A 2D velocity field is given by $\vec{V} = (4 x y)\hat{i} + (2 x^2 - 2 y^2)\hat{j}$. What is the magnitude of the vorticity vector $\vec{\omega} = \nabla \times \vec{V}$ at the point $(x, y) = (2, 3)$?

Answer Key:

0

### Solution: Vorticity in 2D is $\omega_z = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}$. Here $u = 4xy \implies \frac{\partial u}{\partial y} = 4x$. $v = 2x^2 - 2y^2 \implies \frac{\partial v}{\partial x} = 4x$. $\omega_z = 4x - 4x = 0$ everywhere! Thus vorticity magnitude is 0. Correct: 0.

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