integral over a portion of the Von plane. The integral is 
evaluated in equation (10.85) which proves that power is conserved. 
Consider figure 29. The dashed lines bound the area elements 
of A5(h',j',j*) and they form a net over [A,( ¥5 18590") I For 
this particular case 9* equals zero. The shaded area shows the 
area mapped into by Ay (8,1). For this case, j is greater than 
zero and j* is zero, as assumed for a special consideration in 
equation (10.86). Then after Aj(h,j) has been mapped into 
[A, ( ¥598529 1°, the greatest value of y, y, , is then found 
by substituting the smallest value 85 can have, Aaneiy equation 
(10.82), into the upper boundary curve, namely equation (10.81), 
and the result is equation (10.87). Similarly equation (10.88) 
gives the minimum value of v o° 
For the net in the v.,0, plane, A, (hy J) therefore occupies 
part of several area elements given by Ay(h',j',9 4. In fact, there 
exists some value of h’, say K, such that eon a and “oe. are 
sandwiched between »y(2K - 1)/2m and y, (2K + P)/2m as stated by 
equation (10.89). Finally on mapping A,(h,j) into A,(h',j',0*), 
A,(h,j) contributes part of its power to the Aj(h',j',0*) for h' 
ranging from K to [K + (p - 1)/2] and for j' equal to j as stated 
by equation (10.90). 
Consider equation (10.91). The right hand side of the equation 
gives the power in the area element, 4,(h,j) in the m,®© plane. 
Ay (h, j) has the dimensions of om°sec/radian, and the whole term 
has the dimensions of em*. The number B(h,j,h',j',j*) is a number 
which partitions the right hand side into contributions to the 
various elements a,(h',§',J5*) in the v,,0, plane. Equation (10.92) 
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