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Math Help - Green's Theorem Integration Question

  1. #1
    Junior Member CuriosityCabinet's Avatar
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    Green's Theorem Integration Question

    Here's the question:

    Evaluate
    \int_{C} e^x cos y dx - e^xsinydy
    where A = (ln 2, 0) to D =(0,1) and then from D to B = (-ln2, 0). Hint: Apply Green's theorem to the integral around the closed curve ADBA.
    So using Green's Theorem, I got that the integral is equal to

    \int_{C}\frac{\partial}{\partial x}(-e^xsiny) - \frac{\partial}{\partial x}(e^xcosy)dxdy = 0.

    But surely the answer can't be 0? What am I doing wrong?
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  2. #2
    Member Ruun's Avatar
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    Re: Green's Theorem Integration Question

    Look at the curve of your integral


    Your integral is the red one, and the closed curve is the red one plues the green one. The point is that the green one is a integral only over x because y=0 along that line, so you can apply Green's theorem to the closed (red + green) curve. As you have said it gives 0 but what it is 0 is the integral along the closed curve, not the one you asked. So:

    \oint e^x\cos(y)dx-e^{x}sin(y)dy=\int \left( -\frac{\partial}{\partial x}e^{x}sin(y)-\frac{\partial}{\partial y}e^{x}cos(y)\right)dxdy=0

    Now I denote R the red curve and G the green one, so that:

    \oint Mdx+Ndy=\int_{R} Mdx+Ndy+\int_{G} Mdx+Ndy

    Note that both the integral in R and G are line integrals!

    In particular, the G is a easy line integral, the usual integral over some interval in \mathbb{R}
    Last edited by Ruun; April 3rd 2013 at 04:01 AM.
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  3. #3
    Junior Member CuriosityCabinet's Avatar
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    Re: Green's Theorem Integration Question

    Many thanks, you've taught me something vital!
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  4. #4
    Member Ruun's Avatar
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    Re: Green's Theorem Integration Question

    Glad to help, math is better learned in company
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