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Week 1112 - Solution - Tutorial Note Week 11 12 Solution...

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Tutorial Note - Week 11 & 12 - Solution Green’s Theorem 1. (a) c C x 2 y dx + xy 3 dy = c C b x 2 y, xy 3 B [ dx, dy ] So, F ( x, y ) = [ x 2 y, xy 3 ]. i. Figure 27 Γ 1 : straight line from (0 , 0) to (1 , 0) and r ( t ) = [ t, 0] with 0 t 1 Γ 2 : straight line from (1 , 0) to (1 , 1) and r ( t ) = [1 , t ] with 0 t 1 Γ 3 : straight line from (1 , 1) to (0 , 1) and r ( t ) = [1 t, 1] with 0 t 1 Γ 4 : straight line from (0 , 1) to (0 , 0) and r ( t ) = [0 , 1 t ] with 0 t 1 i Γ 1 F d r = i 1 0 b ( t ) 2 (0) , ( t ) (0) 3 B [1 , 0] dt = 0 i Γ 2 F d r = i 1 0 b (1) 2 ( t ) , (1) ( t ) 3 B [0 , 1] dt = i 1 0 t 3 dt = 1 4 i Γ 3 F d r = i 1 0 b (1 t ) 2 (1) , (1 t ) (1) 3 B [ 1 , 0] dt = i 1 0 (1 t ) 2 dt = 1 3 i Γ 4 F d r = i 1 0 b (0) 2 (1 t ) , (0) (1 t ) 3 B [0 , 1] dt = 0 Hence, i C F d r = i Γ 1 F d r + i Γ 2 F d r + i Γ 3 F d r + i Γ 4 F d r = 1 12 . 0 1 0 1 x y D Γ 1 Γ 2 Γ 3 Γ 4 Figure 27 ii. D = { ( x, y ) | 0 x 1 , 0 y 1 } (Figure 27) P ( x, y ) = x 2 y , Q ( x, y ) = xy 3 , ∂P ∂y = x 2 , ∂Q ∂x = y 3 c C F d r = i 1 0 i 1 0 ( y 3 x 2 ) dx dy = i 1 0 p y 3 1 3 P dy = 1 12 1
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(b) F ( x, y ) = [ x, x 2 y 2 ] i. Figure 28 Γ 1 : straight line from (0 , 0) to (1 , 1) and r ( t ) = [ t, t ] with 0 t 1 Γ 2 : straight line from (1 , 1) to (0 , 1) and r ( t ) = [1 t, 1] with 0 t 1 Γ 3 : straight line from (0 , 1) to (0 , 0) and r ( t ) = [0 , 1 t ] with 0 t 1 i Γ 1 F d r = i 1 0 b t, t 4 B [1 , 1] dt = 3 10 i Γ 2 F d r = i 1 0 b 1 t, (1 t ) 2 B [ 1 , 0] dt = 1 2 i Γ 3 F d r = i 1 0 [0 , 0] [0 , 1] dt = 0 Hence, i C F d r = i Γ 1 F d r + i Γ 2 F d r + i Γ 3 F d r = 1 5 . 0 1 0 1 x y D Γ 1 Γ 2 Γ 3 Figure 28 ii. D = { ( x, y ) | 0 x 1 , x y 1 } (Figure 28) ∂Q ∂x ∂P ∂y = ( x 2 y 2 ) ( x ) = 2 xy 2 c C F d r = i 1 0 i 1 x ( 2 xy 2 ) dy dx = 1 5 (c) Figure 29 F ( x, y ) = [ x + 2 y, x 2 y ] i. Γ 1 : curve y = x 2 from (0 , 0) to (1 , 1) and r ( t ) = [ t, t 2 ] with 0 t 1 Γ 2 : straight line from (1 , 1) to (0 , 0) and r ( t ) = [1 t, 1 t ] with 0 t 1 i Γ 1 F d r = i 1 0 b t + 2 t 2 , t 2 t 2 B [1 , 2 t ] dt = 5 6 i Γ 2 F d r = i 1 0 [3 3 t, t 1] [ 1 , 1] dt = 1 2
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Hence, i C F d r = i Γ 1 F d r + i Γ 2 F d r = 1 6 . 0 1 0 1 x y D Γ 1 Γ 2 Figure 29 ii. D = { ( x, y ) | 0 x 1 , x 2 y x } (Figure 29) ∂Q ∂x ∂P ∂y = ( x 2 y ) ( x + 2 y ) = 1 c C F d r = i 1 0 i x x 2 ( 1) dy dx = 1 6 (d) F ( x, y ) = [ x 2 + y 2 , 2 xy ] i. Figure 30 Γ 1 : curve y = x 2 from (0 , 0) to (2 , 4) and r ( t ) = [ t, t 2 ] with 0 t 2 Γ 2 : straight line from (2 ,
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Week 1112 - Solution - Tutorial Note Week 11 12 Solution...

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