If a flow velocity field is given by \(V = 2{x^3}\hat i + 6{x^2}y\hat j\) :

Concept:

Any field which satisfied the continuity equation given below is considered as flow field

\(\frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} + \frac{{\partial w}}{{\partial z}} = 0\)

Where u, v, ω are x, y and z components of velocity field of flow.

Further ; The flow is said to be rotational is curl of velocity vector (i.e. \(\nabla \times \vec U\) is not equal to zero, otherwise flow is irrotational.

i.e. If \(\nabla \times \vec U = 0\) then flow is irrotational.

Calculation:

Given:

\(\vec U = \left( {2{x^3}} \right)i + \left( {6{x^2}y} \right)\hat I\)

Since given  has component in x and y direction and no component in ‘z’ direction, so it is a 2D velocity field.

Further \(\vec u = 2{x^3}\;\& \;U = 6{x^2}y\) 

\(\frac{{\partial u}}{{\partial x}} = 6{x^2}\;;\frac{{\partial u}}{{\partial y}} = 6\;x^2\)

Since \(\frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} = 6{x^2} + 6\;x^2 \ne 0 \Rightarrow \) not a flow.

Since, the given velocity vector did not satisfied the continuity equation, so it did not represent the flow or flow is not possible.

Also, is there is no flow, hence is no sense of saying whether the flow is rotational or irrotational flow.