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1915]
THE  THEORY  OF  THE  HELMHOLTZ  RESONATOR
373
of order 0- in comparison with the principal term. Reference to (39) shows that as regards the numerator of the integrand we have to deal with terms in 6°, d\ and &\
For the principal term we have
4 ff_______^_______...................(43)
J j V(^2 —• &2 — &2 — 26$ cos <b\
T.T                f     dd             fd (6 -\-l) cos c6)      .            6 H- b cos <h
.NOW            I----------------r- =    I----------------------—  — gin"1 -------------------—------ ,
For a given <£ the lower limit of d is 0 and the upper limit 6l is such as to make a2 = ₯ + d^ + %bdl cos <£,
or                                   d1 + b cos c/> = \/(tt2 — &2 sin2 <£>)......................(44)
mi                       "6l      c^         ^                    6cosrf>                            ,.„,.
Thus                       -yi--------r = ^--sm-1   // 2_72  • o ,s.............(45)
When this is integrated with respect to <£, the second part disappears, and we are left with 7r2*simply, so that the principal term (43) is 47r2. That this should turn out independent of b, that is the same at all points of the aperture, is only what was to be expected from the known theory respecting the motion of an incompressible fluid.
The term in d, corresponding to the constant part of 2 (in)-S(oufc), is represented by
Here d dd d(f> is merely the polar element of area, and the integral is, of course, independent of b. To find its value we may take the centre G as the pole of d. We get at once
so that this part of (42) is
\ v*^/r'2         PV                   /                                                                    ^
For the third part (in d2), we write
d2 = - (ft2 - b2 - 2b9 cos cj)-d*}- Zb cos <l>(d+b cos 0) + a2 - 62 + 26s cos2 <£, giving rise to three integrals in d, of which the first is cos <f) — d2}
•)- b cos <i) \/{ct2 — bz sin2 d> — ($ H- b cos c6)2} a2 — b* sin2 <jf> . ^     #-t-&cos</>                                           ( Q.
2          8m   V^-^sm2^'   "•'..................(    }
The second integral is
cos (t> I -—77—„—p^o—= 26 cos <£ \/{a2 — ₯ — 2bd cos </> — d2},... (50)s may be verified :