OF ROTATING LIQUID CYLINDERS. 
71 
Lastly, by actual integration, we find as the value of from equation (17), 
V„ = ,rp{-6,(logf+log,) + 6,(^ +|) + P 3 (-p +Y) + --.} (18). 
The term — h-^ (log ^ + log rj) may be replaced by — 2h^ log ?*; hence V vanishes at 
infinity, except for a term proportional to log r. 
We have now seen that V satisfies all the conditions which must be satisfied by a 
potential; hence the potential will be given by equations (16) and (17). 
§ 5. Calculated by direct integration, the value of Vq is 
Vo = - p II log ^h^'dr'dd', 
where the integration extends throughout the cross-section of the cylinder, and R is 
the distance of the point 6 at vdiich the potential is being evaluated, from the 
point r', 6' of the cylinder. We have 
R2 = - 2rr' cos {9 - 6') = [r - (r - 
Now if I I > I t'' I , we have 
log (/’ — = log r + log (1 
and hence 
= log r 
j ^ 
r'd^' 
i 
n 
rc 
■ue 
log Pt" = 2 log r 
O O 
I'r'd^' ^ r'e 
.,,ie 
rt' 
re 
60 ' 
i6 J 
5 
Upon integration we obtain 
V,= -p 2logr r'd/dO' - 
(-i 
r'-d^'d/dd' -p ^|| r'V-'U/r'd^M - 
. . .J 
(19). 
The cylinder being symmetrical about the plane ^ =: 0, we have 
II d'^d^'dr'de' = II = f| d- cos O'd'r'dB', 
and hence equation (19) becomes 
Yq = — p (log f -f log Tj) II r'dr'dd' + p || r'" cos 6'dr'd6' + . . . (20). 
Let the area of the cross-section be A, and its centre of gravity be at X‘ = ot, 
so that 
A = [| r'dr'de' Aa = II d- cos d'ddde ; 
then equation (15) becomes 
Yq= - pA (log ^ -f log rj) + pAa + -Ij + . . . . 
Comparing this with equation (18), we find that 
A = 7r6j.(21), Aoc = 7760 ........ ( 22 ). 
These last two equations enable us to find, by a process of algebraical transfor¬ 
mation, the cross-section and the centre of gravity of the cylinder whose boundary is 
