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Math Help - Vector field

  1. #1
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    Vector field

    Calculate the circulation of the vector field F(x,y) = (x,xy) around the closed curve x^2+y^2 = 25 situated above the axis OX
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  2. #2
    MHF Contributor Calculus26's Avatar
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    we have talked before if F is a force field the line integral is the Work done

    If F is a velocity field then the line integral is the circulation

    so just compute the line integral of F over C

    Her C consists of 2 parts the circle and the line segment along the x axis from -5 to 5

    But of course you have Green's thm in your tool box as well
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  3. #3
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    Quote Originally Posted by Calculus26 View Post
    we have talked before if F is a force field the line integral is the Work done

    If F is a velocity field then the line integral is the circulation

    so just compute the line integral of F over C

    Her C consists of 2 parts the circle and the line segment along the x axis from -5 to 5

    But of course you have Green's thm in your tool box as well
    Above the axis ox:

    \int_0^{ \pi} (-costsent + cos^2tsent)dt = - \frac{2}{3}

    Correct ?
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  4. #4
    MHF Contributor Calculus26's Avatar
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    Not quite

    x= 5cos(t) y = 5sin(t)

    dr/dt = -5*sin(t) +5*cos(t)

    F = 5cos(t)i +25cos(t)sin(t)

    go from here
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  5. #5
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    Quote Originally Posted by Calculus26 View Post
    Not quite

    x= 5cos(t) y = 5sin(t)

    dr/dt = -5*sin(t) +5*cos(t)

    F = 5cos(t)i +25cos(t)sin(t)

    go from here
    \int_0^{ \pi} (-25costsint + 125 cos^2tsint)dt = \frac{250}{3}

    It is correct ?
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  6. #6
    MHF Contributor Calculus26's Avatar
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    Ok--now you are correct----Again I've mentioned this before--slow down

    and take your time
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  7. #7
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    Quote Originally Posted by Calculus26 View Post
    Ok--now you are correct----Again I've mentioned this before--slow down

    and take your time
    thanks
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