Wave Motion due to a Submerged Obstacle. 528 
Then the complete expression for Xq is 
TX3 = en? { 
oo wT 
oF edi | ar ex" cos (Ka?y sin h/ 7?) cos d 
0 0 
x {A sin (xa*x cos p/7*) + B cos (xa?x cos b/r) }db 
— eat (oP wa { "3 (AAn+BBn)(mn-+1)-1(ajr)™1Vq cos b dd. 
0 0m 
(44) 
We have now 
@ = D—D,+ X%i+X, = D+X, (45) 
and the pressure equation is 
plp = —cep/ox—gz+ wo—} @. (46) 
The wave resistance, or the resultant horizontal pressure on the sphere, is 
7 2 
1 = | da ap sin « cos a dB. (47) 
0 0 
Omitting terms which, from symmetry, will give no contribution to R, we 
have 
2 he MOOR MO aOR 
p Ge "op Gp mS thakh mane 08 0B 
But when r = a, we have 
oD/oB = 0; OD/da = —heasin a; oX/or = 0, 
hence p/p = (8c/2a) sin «0X /da + wX. (49) 
We must now substitute (49) in (47) and use the value of X given by the 
sum of (37) and (44) on the sphere ; it is clear that we may omit the doublet 
D, as it will not affect the limiting value of R when w is zero. 
6. Consider, in the first place, the contribution of the first term in the 
value of p given in (49). In the repeated integrals which are obtained, we 
may change the order of integration, and we shall carry out first the summa- 
tion over the surface of the sphere. We notice that, when r = a, the first 
term in the value of Xz in (44) is equal to the value of X,; the additional 
part of X, is the term involving the expansions in spherical surface 
harmonics. Choose a typical term from the latter part, and we find we have 
to evaluate 
(48) 
f#sin « cos a (OY m/0x) dS, (50) 
taken over the surface of unit sphere. 
But this integral is equal to 
—3)P2(cos «) Ym(«, 8) dS. (51) 
Hence, the only term which has a value different from zero is the term in 
Y2, the surface harmonic of the second order. From the manner in which 
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