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

[coverband]

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

Either 1.:
C is boundary of x^2 + y^2 = R^2. Parametrically x= Rcost, y = Rsint
dx = -Rsintdt, dy = Rcostdt
L.H.S. of Stokes becomes \oint_c udx + vdy + wdz
= \oint_C -uRSintdt + vRCostdt
=\int_{0}^{2 \pi} -uRSint dt + vRCost dt
=uRCost + vRSint \right]_{0}^{2 \pi}
= -uR - uR
=-2uR
multiply term outside integral
=-\frac{u}{\pi R}

Or 2:
\frac {1}{2} (\nabla * \vec {u}). \vec {n} = \lim_{R \rightarrow 0}[\frac {1}{2 \pi R^2} \iint_S (\nabla * u) .n ds ], ...

 

[Coto]

[ Nevermind :) .. I was wrong. ]

 

[zeebek]

Start with this

<br /> \vec {\Omega}. \vec {n} = \lim_{R \rightarrow 0}[\frac {1}{2 \pi R^2} \oint_C \vec {u}.dl] <br />

then use Stokes theorem and get in the limit R \rightarrow 0

<br /> \vec {\Omega}. \vec {n} = \lim_{R \rightarrow 0}[\frac {1}{2 \pi R^2} \iint_S (\nabla * u) .n ds ] = \lim_{R \rightarrow 0}[\frac {1}{2 \pi R^2} (\nabla * u) \pi R^2 ] = 0.5 \nabla * u<br />