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

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

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

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