512 Sir W. Thomson on the Division of Space 



determinate and has only one solution. Taking C A for axis 

 of x ; and z perpendicular to the plane BOB': and regarding 

 z as a function of x, y, to be determined for finding the form 

 of the surface, we have, as the analytical expression of the 

 conditions 



dxA ^ dy 2 ) dxdy dxdy* df\ dx 2 )~^ ' K h 



dx 

 and 



dz 2 dz 2 \~*/ . , dz 



when z={a— x) *y2 



} 



(2). 



17. The required surface deviates so little from the plane 

 BOB' that we get a good approximation to its shape by 

 neglecting dz 2 /dx 2 , dz/dx .dz/dy, and dz 2 /dy 2 , in (1) and (2), 

 which thus become 



V 2 *=0 (3), 



and 



j- = ^ - •§ =-094735, when x=a-z/s/2, . (4), 



V 2 denoting [djdx) 2 -\- {d/dy) 2 . The general solution of (3), 

 in polar coordinates (r, 4>) for the plane (x, y) , is 



2(Acosm0 + Bsinm(/>)?' m , .... (5), 



where A, B, and m are arbitrary constants. The symmetry 

 of our problem requires B = 0, and m = 3 . (22 + 1), where i is 

 any integer. We shall not take more than two terms. It 

 seems not probable that advantage could be gained by taking 

 more than two, unless we also fall back on the rigorous equa- 

 tions (1) and (2), keeping dz 2 /dx 2 &c. in the account, which 

 would require each coefficient A to be not rigorously constant 

 but a function of r. At all e vents we satisfy ourselves with 

 the approximation yielded by two terms, and assume 



0=Ar 3 cos 3$ + AV 9 cos9<£ .... (6); 



with two coefficients A, A' to be determined so as to satisfy 

 (4) for two points of the curved edge, which, for simplicity, 

 we shall take as its middle, E (</> = 0); and end, B (0 = 30°). 

 Now remark that, as z is small, even at E, where it is 

 greatest, we have, in (4) , aF== a or r=~ a sec (/>. Thus, and 

 substituting for dz/dx its expression in polar (r, $) coor- 

 dinates, which is 



dz dz , dz . , 



