(adsbygoogle = window.adsbygoogle || []).push({}); 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!!

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

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