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Homework Help: Circulation of a flow field around circle (x-1)^2+(y-6)^2=4

  1. Apr 12, 2010 #1
    1. The problem statement, all variables and given/known data
    A flow field on the xy-plane has the velocity components
    [tex]
    u=3x+y
    [/tex]
    [tex]
    v=2x-3y
    [/tex]

    Show that the circulation around the circle [tex](x-1)^2+(y-6)^2=4[/tex] equals [tex]4\pi[/tex]

    2. Relevant equations
    The circulation [tex]\Gamma[/tex] around a closed contour is:
    [tex] \Gamma=\int_C\vec{u}\cdot d\vec{s}[/tex]


    3. The attempt at a solution
    Because the contour of investigation is a circle, a parametrisation of the form [tex] x=1+4\cos\theta, y=6+4\sin\theta [/tex] should help.
    The circulation then can be calculated as:

    [tex]
    \Gamma=\int_0^{2\pi} \vec{u} \sqrt{\left(\frac{\partial x}{\partial \theta}{\right)^2+\left(\frac{\partial y}{\partial \theta}{\right)^2} d\theta
    [/tex]

    Calculating the root yields 2 so:
    [tex]
    \Gamma=\int_0^{2\pi} 2 \vec{u} d\theta
    [/tex]
    To get [tex]4\pi[/tex] out of this, [tex]\vec{u}[/tex] should 1, but this is the point where I don't know how to show/calculate that. The integration of the vector is the point where I get stuck. Any help is welcome!!
     
    Last edited: Apr 12, 2010
  2. jcsd
  3. Apr 12, 2010 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Careful, the radius of the circle [itex](x-1)^2+(y-6)^2=4[/itex] is [itex]2[/itex][, not [itex]4[/itex].

    [itex]\sqrt{\left(\frac{\partial x}{\partial \theta}\right)^2+\left(\frac{\partial y}{\partial \theta}\right)^2} d\theta[/itex] only gives the magnitude of the vector [itex]d\textbf{s}[/itex].

    [tex]d\textbf{s}=dx\textbf{i}+dy\textbf{j}=\left(\frac{dx}{d\theta}\right)d\theta\textbf{i}+\left(\frac{dy}{d\theta}\right)d\theta\textbf{j}[/tex]

    (Since [itex]x[/itex] and [itex]y[/itex] depend only on [itex]\theta[/itex] for this curve)
     
  4. Apr 12, 2010 #3
    Oops, little to quick in my steps. So instead of 4 i take 2 in those equations.

    I think I got it now:

    [tex]
    d\textbf{s}=dx\textbf{i}+dy\textbf{j}=\left(\frac{ dx}{d\theta}\right)d\theta\textbf{i}+\left(\frac{d y}{d\theta}\right)d\theta\textbf{j}=-2\sin \theta d\theta\textbf{i}+2\cos \theta d\theta\textbf{j}
    [/tex]

    Also performing the transformation for u yields:
    [tex]
    \vec{u}=(3x+y)\vec{i}+(2x-3y)\vec{j}=(9+6\cos\theta+2\sin\theta)\vec{i}+(-16+4\cos\theta-6\sin\theta)\vec{j}
    [/tex]

    Plugging everything into the integral and taking boundaries 'full-circle' I get:
    [tex]
    \Gamma=\int_0^{2\pi}-2\sin(\theta)(9+6\cos\theta+2\sin\theta)d\theta\vec{i}+2\cos(\theta)(-16+4\cos\theta-6\sin\theta)d\theta\vec{j}
    [/tex]

    Taking the integrals there are only 2 terms which don't completely cancel, the terms coming from the integration of [tex]\sin^2(\theta)[/tex] and [tex]\cos^2(\theta)[/tex] which add up to [tex]\Gamma=-4\pi \vec{i}+8\pi \vec{j}[/tex]. Adding up this yields [tex]4\pi[/tex] but how come I can just add those two component up? Because I integrated both parts to the same parameter?
     
  5. Apr 12, 2010 #4
    The integrand is a dot product. If you compute the dot product of [itex]\mathbf{u}\cdot \mathbf{v}[/itex] you should get a scalar, not a vector.

    [tex](u_1 \vec{\mathbf{i}} + u_2 \vec{\mathbf{j}}) \cdot (v_1 \vec{\mathbf{i}} + v_2 \vec{\mathbf{j}}) = u_1 v_1 + u_2 v_2[/tex]

    Everything else you did is fine. You just have extra [itex]\vec{\mathbf{i}}[/itex]'s and [itex]\vec{\mathbf{j}}[/itex] where there should be none.
     
  6. Apr 12, 2010 #5
    ah I see, thanks! My vector algebra has become a bit rusty in the last years.
     
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