220 Mr Greenhill, On the Rotation of a [March 20, 



it is seen that M is negative in region I., positive in region II., 

 and in region III. changes sign from positive to negative in cross- 

 ing a certain curve M = 0, lying between the hyperbola 



xy — x + y — 4 = 



and the straight line y — 4 = 0. 



Writing N in the form 



N=27rp(y-l)(x-y)J Mx-4>)\ + xy + x — y — 4j- — p — , 



it is seen that JV is positive in regions I. and II. A, negative in 

 region II. B, zero when y= 1, and positive in region III. 



Hence g> 2 2 is negative in region I.; w 2 2 is positive in region II. A, 

 and in region II. B up to the curve H = ; a> 2 2 is negative in 

 region II. B beyond the curve H = and in region III. up to the 

 curve M = ; and &> 2 2 is positive in region III. beyond the curve 

 M = 0. 



Also o) 3 2 is positive in regions I. and II. A, and in II. B up to 

 the curve H = ; &> 3 2 is negative in region II. B beyond the curve 

 H=0; and a> 3 2 is positive in region III. 



Hence o> 2 2 and g> 3 2 are only both positive, and the only real 

 solutions are contained in region II. up to the curve H= or D = 0, 

 and in region III. beyond the curve 31= 0. 



The conditions that a free surface can exist require in addition 

 that a should be positive; and therefore, drawing the transcen- 

 dental curve o- = 0, we are restricted in region II B. to the part 

 below the curve <r = ; and in region III. to the part above the 

 transcendental curve M = ; parts represented in figure 2 by 

 double shading. 



Fig. 2. 



OB It 



Similarly we can draw a figure to represent to the eye, as in 



