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
<br /> \eta&#039; = c \xi + d \eta