312 Mr. Knight on the Penetration 
The fluents to be taken, first from ip = o to ^iz=zn<p, and 
then from cp = o, to a? = 22 .. The first operation gives 
A = rf<p sin. ncp, 
S = “/<?>{■“■ sin. ?Kp * 4 ” ~ sin. S n( p} — ‘~/ ( P sin. ncp cos. 'ncp, 
and by the second we get 
A = C — — cos. 
} + c. 
r 3 f cos. 3 ?2<2> ^ I 
J ‘ “ "7 cos - n $ — Jg C0S ‘ 3 n< P 
2 n L 
3 4 
which fluents being taken from n<p = o, to ?np — go 0 , are 
A = — ; S = — x — ; and if these are multiplied by 4 we 
have 
A = 4/; S = -ir*; 
for the whole that remains of the surface and solidity of the 
hemisphere after the subduction of the 2 n cylinders. Thus A 
and S (for the whole hemisphere) do not depend on the num- 
ber of the cylinders with which the penetration is made ; a 
?nost remarkable circumstance , seeing that amongst the bases of 
those cylinders are curves of an infinity of different kinds and 
orders. 
Let fig. 2 represent half the base of one of the cylinders ; 
. b P h Cb the radius of the hemisphere, C 
the centre. From r, any point in the 
curve, let fall the perpendicular rp 
on the axis ; call Cp, x ; rp, y. 
By construction, Cr = s/ x* -f-y 2 = r cos. n . bCr ; now the 
cosine of the simple arc bCr is 
V' 
■, which being put in the 
trigonometrical expression for the cosine of the multiple arc 
