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640          ON THE  KESULTANT  OF  A  NUMBER OF  UNIT VIJBEATIONS            [442
limits coincide. From 6~\a to <9 = 3?r/2  , there are no corresponding values of r. At the latter limit a zero value of r enters, and from 6  37T/2  -|a to 0 = TT, r ranges from 0 to 2 cos (2?r  a  0).
The whole range from 6 = 0 to 6 = TT thus divides itself into four parts. In the first part from 6 = 0 to 6 = -J (a - TT), we get as the chance of (9 from (29)
In the second part from 6> = -|-(a  TT) to $ = -ija, the chance is f2                   dr
/0/1X (34) a                                         v    '
For the third part, from 6 = $a to 0 = 37r/2 - \a, there is no possibility. For the fourth part, from 9  3?r/2  a to (9 = TT, the chance for 6 is
If we integrate (33), (34), and (35) over the (positive) "ranges to which they apply and add the results, we get the correct value, viz. . This part of the question might be treated more simply without introducing r at all.
We have next to consider what in this case, viz. TT < a < 3?r/2, are the probabilities of various rs when Q is allowed to vary. When r is less than its value at 6 = ir, viz. 2 COS(TT  ), the corresponding range for 6 is made up of two parts, the first from #=() to 9 = \OL~- cos"1 (?"), and the second from 6 = 2?r  ^ a  cos"1 (r) to 0 = TT, so that the whole range of 6 is
a  cos"1 (r) + TT  [2-7T  | a  cos""1 (-|r)} = a  TT.
Thus from r = 0 to r = 2 cos (TT  a) the chance of r lying between r and
r + cZr is
9  ............................... V    ;
When r lies between 2 cos (TT  ^-a) and 2, the second part disappears arid we have only the one range of 9, equal to a  cos"1 (|-r), so that the chance of r lying between r and r + dr is
Expressions (36) and (37), obtained on the supposition that 6 is positive, are to be doubled when we allow for the equally admissible negative values of 0.
When (36), (37), as they stand, are integrated over the ranges of r to which they apply and added, the sum is , as it should be under the suppositions made.. 4, where a is supposed to be o?r/4.