## Minimisung Flow Though a Surface Due to a Vortex

Suppose a fluid is rotating with uniform angular velocity
$\mathbf{\omega}$
about the origin. Where should a unit square be placed in the
$z$
plane so that the velocity flux across it is minimized?
We can write
$\mathbf{\omega}=\omega \mathbf{i}, \mathbf{r}= x \mathbf{i} + \mathbf{j} + z \mathbf{k}$
then
\begin{aligned} \mathbf{v} &= \mathbf{\omega} \times \mathbf{r} \\ &= \omega \mathbf{i} \times x \mathbf{i} + \mathbf{j} + z \mathbf{k} \\ &= - \omega y \mathbf{i} + \omega x {j} \end{aligned}

Let
$M$
be the mass of flux passing through the unit square per second. then
$M= \int_S \rho \mathbf{v} \cdot \mathbf{n} dS$
The rotation is uniform around the
$z$
axis so we can pick a point to place the unit square to make the calculations simple. Pick a point on the y axis with coordinates
$(0,a/2,1/2)$
to be the centre of the square, then
$\mathbf{n} = \mathbf{i}$
. Then
\begin{aligned} M= \int_S \rho \mathbf{v} \cdot \mathbf{n} dS &= \rho \omega \int^1_0 \int^{a+1}_a y dy dz \\ &= \rho \omega \int^1_0 [\frac{y^2}{2}]^{a+1}_a dz \\ &= \rho \omega \int^1_0 [\frac{2a=1}{2}]^{a+1}_a dz \\ & =\rho \omega \frac{2A+1}{2} \end{aligned}

$m$
will be zero at
$a=- \frac{1}{2}$
.