M E X S r R A T I N. 



51 



area of a great < 



S. 9. Part I.) and by Rule I. 



,_ _tbesoHdtty. 



What ii the content of a: sphere 2 feet in dia- 

 """' : -er? 



Answer 2* x. 5236=4.1888 cubic feet. 



PROBLEM IX. 



To find the solid content of a pherical segment. 

 RULE. From three times the diameter of the sphere, 

 take twice the height of the segment, then multiply 

 the remainder by the square of the height, and the pro- 

 duct by the decimal .58<i for the content. 



This rule may be derived from Ex. 3. Art. Ml. 

 : IT it may be found in a more elementary 

 form, (GEOMETRY, part ii. Sect. 3, prop. 5.) as follows: 

 M. -a square described about a qnajdrant of 



a circle, and CE toe diagonal drawn to the centre. 

 Draw (il)HU perpendicular to t! F, meeting 



the diagonal in O, the quadiantal arc in H, and the 

 tide of the square in R. Conceive the square to revolve 

 about CF as an axis ; then I'. I., the side of the square, 

 will generate a cylinder, CE, ' 



cone, and the quadrantal a- .--rate a he- 



misphere, having the common axit C ! . Moreover the 

 .11 generate a plane, the sections of which, 

 with . and the cone, will 



. !!. and GO for their radii, 

 dent, from the above quoted 

 in geometry, that any section of the cylin- 





; f 



x* 



Ex. Tf BC, the rndfus of the bae of a paraboloid, be 

 5, anil A !',. i: heiffht, be 12 feet, what U its content ? 

 Fi r - 785.+ the area of the base ; 



NV . V feet the solidity. 



PROBLEM XI. 



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

 loid. 



RCLE. Add together the areas of the circular ends ; Fig. 21. 

 then multiply the sum by the height of the frustum, 

 and take half the product for its solid content. 



To prove thi* rule, put A and a for the ends, A for 



the height of the frustum, and c for the height wanted 



liaralxjloid. By last problem, the con- 



solid would be ^ A (A+c) and 



that of the part cut off J ac, therefore the content of the 

 frustum is 



will be equal to the turn of the corresponding sec- 

 tion? of the cone and sphere. Hence if we conceive 

 the three solid* to be made up of very thin cylinders 

 having the**. sections for their bav- <. s that any 



portion of the cylinder comprehended between two 

 plane*, parallel to it* base, will be equal to the sum of 

 the correanondipg portion* of the hesniapher* and cone. 

 Put d fat the diameter of the sphere, A for FG the 

 cowmen height of the cy 1m ' frus- 



tum KE'O'O, ead the spherical segment I II 1 1 , and 

 for the number -7 *.'!. Then the area of the com- 

 mon beae of the cylinder EE'R'R and conic frustum 

 .,;, and because GO=GC, the diameter 

 of the top of the frustum will be di k, and its area 

 n(d * A)* ; also the mean proportional between the top 

 **4 bottom will be *(<*_ A) Therefore ( by Prub. 

 3,) the solid content of the frustum is 



=(</* _tiA+*A>) 

 Now the solid content of the cylinder is nf A (by Prob. 

 Therefore the spherical segment, (which is the 

 difference of the cylinder and conic fruttum) is 



(fk rf' 



Thi* last formula is the analytic expression of the rule. 



In a sphere whose diameter it 21 inches, what 

 is the solid content of a segment whose height is 4.5 

 inches? 



First, 3 X*l X 4.5=54. 



Then 54 x * 5 X -5 x .4 136=57.25566 inches the 

 cosstent required. 



PROBLEV 

 To find the solid content of a ar.. 



duced by th 



and take ha 

 I hit rule 



rotation of a parabola about i 

 'ie area of the base by 

 the product for the cont- 

 a* been investigated in Flaxioiu, 



I'.nt from the nature of the parabola A : n :: A-f-c : c 

 \ a : a :: A : c, hence c (A o)=n A, and the con- 

 tent of the SOi: 



4(AA+aA)=lA(A+.) 



The diameter . e greater end of a para- 



bolic frustum, is 58, that of 1)1) the less 30, and the 

 height 18 inches. 1 I the rontint. 



"854 (30 + 58 t )=S348 9456. 

 The content=3S4fi !U5i>X!>=30140.51(Hcub. in. 



PROBLEM XII. 



To find the solid content of a parabolic spindle, or 

 olid generated by the rotation of an arc ALB of a pa- 

 rabola about AB an ordinate to the axis. 



.Itiply toe area uf the middle section by its Fig. 1 1. 

 length, and take ,* r of the product fur the content of 

 the solid. 



For the investigation of this rale, tec Fluxions, Art. 

 x.*. 



PBOBLEM XIII. 



To find the soKd content of a frustum of a naralx>lic 

 spindle, one of the ends of the fruitum passing through 

 the centre of the spindle. 



Add into one um eight time* the square of the ilia- pig. gg. 

 meter of the greater end, and three time* the square of 

 the diameter of the lesser end, anil four times the 

 product of the dinten. multiply the sum by the 

 length, anil thi* prod act again by .05930 (viz. J T of 

 7.854.) and the result will be the content, 



Potting p=CL>, the abcissaof the generating curve, 

 which i* also the radius of the greater end of the frus- 

 tum, 9= AC the temiordinate of the curve, i v=('lt=PO 

 the radio* of the lesser end of the frustuin, T=PR=r 

 (,' ;t length, and =.7854 ; it ha* been found (1 



Art. 161, Ex. *.) that the content of the frus- 

 tum it 



Now, from the nature of the curve, PR 1 : AC 1 : : DR 



lid pro- 

 height, 



1 16', 



: DC, that i* * : o* : :py:p, hence g*= *-- . 



Let the values of 9' and 9* be substituted in the above 

 formula for the content of the frustum, it then become* 



r ( 

 an expression from which the rule is derived. 



