CHAP. 6]
MULTIPLE, LINE AND SURFACE INTEGRALS. INTEGRAL THEOREMS
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(b) Along the straight line from (0,1) to (1,1), y = 1, dy = 0 and the line integral equals
Along the straight line from (1,1) to (1,2), x = 1, dx = 0 and the line integral equals
Then the required value = -2/3 +10/3 = 8/3. (c) Since í = 0 at (0,1) and f = 1 at (1, 2), the line integral equals
6.11.
If A = (3a; 2 -6ye)i + (2y + 3xz)j + (1 -4xyz 2 )k, evaluate k-dr from (0,0,0) to (1,1,1) along the following paths C: (a) x-t, y = t2, z = t3. (b) the straight lines from (0,0,0) to (0,0,1), then to (0,1,1), and then to (1,1,1). (c) the straight line joining (0,0,0) and (1,1,1).
(a) If x = t, y = t2, z = t3, points (0, 0,0) and (1,1,1) correspond to t = 0 and í = 1 respectively. Then
Another method. Along C, A = (3 «2 - 6i«)i + (2«2 + 3t*)j + (1 - 4i")k and r = x\ + yj + zk = tí. + t*j + t%, dr = (i + 2ij + 3f%) dt. Then
(b) Along the straight line from (0,0,0) to (0,0,1), x = O, y = 0, dx - 0, dy = 0 while z varies from 0 to 1. Then the integral over this part of the path is
Along the straight line from (0, 0,1) to (0,1,1), x = 0, z - 1, dx = 0, dz = 0 while y varies from 0 to 1. Then the integral over this part of the path is
Along the straight line from (0,1,1) to (1,1,1), y = 1, z = 1, dy = 0, dz = 0 x varies from 0 to 1. Then the integral over this part of the path is
Adding,
while