I am trying to show that the kinetic energy for an incompressible and irrotational fluid with no sources and no sinks is given by
$$\frac{\delta}{2} \iint_{S} \psi \frac{\partial \psi}{\partial n} dS $$
I tried to use that Kinetic Energy$= \frac{1}{2} \iiint_{V} \delta v^{2} d V $ where $\psi$ is the velocity potential, $v= \nabla \psi$
and that $v^2= \nabla^2 \psi ^2$ and then tried using divergence theorem but I cant get it to work .
Use the vector calculus identity
$$\nabla \cdot (\psi \nabla \psi)= \nabla \psi \cdot \nabla \psi + \nabla \cdot\nabla \psi.$$
Then with $\mathbb{v} = \nabla \psi$ we have
$$\nabla \cdot (\psi \nabla \psi)= \mathbb{v}\cdot\mathbb{v} + \nabla \cdot \mathbb{v} = v^2,$$
since $\nabla \cdot \mathbb{v} = 0$ in incompressible flow.
Hence, by the divergence theorem
$$\int_V v^2 dV = \int_V \nabla \cdot (\psi \nabla \psi) dV= \int_S\psi \nabla \psi \cdot \mathbb{n}dS,$$
where $\mathbb{n}$ is the outward unit normal vector. The dot product of the gradient with the normal vector is by definition the normal derivative of the potential.
Thus,
$$\int_V v^2 dV = \int_S\psi \frac{ \partial \psi}{\partial n} dS$$