: 
Let ABDC be a 
-BFKG, having the same base and 
; 
GEOM 
t Por: Tl. Tutor.) | 
A cylinder and a parallelopiped having equal bases 
and altitudes, are equal to one another, 
‘If the cylinder ABCD and parallelopiped EF, which 
have wat bases, (viz. the circle AGB and parallelo- 
gram EH), and the same altitude, be not equal, let us 
suppose that they are pk and _ first, let the cylin- 
der be less than the parallelopiped. From EF, let a 
parallelopiped'EQ, equal to the cylinder, be cut. off by 
a plane Po paraliel to NF.. Leta polygon AGKBLM 
be inscribed in the circle AGB so as to differ from it by 
aless Space than the parallelogram PH (6.5. Part L.) 
and let the parallelogram RO be equal to the polygon 
AGKBLM; the point R will manifestly fall between 
PandN., Now, if an upright prism, having the same 
altitude as the cylinder, be formed on the polygon 
AGKBLM as a base, and a solid ES be cutoff from 
the piorion! EF by a plane RS 1 to NF ; 
the prism and solid ES will ual (11.2.)) But the 
rism being entirely contained within. the :cylinder,. is 
ess than it; therefore the oo less thamthe oe 
; and consequently the solid ES is equal to:a soli 
whch is less than EO, now this is impossible ; there- 
fore the cylinder is not less than the parallelopiped EF, 
In the same way it may be shewn not to be greater, 
therefore the cylinder and. parallelopiped having equal 
bases and altitudes, are equal or equivalent: toone:ano- 
‘ther. Ly¥ rif 
Saul? Jey ’ Prop. III.) Taror. 
" If 2 cone and.cylinder have the same base and_alti- 
tude, the cone ‘o-the third part of the cylinder, 
. If acone A-BCD be not the third of a, cylinder 
titude, it will be 
the third part of a cylinder LMNO, having the same 
altitude as the other, but a base either less or greater ; 
-and first, let the base LIM be less than the: base BCD; 
' then, because the circle LIM is less than the. circle 
BCD, a polygon BECFD may be described in the lat- 
ter, which « shall differ from it by less .than. its. excess 
above the circle- LIM-(6. 5. Part I.).; wherefore this 
polygon will be ee than the circle LIM. Let an 
upright prism and pyramid be constituted on the poly- 
oon Be FD as a base, and having the same altitude 
. as the cylinders; and because’ the cone A-BCD is the 
third part of the cylinder LMNO, and this cylinder is 
less than the prism BCD-GHK, because ‘it es ‘a less 
base and the same altitude, therefore the cone A-BCD 
is less than the third of the prism BCD-GHK; but 
‘the pyramid A-BECFD is the third part of the prism 
(17. 2.) ; therefore the cone A-BCD is less than the 
pyramid A-BCD : Now this is impossible, because the 
id being contained entirely within the cone, ‘the 
cone must be greater than the hin Therefore the 
cone A-BCD is not less than the third part of the cy- 
linder BFKG. In the same manner, by circumscrib- 
ing ® polygon about the base of the cylinder, it may 
be shewn that the cone is not greater than the:third 
‘part of the cylinder ; ‘therefore it is equal to the third 
part-of the cylinder. — 
‘Prop. 1V. Lemma. 
F plane figure, bounded by a-straight 
line.CD, a line of any kind AB, which is ieeeeteaeal by 
‘perpendiculars at the extremities of CD, and by these 
perpendiculars AC, BD. Let AB da be a solid gene- 
rated by the revolution of this figure~about'CD as an 
VOL, X. PART 1, 
ETRY. 933 
‘axis ; a series of cylinders may be described about the 
solid, and arfother series may be inscribed in it, having 
all the same altitude, and such that the sum of the 
circumscribed ‘cylinders shall exceed the sum of the in- 
seribed cylinders by less than any given solid S. 
Let the solid S be a cylinder, having Bd for the dia- 
meter of its base, and DP for its height. Suppose the 
fixed axis CD to be divided into a number of equal 
parts DK, KG, GE, EC, each less than DP. In the 
plane of the figure ABDC, draw ndiculars EF, 
GH, KL to meet the line AB in F, H, L. . Construct 
the inscribed rectangles AE, FG, HK, LD, also the 
circumscribed rectangles CF, EH, GL, KB.) -By'the ro- 
tation of the plane figure about the axis CD, these vect- 
angles will evidently generate a series of cylinders in- 
scribed, in the solid, and another series described about 
it... Let the circumscribed cylinders, reckoned from the 
bottom of the. solid to the top, be denoted. by V, X, 
Y, Z, and the inscribed cylinders by v, 2, ¥, 2% then the 
sums of the circumscribed and inscribed cylinders will be 
Vi he IK es ¥ te Zs 
and Vf ety tn i : 
Now by the nature of the figure, each circumscribed 
cylinder is equal to the inscribed cylinder next below it.; 
therefore X=v, Y=a, and Z=y, and hence the excess 
of the sum of all the circumscribed above the inscribed 
cylinders will be the'same as the excess of the greatest 
circumscribed above the. least inscribed cylinder :, that 
is, it will be equal to V—z, and consequently will be 
less than V ;,.but. the lowest, circumscribed cylinder V 
is less than the solid. S, because it has the same base, 
(viz. the circle having for its.diameter B 4), and a less 
altitude KD, by construction ; therefore the excess of 
the series of circumscribed above.the series of inscribed 
cylinders is less than the given solid S. 
Cor. The difference between the solid ABda and 
either of the two series of cylinders will be less than 
the greatest circumscribed cylinder: For the solid 
‘AB? a is greater than the one series of cylinders,and Jess 
than the other, therefore it will differ from either series 
by a quantity less than the difference between the two. 
Peer Vv. THEOR. 
Ifa cone and hemisphere have equal bases and alti- 
tudes, and if a series of cylinders be described about 
the cone, and another series be'inscribed in the /hemi- 
Ea and the cylinders have all the same altitude, 
e sum of. the two series will be equal to a cylinder 
having the same base and altitude as the hemisphere. 
Let AFB bea semicircle, and CFDA, CFEB, squares 
described on the radius CF, and Jet CE be the diagonal of 
one of the squares BF: Let CF be divided into any 
number of equal parts CG, GK, KM, MF; and let per- 
pendiculars be drawn through the points of division, 
meeting the-diagonal CE, in the points O, P, Q; the 
quadrantal are BF in the points H, L, N; and the side 
of the square in the points R,S, T : Construct the rec- 
tangles CO,.GP, KQ, ME, which will circumscribe the 
triangle CFE ; construct alsothe rectangles CH, GL, 
KN, which will be inscribed;in the quadrant CFB. 
Suppose now the plane of the square to revolve about 
its side CF as an axis ;.the triangle CFE will then ge- 
nerate a cone, which will have DE for the diameter of 
its base, and C for its vertex ; the quadrant CFB will 
generate a hemisphere, hayin 
which AB is a diameter ; and ;the square CBEF. will 
generate a cylinder, having the same base and altitude 
as the hemisphere : Also; the rectangles described about 
2G 
Of the three 
Yound So- 
lids. 
Fig. 169, 
for its base a circle. of - 
