Green's Theorem

  • Thread starter mansi
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  • #1
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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.
i
 

Answers and Replies

  • #2
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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:
  • #3
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thanks...but just to make it clearer,can you give me another set of of values of P and Q?
 
  • #4
HallsofIvy
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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]?
 
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  • #5
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thanks...trust me,that DID help...
 

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