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

[MichielM]

Homework Statement

A flow field on the xy-plane has the velocity components
$$u=3x+y$$
$$v=2x-3y$$

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

Homework Equations

The circulation $$\Gamma$$ around a closed contour is:
$$\Gamma=\int_C\vec{u}\cdot d\vec{s}$$

The Attempt at a Solution

Because the contour of investigation is a circle, a parametrisation of the form $$x=1+4\cos\theta, y=6+4\sin\theta$$ 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 $$4\pi$$ out of this, $$\vec{u}$$ 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!

 

[gabbagabbahey]

Because the contour of investigation is a circle, a parametrisation of the form $$x=1+4\cos\theta, y=6+4\sin\theta$$ should help.

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

The circulation then can be calculated as:

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

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

$$d\textbf{s}=dx\textbf{i}+dy\textbf{j}=\left(\frac{dx}{d\theta}\right)d\theta\textbf{i}+\left(\frac{dy}{d\theta}\right)d\theta\textbf{j}$$

(Since $x$ and $y$ depend only on $\theta$ for this curve)

 

[MichielM]

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

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:
$$\vec{u}=(3x+y)\vec{i}+(2x-3y)\vec{j}=(9+6\cos\theta+2\sin\theta)\vec{i}+(-16+4\cos\theta-6\sin\theta)\vec{j}$$

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

Taking the integrals there are only 2 terms which don't completely cancel, the terms coming from the integration of $$\sin^2(\theta)$$ and $$\cos^2(\theta)$$ which add up to $$\Gamma=-4\pi \vec{i}+8\pi \vec{j}$$. Adding up this yields $$4\pi$$ but how come I can just add those two component up? Because I integrated both parts to the same parameter?

 

[rs1n]

$$\Gamma=\int_0^{2\pi}-2\sin(\theta)(9+6\cos\theta+2\sin\theta)d\theta\vec{i}+2\cos(\theta)(-16+4\cos\theta-6\sin\theta)d\theta\vec{j}$$

The integrand is a dot product. If you compute the dot product of $\mathbf{u}\cdot \mathbf{v}$ you should get a scalar, not a vector.

$$(u_1 \vec{\mathbf{i}} + u_2 \vec{\mathbf{j}}) \cdot (v_1 \vec{\mathbf{i}} + v_2 \vec{\mathbf{j}}) = u_1 v_1 + u_2 v_2$$

Everything else you did is fine. You just have extra $\vec{\mathbf{i}}$'s and $\vec{\mathbf{j}}$ where there should be none.

 

[MichielM]

ah I see, thanks! My vector algebra has become a bit rusty in the last years.