Since t is a counterclockwise parametrization of a

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Unformatted text preview: change to 2 (1 − y 2) dx + (x + ey ) dy = 0 B ¨ =0 ¨ 2 2𠨨 ¨ θ dθ dθ + sin 3 ¨0¨ ¨ = 1 θ 2 2π = π. 0 ∂ (P Q) ∂ (P Q) + dA = Theorem ∂y ∂x ∂D D ∂Q ∂Q ∂P ∂P +P +P −Q +Q dA = ∂y ∂y ∂x ∂x D ∂ Q ∂Q ∂P ∂P +P dA. Q − − ∂x ∂y ∂x ∂y D ∂P ∂Q Green s ∂Q ∂P Q dx + P dy −P −Q = Theorem ∂x ∂x ∂y ∂y ∂D ∂ P ∂Q − Q ∂P ∂ Q ∂P − P ∂Q ∂y ∂y ∂ P ∂Q ∂x ∂x dA = − + ∂x ∂y ∂x ∂y D D P Q dx + P Q dy Green s − = ∂ 2 Q ∂Q ∂P ∂ 2P ∂Q ∂P ∂ 2P ∂P ∂Q ∂ 2Q − −Q − −Q + +P dA = ∂x ∂y ∂x ∂y ∂x ∂y ∂y ∂x ∂y ∂x ∂y ∂x ∂y ∂x ∂ 2Q ∂ 2P ∂ 2Q ∂ 2P P −Q dA = + + ∂x∂y ∂y ∂x ∂x, ∂y ∂y ∂x D ∂ 2Q ∂ 2P dA, since P, Q are assumed to be of class C 2 . 2 P −Q ∂x ∂y ∂x ∂y D P (b) ∂D − ∂f ∂f Green s dx − dy = Theorem ∂y ∂x ∂ 2f ∂ 2f + dA = − ∂x2 ∂y 2 D D ∂ ∂x − ∂f ∂x −...
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