( 554 ) 



The equations {VII) become 



du du ö'm dv dv d'v 



which are satisfied by a function and its opposite. From this we 

 deduce -. 



d_ rdjiu)\ ^ ^ 



Therefore e. g. 



u = é'X-i)+?(0 , v = — e'K-i) + ?(?). 



By quadratures we find out of {VI), 



2Qiz = - if.(7i) + if {I), 



4QiA- = é^ + ?(f) + e- •K-'i) - ?(f ), 



4Q?/ = - «■X-^) + ?(0 + e-'X-';)-?(0. 

 The surface is a cylinder of revolution. Its section with the plane 

 XOY is a circle, as we find 



^' + '^' = -4^^^- 



1 



The radius of the circle is therefore — , as it has to be. 



H 



We can furthermore easily show that our solution agrees with (he 



differentia] equations (/vY), when we put 



AG=)=/,(»i)=--l- 

 We find namely that the second member becomes zero, so that 

 1 _H _ \ n \ 



1 1/1 1\ 



As moreover — = - , as we saw before,?' is therefore = oo. 



F 2 \r^ rj 



§ 8. We can now investigate what in the equations (F7/) the 

 significance would be of a solution m = x (^)' if it were possible. 



h — +3 — 4-2 — 0. 



2 dv" \dvj ^ \dv ) ^ 



dw d^to dp 



Lel~-=p, so ^r^ = p-^, 

 dv dv dw 



then : 



IV dp 



2 div 



{P 4- 2)"- 



from which ensues, — == kw^ (k = const.) . 



p -]- I 



For A; = this solution gives^the one used in the text. 



