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 (35) 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 -Tr) 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.