220 Prof. Challis on the Solution of a Problem in the 



essentially the same, and the investigation by polar coordinates 

 is here adverted to only because it first suggested to me the idea 

 that the introduction of the factor has an important bearing on 

 the solution of the problem. Since the differential equation 

 from which the subsequent inferences are drawn is incomplete 

 without the factor, and as the factor equated to zero satisfies 

 the complete equation, this solution must certainly be taken as 

 an answer, in part, to the proposed question. The integral of 



r cos 6 + r 1 sin 0=0, or -^ = 0, is r sin 6 = const., or y = const. ; 



and, so far as this integration indicates, the required surface is 

 cylindrical. 



We have next to inquire what inferences may be drawn from 

 the integral of the complete equation. The conditions of the 

 problem being expressed by 



the usual process gives 



2ydx + \dx x /\^-d.-^L=,=0 } 

 1 v 1 + p* 



p being put for -~- and q for -J-. After multiplying this equa- 

 tion by p and integrating, the result is 



y'+ X l =C, 



C being the arbitrary constant introduced by the integration. 

 We may now suppose that one of the points through which the 

 surface has to pass is the origin of coordinates, in which case 

 the above equation is satisfied if y = 0. Thus on this suppo- 

 sition C = 0, and we have 



Removing the factor y, the effect of which has just been taken 

 into account, and attaching to the radical the double sign, as the 

 theory of equations requires, we obtain the two equations 



Each of these equations gives by integration an equation of the 

 form 



(ff + c)«+y*=X 8 S 



