{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Week 1112 - Solution

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

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
(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
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 ,

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 10

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online