How Does Stokes' Theorem Relate to Vorticity in Fluid Dynamics?

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Homework Statement

Consider an imaginary circular disc, of radius R, whose arbitrary orientation is described by the unit vector, $$\vec {n}$$, perpendicular to the plane of the disc. Define the component, in the direction $$\vec {n}$$, of the angular velocity, $$\vec {\Omega}$$, at a point in the fluid by $$\vec {\Omega}. \vec {n} = \lim_{R \rightarrow 0}[\frac {1}{2 \pi R^2} \oint_C \vec {u}.dl]$$, where C denotes the the boundary (rim) of the disc. Use Stokes' theorem, and the arbitrariness of $$\vec {n}$$, to show that $$\vec {\Omega}= \frac {1}{2} \vec {\omega}$$, where $$\vec {\omega} = \nabla * \vec {u}$$ is the vorticity of the fluid at R=0. [This definition is based on a description applicable to the rotation of solid bodies. Confirm this by considering $$\vec {u} = \vec {U} + \vec {\Omega}* \vec{r}$$, where $$\vec {U}$$ is the translational velocity of the body, $$\vec {\Omega}$$ is its angular velocity and $$\vec {r}$$ is the position vector of a point relative to a point on the axis of rotation.]

Homework Equations

Stokes' theorem : $$\oint_c u.dl = \iint_S (\nabla * u) .n ds$$

The Attempt at a Solution

Answer it gives in back of book is:
Stokes theorem gives
$$\oint_c u.dl = \iint_S (\nabla * u) .n ds = (\omega.n)\pi a^2$$ (How did they get this !?)so $$\Omega.n = \frac {1}{2} \omega . n$$; but n is arbitrary so $$\Omega = \frac {1}{2} \omega.$$ NB $$\oint_c u.dl = \oint_c U.dl + \oint_c (\Omega * r).dl = \Omega.\oint_c (r*dl) = \Omega.n \int_{0}^{2\pi} a^2 d\theta$$

 

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$$x=\alpha, y=\beta.$$
[
i.e. solution of
ax+by = -e
cx + dy = -f
]

Therefore if you introduce new variables$$\xi=x-\alpha$$ and $$\eta=y-\beta,$$ in the new variables $$\xi$$ and $$\eta$$ you will have a system of equations

$$\xi' = a \xi + b \eta$$
$$\eta' = c \xi + d \eta$$