1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    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:

    \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

    Calculating the root yields 2 so:
    \Gamma=\int_0^{2\pi} 2 \vec{u} d\theta
    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


    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].


    (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:

    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}

    Also performing the transformation for u yields:

    Plugging everything into the integral and taking boundaries 'full-circle' I get:

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook