, since D d, on account of its smallness, may be 



inches. 



To find the solid content of the frustum of a pyra- 

 mid or cone. 



RULE. Add into one sum the areas of the two ends 

 and the mean proportional between them, (that is the 

 square root of their product) and one-third of that sum 

 will be a mean area, which, multiplied by the perpen- 

 dicular height of the frustum will give the solid con- 

 tent. 



Litcstigatioji. Let ABCD be the base of the frustum, 

 EFGH its top, P the vertex of the pyramid, and PM 

 a perpendicular on the base, meeting the top in- L. 



Let S denote the side of a square equal to the base, 

 5 the side of a square equal to the top, put H to denote 

 LM the height of the frustum, and put r for PL the 

 remainder of the perpendicular. By last problem, the 

 content of the whole pyramid is -fS'^H-fO' and the 

 content of the part above the frustum is iiV, hence the 

 frustum, which is their difference, is 



, 



reckoned a straight line, s may be regarded as the sur- 

 face of a frustum of a cone. Hence, by Prob. 1. of 

 Part I., and Prob. 3. of this Part. 



That is, because DE and d e are almost equal, 



=*S-H-ri(S-M) (S Or. 

 Now the base ABCD=S 2 , and the top EFGH=s 2 being 



similar figures (GEOM. Part II. Sect. 2. Prop. 13.) they diameter ? 



In like manner, cr:2 X FE X F / ; 

 Therefore, s : c : : DE X D d : FE X F/ 

 Draw DG perpendicular to d e, and DC to the cen- 

 tre, and because of the similar triangles DG d, DEC, 



D d : DG : : CD : DE ; or D d : F/: : EF : ED 

 Hence DE x D d=FE x Ff, and therefore s=c : 

 Thus it appears, that the corresponding indefinitely 

 small elements of the spherical and cylindric surfaces 

 are always equal, and hence, that any finite portions of 

 them comprehended between planes perpendicular to 

 the axis AB will be equal ; so that the truth of the rule 

 is evident. 



Ex. I. What is the superficies of a globe 6 feet in 



are to one another as the squares of their like sides ; 

 (GEOM. PART I. Sect. 4. Prop. 27.) that is, S 4 : ss : : 

 AB-' : EF^ ; hence S : s : : AB : EF. But because of 

 the similar triangles PAB, PEF, AB : EF : : AP : EP, 

 and, again, (PART II. Sect. 1. Prop. 12.) AP: EP : : 

 (MP : LP : :) H + r : r, therefore, S : s : : H + r : r, and 

 S s : s: : H : r, hence, (S s) r=*H. This value of 

 (S *) r being substituted in the expression given 

 above for the solidity of the frustum, it becomes 

 *S a H+-J(S+s) *H=f(S 1 +Si-|-* S! )H. 



If it be now remarked that the product or rectangle 

 S* is a mean proportional between S-' and 2 , the top 

 and bottom of the frustum, it will appear that the for- 

 mula just found gives the rale. 



Ex. The areas of the ends of a frustum of a pyramid 

 w cone are 41.57 nd 23 88 square feet, and its height 

 i 9 feet : Find its solid content. 



First 6 X 3.14l6=:18.8496=the circumference. 



Then 18.8496x6=113.0976 square feet the super- 

 ficies. 



Ex. 2. What is the convex surface of a segment, 2 

 feet in height, and cut off from the same globe. 

 18.8496 X 2=37.6992 square feet the surface. 



PROBLEM VIII. 



To find the solid content of a sphere. 



RULE I. Multiply the area of a great circle of the 

 sphere by the diameter, and -f. of the product is the so- 

 lid content. 



RULE II. Multiply the cube of the diameter by the 

 decimal .5236, and the product is the content. 



The first rule has been demonstrated in GEOMETRY, 

 Part II. Sect. 3. Prop. 6. The second is deduced from 

 the first thus : put d for the diameter, then .7854 rf ? = 



