Therefore 



Closed and Open Tubes* 525 



fixed by the equations of equilibrium. We arc therefore 

 justified in choosing for f over one of the boundaries 

 where £ is constant. 



Now integrating by parts we get 



=0-r 2 C- I ri 2r£dr. 



Q = nCl rfdd + %i\ \ %rdrd6 



Jo Jr Jo 



= 2nCA + 2n^dxdi/ } (22) 



where A denotes the area enclosed by the inner boundary 

 of the section, and the double integral extends over the area 

 of the section of the material. 



If we had taken f to be zero over the inner boundary 

 and — C over the outer boundary, thus keeping the same 

 difference as before over the two boundaries, we should 

 have got 



Q l = 2nQK 1 + 2n§$dxdy, .... (23) 



where now A x denotes the area enclosed by the outer boundary 

 of the section. The double integral has not, of course, the 

 same value in (22) and (23) because, in the first equation, 

 f varies from to in passing from the inner to the outer 

 boundary, while in the second equation £ varies from 

 to -C. 



There is a slightly better form for Q than either the one 

 in (22) or that in (23). Since f varies continuously from 

 one boundary to the other there must clearly be some closed 

 curve, between the two boundary curves, along which f is 

 constant, and such that, if we make this constant zero, the 

 values of f along the inner and outer boundaries are ^C 

 and — ^C respectively. If A denotes the area enclosed by 

 this midway curve, then the torque is 



Q > =2nVA.+2n§%dxdy. .... (24) 



Equation (24) could have been obtained by adding corre- 

 sponding sides of equations (22) and (23) and then dividing 

 bv 2. It follows from this that A is the arithmetic mean 

 of A and A } . 



If we apply (24) to a thin tube the double integral must 



