10 PROFKSSOR ANDREW GRAY, 



Q suffers in dt a displacement relatively to P of amount (dq/ds' . ds' cos 6 — qd6 . sin 6)dt, 

 parallel to its original length. 



The additional flow due to this displacement is (q' dq/ds' . ds cos 6 — qq'dd . sin 6)dt, or 



(q^ds' cos 2 - q 2 dO . sin cos \dt, 



since q = q cos ». 



But Q has also moved out through the distance (q + dq/ds' . ds')dt . sin (6 + d0) 

 at right angles to the original position of ds'. This is not parallel to q' sin Odt, the 

 displacement of P, but is so to the first order of small quantities. (This degree 

 of approximation is, however, sufficient for the present calculation, as the velocity 

 at Q in the direction at right angles to the plane ds, ds' is a small quantity of 

 the first order. Thus the flow due to the deviation from the plane ds, ds' must 

 be a small quantity of the second order.) We have to the degree of approximation 

 required for the present purpose {(q + dq/ds' . ds') sin (6 + <W) — q sin 9\dt, or 



( q cos dO + --,ds' sin Ojdt 



for the only other relative displacement that need be considered. 



The velocity in the direction of this displacement is approximately q sin 6, and 

 therefore the component of flow along the element due to the displacement is 



( q* sin 6 cos 6d0 + q&ds' sin 2 0\ dt. 



Thus the whole change in the rate of flow brought about by the two displacements is 



<?-i a!s'(sin 2 + cos 2 6) = q^,ds'. 



But the element ds', carried along by the motion of the fluid, is displaced as a whole 

 in time dt through a certain distance parallel to its own direction, and this, apart from 

 alteration of length or direction of the element, causes an alteration of the rate of flow 

 along it. The amount of this is easily shown to be qdq/ds'.ds'. For the total growth 

 in time dt of flow along ds', with its given length and direction, is (dq'/dt + qdq'/ds)ds'dt. 

 This is due to the longitudinal and lateral displacements in dt taken together. But 

 by (15), dq' /ds = dq/ds' + 2ft> gs . sin 6. Thus for the element ds', with its given length and 

 direction, the rate of increase of flow is (dq'/dt + qdq/ds' + 2oo s /j sin 6)ds'. 



17. We therefore have finally 



(% - 2 q d l)ds' = - 2r 7w „ sin Ode' - 2<hds - %U . . (22). 



yds as / ds dt 



The expression on the right is twice the total time-rate of change of the surface- 

 integral of rotation caused by the motion of ds', together with twice the time-rate 

 of change of the same integral due to variation of velocity while ds' remains at rest. 

 Integrating along AB we get 



* A -<flJ-flJ>- -2Jqo^ S m6ds'-2fq^ds'-f^ds' . (23). 



