# Why that Bernoulli function contains many streamlines and vortex lines but not only singular streamlines?

I know that for a stream line inside the gradient part equal to a constant. But when we consider the fluid is rotational which the velocity cross product vorticity not equal to zero. So how can we prove that this bernouli surface contains many stramlines and vortex lines but not only one singular streamlines?

If $$H\equiv p/\rho+u^2/2$$ is the Bernoulli constant in an inviscid flow then Bernoulli equation says $$\nabla H=\vec{u}\times\vec{\omega}$$ where symbols have their usual meaning ($$\vec{\omega}$$ is the vorticity vector). It says that $$H$$ is constant along streamlines and vortexlines, although its value can differ between streamlines/vortexlines.
Suppose on a particular vortexline, Bernoulli constant has value $$H_0$$. Then every streamline intersecting that vortexline must have the same Bernoulli constant, because that value is common at the intersection.
Using the same argument, the surface generated by intersection of vortexlines and streamlines must have the same Bernoulli constant. Thus where vorticity is present we can have surfaces (rather than simply one-dimensional curves) of constant $$H$$.
This doesn't hold for Beltrami flows, a (hypothetical) degenerate class of flows, in which vorticity is aligned with velocity, so $$\vec{u}\times\vec{\omega}=0$$. There $$H$$ is constant along one-dimensional curves only.