and assume the slow variation of C?(x) , such as 



a(k) = 0(l/k) 



and Ky = 0(1), Kz = 0(1), because the wavelength is comparable with the 

 breadth of the body. Then the asymptotic expression of the above integrals 

 for large K leads to the following expression. 



x 



/. Kz-iKx-i tt/4 j a'(g) , F 

 x = /2ttK e — — — at, 



-I /x " ? 



, . i(x) e -iKx| 2 | v cos(Kzv) + sin(Kzv) e ~K [ y | // 2 +l vdv 



2 / 2 



(v +1) /v +1 



J* 



I 



- (2TTK|y| + if)e Kz 



(227) 



where x = - I is the forward end of the body. The second term is the two- 

 dimensional part of the potential, while the first term gives the three- 

 dimensional effect. Because of the presence of the factor \fe in the first 



1/2 

 term, the order of the second term is higher by £ .If we take the 



lowest order term only, we can omit the second term. Then the boundary 



value problem becomes extremely simple. The boundary condition on the 



hull surface is 



*? - - *r (228) 



If we write 



n . = |2L. „• = |5_ (229) 



y 3n z 9n 



81 



