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Math Help - Green's Theorem

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    Super Member Aryth's Avatar
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    Green's Theorem

    Use Green's Theorem to evaluate the line integral along the given positively oriented curve.

    \int_C e^y ~dx + 2xe^y ~dy

    C is the square with sides x=0, x=1, y=0, and y=1.
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    Quote Originally Posted by Aryth View Post
    Use Green's Theorem to evaluate the line integral along the given positively oriented curve.

    \int_C e^y ~dx + 2xe^y ~dy

    C is the square with sides x=0, x=1, y=0, and y=1.
    Green's Theorem: \oint_C f(x, y) \, dx + g(x, y) \, dy = \int \int_{R} \left( \frac{\partial g}{\partial x} - \frac{\partial f}{\partial y} \right) \, dx \, dy where R is the region bounded by C.

    In your case, f(x, y) = e^y \Rightarrow \frac{\partial f}{\partial y} = e^y and g(x, y) = 2x e^y \Rightarrow \frac{\partial g}{\partial x} = 2 e^y.

    Therefore \oint_C e^y ~dx + 2xe^y ~dy = \int_{y=0}^{1} \int_{x = 0}^{1} e^y \, dx \, dy = \int_{y=0}^{1} e^y \, dy = e - 1.
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    Super Member Aryth's Avatar
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    Would you mind showing the iterations for the double integral?

    It's just for my verification.
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    Quote Originally Posted by Aryth View Post
    Would you mind showing the iterations for the double integral?

    It's just for my verification.
    You mean this one :

    \int_{y=0}^{1} \int_{x = 0}^{1} e^y \, dx \, dy

    ?

    Because e^y doesn't contain any x, we can write it this way :

    =\underbrace{\left(\int_{y=0}^1 e^y \, dy\right)}_{=e-1} \cdot \underbrace{\left(\int_{x=0}^1 \, dx\right)}_{=1}
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