Week 1112 - Solution

# d d 4 2 2 r 2r sin r dr d 0 0 8 22 3 y c 1 d x 0 0 2

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Unformatted text preview: ∂x ∂y dy dx = 1 3 y 1 D C x 0 0 1 Figure 31 (b) D = {(x, y ) | − 2 ≤ x ≤ 2, x2 ≤ y ≤ 4} (Figure 32) C 2 y 2 − tan−1 x dx + (3x + sin y ) dy 4 = −2 x2 ∂ ∂ (3x + sin y ) − y 2 − tan−1 x ∂x ∂y 4 dy dx = − y D C x −2 0 2 Figure 32 (c) D = {(r, θ) | 0 ≤ r ≤ 1 + cos θ, 0 ≤ θ ≤ 2π } (Figure 33) ∂ ∂ x2 − (2xy ) dA = 0 ∂x ∂y 2xy dx + x2 dy = C D 4 96 5 y C 1 D x 2 −1 Figure 33 π 4 (d) D = (r, θ) | 0 ≤ r ≤ 2, 0 ≤ θ ≤ C (Figure 34) ∂ ∂ xy 2 − y 2 − x2 y ∂x ∂y F • dr = y 2 − 2y + x2 dA dA = D D π 4 2 2 r − 2r sin θ (r dr dθ) = π + = 0 0 8 √ 2−2 3 y C 1 D x 0 0 2 Figure 34 √ √ (e) D = (x, y ) | − 1 ≤ x ≤ 1, − 4 1 − x4 ≤ y ≤ 4 1 − x4 (Figure 35) C 1 F • dr = √ 4 −1 ∂ ∂ x4 − x3 y ∂x ∂y √ 4 dA = D − 1−x4 3x3 dy dx = 0 1−x4 y 1 C D x −1 1 −1 Figure 35 3. D = {(x, y ) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x} (Figure 36) 1−x 1 C F • dr = 0 0 ∂ ∂ 1 [x (x + y )] dy dx = − xy 2 − ∂x ∂y 12 5 y 1 C D x 0 0 1 Figure 36 4. D = {(r, θ) | 0 ≤ r ≤ 2, 0 ≤ θ ≤ π } (Figure 37) C ∂ ∂ x3 + 3xy 2 − (x) dA = ∂x ∂y F • dr = π 2 3r2 (r dr dθ) = 12π 0 0 D y 2 C D x −2 0 2 Figure 37 y x and Q (x, y ) = 2 are undeﬁned at the origin and the origin 2 +y x + y2 lies in region D, then both P (x, y ) and Q (x, y ) are discontinuous at the origin and the partial derivatives do not exist. Therefore, we could not evaluate th...
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## This document was uploaded on 01/23/2014.

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