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Thread: Circulation

  1. #1
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    Circulation

    Calculate the circulation of vector field F arrond $\displaystyle \alpha$

    1) $\displaystyle F(x,y) = (x-y)i + (x+y)j$ and $\displaystyle \alpha$ is the closed polygonal ABCDA traveled once in anticlockwise joining the points $\displaystyle A(1,0)$, $\displaystyle B(1,2)$, $\displaystyle C(0,2)$ and $\displaystyle D(0,1)$.

    My solution:
    Using green i have 2
    The circulation is $\displaystyle \frac{1}{2}(A1+A2) = \frac{3}{4}$
    A1 = $\displaystyle 1*1=1$
    A2 = $\displaystyle \frac{1*1}{2}$

    Is correct ?

    2) $\displaystyle F(x,y) = (x+2xy)i + (y+x^2)j$ and $\displaystyle \alpha : |x| + |y| = 1$
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  2. #2
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    Quote Originally Posted by Apprentice123 View Post
    Calculate the circulation of vector field F arrond $\displaystyle \alpha$

    1) $\displaystyle F(x,y) = (x-y)i + (x+y)j$ and $\displaystyle \alpha$ is the closed polygonal ABCDA traveled once in anticlockwise joining the points $\displaystyle A(1,0)$, $\displaystyle B(1,2)$, $\displaystyle C(0,2)$ and $\displaystyle D(0,1)$.

    My solution:
    Using green i have 2
    The circulation is $\displaystyle \frac{1}{2}(A1+A2) = \frac{3}{4}$
    A1 = $\displaystyle 1*1=1$
    A2 = $\displaystyle \frac{1*1}{2}$

    Is correct ?

    2) $\displaystyle F(x,y) = (x+2xy)i + (y+x^2)j$ and $\displaystyle \alpha : |x| + |y| = 1$
    I think that your $\displaystyle \frac{1}{2}$ is supposed to be a 2

    for number 2

    $\displaystyle \frac{\partial }{\partial x}(y+x^2)-\frac{\partial }{\partial y} =2x-2x=0$

    Since the above is a closed curve (a Dimond) Greens theorem says it is zero.
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    I think that your $\displaystyle \frac{1}{2}$ is supposed to be a 2

    for number 2

    $\displaystyle \frac{\partial }{\partial x}(y+x^2)-\frac{\partial }{\partial y} =2x-2x=0$

    Since the above is a closed curve (a Dimond) Greens theorem says it is zero.
    If green = 2 of the area nos is $\displaystyle \frac{1}{2}$ ?
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  4. #4
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    Quote Originally Posted by Apprentice123 View Post
    Calculate the circulation of vector field F arrond $\displaystyle \alpha$

    1) $\displaystyle F(x,y) = (x-y)i + (x+y)j$ and $\displaystyle \alpha$ is the closed polygonal ABCDA traveled once in anticlockwise joining the points $\displaystyle A(1,0)$, $\displaystyle B(1,2)$, $\displaystyle C(0,2)$ and $\displaystyle D(0,1)$.
    $\displaystyle \iint \left( \frac{\partial }{\partial x}(x+y)-\frac{\partial }{\partial y}(x-y) \right)dA=\iint 2 dA=2\iint dA$

    The area of the region you calculated is correct $\displaystyle A=\frac{3}{2}$

    $\displaystyle \iint 2 dA=2\iint dA=2\cdot \frac{3}{2}=3$
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  5. #5
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    Quote Originally Posted by TheEmptySet View Post
    $\displaystyle \iint \left( \frac{\partial }{\partial x}(x+y)-\frac{\partial }{\partial y}(x-y) \right)dA=\iint 2 dA=2\iint dA$

    The area of the region you calculated is correct $\displaystyle A=\frac{3}{2}$

    $\displaystyle \iint 2 dA=2\iint dA=2\cdot \frac{3}{2}=3$
    I calculate $\displaystyle \frac{1}{2}(1+\frac{1}{2}) = \frac{3}{4}$
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  6. #6
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    Quote Originally Posted by Apprentice123 View Post
    I calculate $\displaystyle \frac{1}{2}(1+\frac{1}{2}) = \frac{3}{4}$
    How??

    Circulation-capture.jpg

    Where does the 1/2 come from??
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  7. #7
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    Quote Originally Posted by TheEmptySet View Post
    How??

    Click image for larger version. 

Name:	Capture.JPG 
Views:	9 
Size:	45.9 KB 
ID:	11235

    Where does the 1/2 come from??

    Ohhh yes, thank you
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