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Green's Theorem

  1. Apr 30, 2005 #1
    find the double integral of the function e^(x^2) over the region where
    y/2 <= x<= 1 and 0<=y <=2 USING GREEN's THEOREM.

    I can't imagine how we'd use green's theorem here...if F=(P,Q) is the function, are we supposed to find P and Q using green's theorem and then parametrize the boundry of the region. I've tried it...but it's too hard and besides,doesn't seem to work.
  2. jcsd
  3. Apr 30, 2005 #2
    well, to use Green's Theorem, you need P, Q s.t.

    [tex]e^{(x^2)} = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}.[/tex]

    What you need to realise is that you can take ANY P,Q satisfying this requirement. In this case, I would advise using

    [tex]Q = 0, \ P = -ye^{(x^2)}.[/tex]

    By Green's Theorem, you then get

    [tex]\int_0^2 \int_{y/2}^1 e^{(x^2)} \ dx \ dy = \oint_C -ye^{(x^2)} \ dx[/tex]

    where C is the boundary of the region you specified in the counter-clockwise direction. Now just parameterize it and see where you get!
    Last edited: Apr 30, 2005
  4. May 1, 2005 #3
    thanks...but just to make it clearer,can you give me another set of of values of P and Q?
  5. May 1, 2005 #4


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    Do you have any reason to think that other values of P and Q will make this clearer?

    Fine, take [tex]P(x,y)= cos(x)log(x)- ye^{x^2}[/tex] and Q(x,y)= [tex]e^{\frac{sin(y)}{e^{y^2}}[/tex]!

    Do you see why that would give exactly the same answer?
    (What are [tex]\frac{\partial P}{\partial y}[/tex] and [tex]\frac{\partial Q}{\partial x}[/tex]?
    Last edited by a moderator: May 1, 2005
  6. May 1, 2005 #5
    thanks...trust me,that DID help...
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