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Thread: Green's Theorem

  1. #1
    Senior Member chella182's Avatar
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    Green's Theorem

    I only have a minor problem with this really... the integral is...
    $\displaystyle \oint_{\Gamma}\{(y\exp{[x^2y^2]}-y)dx+(x\exp{[x^2y^2]}+x)dy\}$
    ...where $\displaystyle \Gamma$, mapped counter-clockwise, is the circle of radius $\displaystyle a$ with centre at the origin.

    So I identified $\displaystyle P=y\exp{[x^2y^2]}-y$ and $\displaystyle Q=x\exp{[x^2y^2]}+x$ and similarly $\displaystyle \frac{\partial P}{\partial y}=2y^2x^2\exp{[x^2y^2]}+\exp{[x^2y^2]}-1$ and $\displaystyle \frac{\partial Q}{\partial x}=2y^2x^2\exp{[x^2y^2]}+\exp{[x^2y^2]}+1$.

    Putting that into the double integral you end up with $\displaystyle 2\iint_{A}dA$, but I'm not sure where to proceed from here. Is $\displaystyle \iint_{A}dA$ the area of the circle?
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  2. #2
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    Quote Originally Posted by chella182 View Post
    I only have a minor problem with this really... the integral is...
    $\displaystyle \oint_{\Gamma}\{(y\exp{[x^2y^2]}-y)dx+(x\exp{[x^2y^2]}+x)dy\}$
    ...where $\displaystyle \Gamma$, mapped counter-clockwise, is the circle of radius $\displaystyle a$ with centre at the origin.

    So I identified $\displaystyle P=y\exp{[x^2y^2]}-y$ and $\displaystyle Q=x\exp{[x^2y^2]}+x$ and similarly $\displaystyle \frac{\partial P}{\partial y}=2y^2x^2\exp{[x^2y^2]}+\exp{[x^2y^2]}-1$ and $\displaystyle \frac{\partial Q}{\partial x}=2y^2x^2\exp{[x^2y^2]}+\exp{[x^2y^2]}+1$.

    Putting that into the double integral you end up with $\displaystyle 2\iint_{A}dA$, but I'm not sure where to proceed from here. Is $\displaystyle \iint_{A}dA$ the area of the circle?
    yes, it's the area of the circle. so your integral is equal to $\displaystyle 2 \pi a^2.$
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  3. #3
    Senior Member chella182's Avatar
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    Thankyou
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