Streamlines of $\psi = x^2 + y^2$

Given: $\psi = x^2 + y^2$, with $u = \partial\psi/\partial y$, $v = -\partial\psi/\partial x$.

Streamlines

A streamline is $\psi =$ constant, so

$$x^2 + y^2 = C = r^2$$

These are concentric circles about the origin with radius $r = \sqrt{\psi}$. For example, $\psi = 1, 4, 9, 16, 25$ give circles of radius $1, 2, 3, 4, 5$ m. For equal steps of $\psi$ (say $\psi = 5, 10, 15, \ldots$) the circles get closer together as $r$ increases, which shows that the velocity increases outward.

             y
             |
         .-"""""-.
       .'  .---.  '.
      /   /     \   \
 ----|---|---O---|---|---- x
      \   \     /   /
       '.  '---'  .'
         '-.....-'
             |
      (flow is clockwise)

Velocity at (3, 4)

$$u = \frac{\partial\psi}{\partial y} = 2y = 8\ \text{m/s},\qquad v = -\frac{\partial\psi}{\partial x} = -2x = -6\ \text{m/s}$$ $$|V| = \sqrt{8^2 + 6^2} = 10\ \text{m/s},\qquad \theta = \tan^{-1}\frac{-6}{8} = -36.87^\circ$$

The velocity is $10$ m/s at $36.9^\circ$ below the +x axis. It is tangent to the circle $r = 5$ m through $(3, 4)$ (because $\psi = 25$ there), and the flow is clockwise. In general $|V| = 2r$, so the flow is rotational, like a rigid-body rotation.

Answer: the streamlines are concentric circles $x^2 + y^2 = \psi$; at (3, 4) the velocity is $(8\hat i - 6\hat j)$ m/s, i.e. 10 m/s at $-36.9^\circ$ to the x-axis.