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Week 13 - Solution - Tutorial Note Week 13 Solution...

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Tutorial Note - Week 13 - Solution Divergence Theorem 1. (a) i. Let S 1 , S 2 , . . . , and S 6 be the six faces of the cube. (Figure 41a and 41b) Top (yellow) S 1 : r ( x, y ) = [ x, y, 1] with - 1 x 1 and - 1 y 1 Bottom (yellow) S 2 : r ( x, y ) = [ x, y, - 1] with - 1 x 1 and - 1 y 1 Left (green) S 3 : r ( y, z ) = [1 , y, z ] with - 1 y 1 and - 1 z 1 Right (green) S 4 : r ( y, z ) = [ - 1 , y, z ] with - 1 y 1 and - 1 z 1 Front (pink) S 5 : r ( x, z ) = [ x, 1 , z ] with - 1 x 1 and - 1 z 1 Back (pink) S 6 : r ( x, z ) = [ x, - 1 , z ] with - 1 x 1 and - 1 z 1 ii S 1 F d S = i 1 1 i 1 1 b 3 y 2 , 9 x 2 y, - 4 xy 2 B ([1 , 0 , 0] × [0 , 1 , 0]) dx dy = 0 S 2 F d S = i 1 1 i 1 1 b - 3 y 2 , 9 x 2 y, - 4 xy 2 B ([0 , 1 , 0] × [1 , 0 , 0]) dx dy = 0 S 3 F d S = i 1 1 i 1 1 b 3 y 2 z 3 , 9 yz 2 , - 4 y 2 B ([0 , 1 , 0] × [0 , 0 , 1]) dy dz = 0 S 4 F d S = i 1 1 i 1 1 b 3 y 2 z 3 , 9 yz 2 , 4 y 2 B ([0 , 0 , 1] × [0 , 1 , 0]) dy dz = 0 S 5 F d S = i 1 1 i 1 1 b 3 z 3 , 9 x 2 z 2 , - 4 x B ([0 , 0 , 1] × [1 , 0 , 0]) dx dz = 4 S 6 F d S = i 1 1 i 1 1 b 3 z 3 , - 9 x 2 z 2 , - 4 x B ([1 , 0 , 0] × [0 , 0 , 1]) dx dz = 4 Therefore, S F d S = S 1 F d S + S 2 F d S + . . . + S 6 F d S = 8. ii. E = { ( x, y, z ) | - 1 x 1 , - 1 y 1 , - 1 z 1 } div F = ∂x ( 3 y 2 z 3 ) + ∂y ( 9 x 2 yz 2 ) + ∂z ( - 4 xy 2 ) = 9 x 2 z 2 S F d S = i 1 1 i 1 1 i 1 1 9 x 2 z 2 dx dy dz = 8 1
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x y z x y z Figure 41a Figure 41b (b) i. Let S 1 , S 2 , S 3 , and S 4 be the four faces of the tetrahedron. (Figure 42a and 42b) Top (yellow) S 1 : r ( x, y ) = [ x, y, 1 - x - y ] with 0 x 1 and 0 y 1 - x Left (green) S 2 : r ( x, z ) = [ x, 0 , z ] with 0 x 1 and 0 z 1 - x Back (blue) S 3 : r ( y, z ) = [0 , y, z ] with 0 y 1 and 0 z 1 - y Bottom (pink) S 4 : r ( x, y ) = [ x, y, 0] with 0 x 1 and 0 y 1 - x ii S 1 F d S = i 1 0 i 1 x 0 b 3 xy, y 2 , - x 2 y 4 B ([1 , 0 , - 1] × [0 , 1 , - 1]) dy dx = 29 140 S 2 F d S = i 1 0 i 1 x 0 [0 , 0 , 0] ([1 , 0 , 0] × [0 , 0 , 1]) dz dx = 0 S 3 F d S = i 1 0 i 1 y 0 b 0 , y 2 , 0 B ([0 , 0 , 1] × [0 , 1 , 0]) dz dy = 0 S 4 F d S = i 1 0 i 1 x 0 b 3 xy, y 2 , - x 2 y 4 B ([0 , 1 , 0] × [1 , 0 , 0]) dy dx = 1 840 Therefore, S F d S = S 1 F d S + S 2 F d S + S 3 F d S + S 4 F
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