Week 1112 - Solution

# 5 3 2 1 d 1 3 x 0 0 1 figure 28 ii d x y 0

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Unformatted text preview: • [1, 1] dt = 0 3 10 1 1 − t, − (1 − t)2 • [−1, 0] dt = − 0 1 2 1 0 [0, 0] • [0, −1] dt = 0 Hence, C F • dr = Γ1 F • dr + y Γ2 F • dr + 1 F • dr = − . 5 Γ3 Γ2 1 D Γ1 Γ3 x 0 0 1 Figure 28 ii. D = {(x, y ) | 0 ≤ x ≤ 1, x ≤ y ≤ 1} (Figure 28) ∂Q ∂P ∂ ∂ − = −x2 y 2 − (x) = −2xy 2 ∂x ∂y ∂x ∂y 1 C F • dr = 0 1 x −2xy 2 dy dx = − 1 5 (c) Figure 29 F (x, y ) = [x + 2y, x − 2y ] i. Γ1 : curve y = x2 from (0, 0) to (1, 1) and r (t) = [t, t2 ] with 0 ≤ t ≤ 1 Γ2 : straight line from (1, 1) to (0, 0) and r (t) = [1 − t, 1 − t] with 0 ≤ t ≤ 1 1 Γ1 F • dr = Γ2 F • dr = t + 2t2 , t − 2t2 • [1, 2t] dt = 0 5 6 1 0 [3 − 3t, t − 1] • [−1, −1] dt = −1 2 Hence, C F • dr = Γ1 F • dr + 1 F • dr = − . 6 Γ2 y 1 Γ1 Γ2 D x 0 0 1 Figure 29 ii. D = {(x, y ) | 0 ≤ x ≤ 1, x2 ≤ y ≤ x} (Figure 29) ∂ ∂ ∂Q ∂P − = (x − 2y ) − (x + 2y ) = −1 ∂x ∂y ∂x ∂y 1 C 2 F • dr = x x2 0 (−1) dy dx = − 1 6 2 (d) F (x, y ) = [x + y , 2xy ] i. Figure 30 Γ1 : curve y = x2 from (0, 0) to (2, 4) and r (t) = [t, t2 ] with 0 ≤ t ≤ 2 Γ2 : straight line from (2, 4) to (0, 4) and r (t) = [2 − 2t, 4] with 0 ≤ t ≤ 1 Γ3 : straight line from (0, 4) to (0, 0) and r (t) = [0, 4 − 4t] with 0 ≤ t ≤ 1 2 Γ1 F • dr = Γ2 F • dr = Γ3 F • dr = t2 + t4 , 2t3 • [1, 2t] dt = 0 104 3 1 4t2 − 8t + 20, 16 − 16t • [−2, 0] dt = − 0 1 0 (4 − 4t)2 , 0 • [0, −4] dt = 0 Hence, C F • dr = Γ1 F • dr + y Γ2 F • dr + Γ3 Γ2 4 D Γ1 Γ3 x 0 0 2 Figure 30 3 F • dr = 0. 104 3 ii. D = {(x, y ) | 0 ≤ x ≤ 2, x2 ≤ y ≤ 4} (Figure 30) ∂Q ∂P ∂ ∂ x2 + y 2 = 0 − = (2xy ) − ∂x ∂y ∂x ∂y 2 C F • dr = 2. (a) D = {(x, y ) | 0 ≤ x ≤ 1, x2 ≤ y ≤ √ y+e C 1 √ = x2 0 x x √ 4 (0) dy dx = 0 x2 0 x} (Figure 31) dx + 2x + cos y 2 dy √ ∂ ∂ 2x + cos y 2 − y+e x...
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## This document was uploaded on 01/23/2014.

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