Proof That Integral of Conjugal Veclocity Around a Contour is Zero

Theorem
Let  
\[\mathbf{v} = f \mathbf{i} = g \mathbf{j}\]
  be the velocity fluid of an incompressible vector field.
Consider the irrotational vector field  
\[\mathbf{u} = g \mathbf{i} + f \mathbf{j} \]
  where  
\[f, \: g\]
  are differentiable.
\[\oint \mathbf{u} \cdot \mathbf{r} =0\]
.
Proof
Since  
\[\mathbf{u}\]
  is irrotational  
\[\mathbf{\nabla} \times \mathbf{u} ==(\frac{\partial f}{\partial x} - \frac{\partial g}{\partial y}) \mathbf{k} =0\]

Since  
\[\mathbf{v} = f \mathbf{i} - g \mathbf{j}\]
  is incompressible  
\[\mathbf{\nabla} \cdot \mathbf{v} = (\frac{\partial }{\partial x} \mathbf{i} + \frac{\partial }{\partial x} \mathbf{i}) \cdot ( f \mathbf{i} - g \mathbf{j}) = \frac{\partial f}{\partial x} - \frac{\partial g}{\partial y}\]

Bu Green's Theorem,  
\[\oint_C g dx + f dy = \int \int_A (\frac{\partial f}{\partial x} - \frac{\partial g}{\partial y}) dx dy\]

Hence  
\[\oint \mathbf{u} \cdot \mathbf{r} = \oint_C g dx + f dy = \int \int_A (\frac{\partial f}{\partial x} - \frac{\partial g}{\partial y}) dx dy = 0\]
.

Add comment

Security code
Refresh