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Green's theorem

  1. May 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Use greens theorem to calculate.
    [itex]\int_{c}(e^{x}+y^{2})dx+(e^{x}+y^{2})dy[/itex]

    Where c is the region between y=x2y=x

    2. Relevant equations

    Greens Theorem

    [itex]\int_{c}f(x.y)dx+g(x,y)dy= \int_{R}\int (\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y})dA[/itex]


    3. The attempt at a solution

    [itex]\frac{\partial g}{\partial x}= 2x[/itex]
    [itex]\frac{\partial g}{\partial x}= 2y[/itex]
    Calculate the integral

    [itex]\int_{0}^{x}\int_{0}^{\sqrt{y}}2x-2y\text{ }dy dx[/itex]

    [itex]=\frac{x^2}{2}-\frac{4x^{5/2}}{5}[/itex]

    Does this look right?
    regards
     
  2. jcsd
  3. May 24, 2009 #2
    with f(x,y)=g(x,y)=exp(x)+y*y, dg/dx=exp(x), the second dg/dx is a typo.

    if you want the region bounded by y=x^2 and y=x, the inside integral must be from x^2 to x and the outside 0 to 1 with area element dydx, the result needs to be a value rather than a function, just something to get use to with multiple integrals.
     
  4. May 24, 2009 #3
    Thanks

    [itex]\int_{0}^{1}\int_{x}^{x^2}2x-2y\text{ }dy dx[/itex]

    [itex]=\frac{1}{30}[/itex]

    With the outside limits of double integrals eg 0 to 1 do they always have to be constants?
    regards
     
  5. May 24, 2009 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    If the result is supposed to be a constant, then, yes, the limits of the integral have to be numbers, not variables!
     
  6. May 25, 2009 #5
    Thanks
     
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