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Math Help - Theorem of Green

  1. #1
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    Theorem of Green

    If C is the triangle of vertices (0,0), (1,0) and (1,1) oriented in direction anti-clockwise

    \int_C 2xydx + (x^2+x)dy = ?

    Solution:

    \frac{ \partial (x^2+x)}{ \partial x} - \frac{ \partial (2xy)}{ \partial y} = 1

    1 \int \int_D dxdy = \int_0^1 \int_0^1 dxdy = 1

    Correct ????
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Not quite

    Green's Thm --- Line Integral = integral(dg/dx-df/dy) over the region enclosed by the C

    you computed dg/dx-df/dy correctly = 1

    so the line integral is the area of the triangle then which is 1/2
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  3. #3
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    Quote Originally Posted by Calculus26 View Post
    Not quite

    Green's Thm --- Line Integral = integral(dg/dx-df/dy) over the region enclosed by the C

    you computed dg/dx-df/dy correctly = 1

    so the line integral is the area of the triangle then which is 1/2
    \int_a^b
    How do I determine a and b ?
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  4. #4
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     <br />
\int_0^1\int_0^x dy\,dx<br />
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  5. #5
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    Quote Originally Posted by danny arrigo View Post
     <br />
\int_0^1\int_0^x dy\,dx<br />
    How did you know \int_0^1\int_0^x dy\,dx ?
    Last edited by mr fantastic; April 27th 2009 at 02:08 AM. Reason: Added clarification
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  6. #6
    MHF Contributor Calculus26's Avatar
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    I know you know how to set up a double integral otherwise you wouldn't be doing line integrals and green's theorem.

    the equation of the line from (0,0) to (1,1)

    is y = x

    so y varies from 0 to x as x varies from 0 to 1
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  7. #7
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    Of course. thanks
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  8. #8
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    Quote Originally Posted by Apprentice123 View Post

    How did you know \int_0^1\int_0^x dy\,dx ?
    Here is the region of interest. In order to set up

    \iint_R f(x,y)\,dA then we would need the following

     <br />
\int_{\text{point}}^{\text{point}} <br />
\int_{\text{curve}}^{\text{curve}} f(x,y)\,dy\,dx.

    Since the inner integral involves dy then the curve to cuve in in the y direction and in your case y=0 \to x and hence the inside integral \int_0^x. Then outer integral is point to point and since the outer involves dx then we are moving in the x direction and from the picture of the region it x = 0 \to 1 and hence the outer integral of \int_0^1.

    Hope that helps.
    Attached Thumbnails Attached Thumbnails Theorem of Green-pic.jpg  
    Last edited by mr fantastic; April 27th 2009 at 02:10 AM. Reason: Edited quote to reflect moderation of thread.
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  9. #9
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    Quote Originally Posted by danny arrigo View Post
    Here is the region of interest. In order to set up

    \iint_R f(x,y)\,dA then we would need the following

     <br />
\int_{\text{point}}^{\text{point}} <br />
\int_{\text{curve}}^{\text{curve}} f(x,y)\,dy\,dx.

    Since the inner integral involves dy then the curve to cuve in in the y direction and in your case y=0 \to x and hence the inside integral \int_0^x. Then outer integral is point to point and since the outer involves dx then we are moving in the x direction and from the picture of the region it x = 0 \to 1 and hence the outer integral of \int_0^1.

    Hope that helps.

    Thank you
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