Let C be the positively oriented square with vertices (0,0), (2,0), (2,2), (0,2). Use Green's Theorem to find∫C4y2xdx+3x2ydy.?

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  • 5 months ago
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    Green says: ∮_C (u dx + v dy) = ∫∫_R (∂v/∂x - ∂u/∂y) dx dy.

    You have u=4y²x and v=3x²y, so, the right side becomes:

        =  ∫∫_R [∂(3x²y)/∂x - ∂(4xy²)/∂y] dx dy

        =  ∫∫_R (6xy - 8xy) dx dy

        =  -2 ∫ ∫ xy dx dy     [from x=0 to 2 and y=0 to 2]

        =  -2 ∫ (∫ x dx) y dy

        =  -2 ∫ (2²/2 - 0²/2) y dy

        =  -4 ∫ y dy

        =  -8

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