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Math Help - ASAP help Test in 2days. GReens theorem.

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    Unhappy ASAP help Test in 2days. GReens theorem.

    use the Green's Theorem to calculate F dr, where C is the boundary of the region enclosed by y=x^2 and y=cube root of x, and F(x,y)=<x^2-y^3,x^3-y^2>. Please help me.
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    Quote Originally Posted by ottmar0 View Post
    use the Green's Theorem to calculate F dr, where C is the boundary of the region enclosed by y=x^2 and y=cube root of x, and F(x,y)=<x^2-y^3,x^3-y^2>. Please help me.
    Greens theorem tells us that if C is any peicewise simple closed curve then

    \oint _C \vec F \cdot d\vec r=\iint_D \left(\frac{\partial }{\partial x}(x^3-y^2)-\frac{\partial }{\partial x}(x^2-y^3) \right)dA=\iint_D(3x^2+3y^2)dA

    Can you finish from here
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by ottmar0 View Post
    use the Green's Theorem to calculate \int_C F\,dr, where C is the boundary of the region enclosed by y=x^2 and y=\sqrt[3]{x}, and F(x,y)=\langle x^2-y^3,x^3-y^2\rangle. Please help me.
    F(x,y) = \langle P,Q\rangle = \langle x^2-y^3,x^3-y^2\rangle

    \frac{\partial Q}{\partial x} = 3x^2

    \frac{\partial P}{\partial y} = -3y^2

    So, \int\int_A\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,dA = 3\int\int_A(x^2+y^2)\,dA

    Keeping in mind that \sqrt[3]{x}\geq x^2 on [0,1], the bounds are:

    x=0...1
    y=x^2...\sqrt[3]{x}

    So you need to integrate:

    3\int_0^1\int_{x^2}^{\sqrt[3]{x}}(x^2+y^2)\,dy\,dx
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