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  1. #1
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    Circulation

    Calculate the circulation of vector field F arrond \alpha

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

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

    Is correct ?

    2) F(x,y) = (x+2xy)i + (y+x^2)j and \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 \alpha

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

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

    Is correct ?

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

    for number 2

    \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 \frac{1}{2} is supposed to be a 2

    for number 2

    \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 \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 \alpha

    1) F(x,y) = (x-y)i + (x+y)j and \alpha is the closed polygonal ABCDA traveled once in anticlockwise joining the points A(1,0), B(1,2), C(0,2) and D(0,1).
    \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 A=\frac{3}{2}

    \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
    \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 A=\frac{3}{2}

    \iint 2 dA=2\iint dA=2\cdot \frac{3}{2}=3
    I calculate \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 \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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