Discussion
If the stream function (\(\psi(x, y)\)) for a two-dimensional incompressible flow field is given as \(2y(x^{2} - y^{2})\), the corresponding velocity field is
Answer: B
The velocity field is \(\vec{V} = 2(x^{2} - 3y^{2})\hat{i} - 4xy\,\hat{j}\).
With \(\psi = 2x^{2}y - 2y^{3}\) and the usual convention \(u = \partial\psi/\partial y\), \(v = -\partial\psi/\partial x\):
\[ \begin{aligned} u &= 2x^{2} - 6y^{2} = 2(x^{2} - 3y^{2}) \\ v &= -4xy \end{aligned} \]
Check: \(\partial u/\partial x + \partial v/\partial y = 4x - 4x = 0\), so the flow is incompressible, as it must be. The option with \(+4xy\) has the wrong sign on v and fails this continuity check. The two options with \(2x^{2}y\) for u are just the stream function's first term, not its derivative.
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