From (1) and (3)

u = Wc[tau]/p[Omega];

dl/u = g d[Omega]³p² dp/2[zeta]W³c²[tau]²;

_

/p0

t = | g d[Omega]³p² dp/2[zeta]W³c²[tau]²,

_/p1

= gd[Omega]³(p0³ - p1³)/6[zeta]W³c²[tau]². (14)

Encyclopaedia Britannica, 11th Edition, "Husband" to "Hydrolysis" Volume 14, Slice 1 · 第 115 / 207 题