48 Prof. Challis on an Extension of the 



eyt] 

 dx 

 dy ~~ " dy 



Again, in the expression for A let us make y the independent 



dx d x 



variable instead of x ; so that, putting p 1 for -^ and q' for -^-^ } 



we h 



ave 



A^+^-f 



(i+p'*r> (i+y 2 )^ 



Supposing now the line to be also symmetrical with respect to 

 the axis of y, we shall have p* and v 1 +p 12 , or — and -7-, of dif- 

 ferent signs on the two sides of the axis, and the radius of cur- 

 vature -, of the same sign. Thus Ady neither changes 



sign nor magnitude by the change of sign of x. Consequently 

 the integral of Ady = embraces the two values of x correspond- 

 ing to each value of y, and thus satisfies the hypothesis of sym- 

 metry with regard to the axis of y. 



But the integral of Adx = does not satisfy the supposition of 

 symmetry with regard to the axis of x, because on that supposi- 

 tion A would change sign with the change of sign of y. 



The above considerations enable us to draw a distinction be- 

 tween the equations A = and Ap = 0. If it were proposed to 

 find, of all the surfaces of given extent generated by lines sym- 

 metrical with respect to the axis of x and also with respect to a 

 perpendicular axis, that which encloses the greatest solid, the 

 answer could not be given by the equation A = 0, but might be 

 given by Ap = 0. In fact the solid deduced from the latter 

 equation fulfils, as I am about to show, those conditions, and 

 may therefore be regarded as the solution of a case of the general 

 problem which involves the limitation of symmetry. 



The differential equation Ap = is satisfied if p = ; and this 

 mode of satisfying it is, by the rules of analysis, as much entitled 

 to consideration as any other. By integration, y = c, which 

 result, it will presently appear, is part of the answer to the pro- 

 blem. The same equation is satisfied by the integral obtained 

 above, viz. 



the meaning of which is next to be considered. Since the 

 straight line joining the two given points, designated hereafter 

 as P and Q, coincides with the axis of x, and the required line 

 has to pass through the points, that integral must be verified by 

 y — 0. Hence C = 0. Consequently, taking account of the 

 double sign of the radical, as the theory of equations demands, 



