Show that$\int_V{\left(\vec{v}\cdot \nabla\right)\vec{v}} dV = \int_B{\vec{v}\cdot (\vec{v} \cdot \vec{n}) dS}$ i was revisting some old documents from my fluid-dynamic courses and tried to derive the finite-volume discretization of the navier-stokes equations myself.
The Navier-Stokes equation describes the momentum-equilibrium in moving fluids and is a partial differential equation. It has the basic form:
$$\dfrac{d\vec{u}}{dt} = \dfrac{\partial \vec{u}}{\partial t} + \left(\vec{v}\cdot \nabla\right)\vec{v} = f$$
For a finite-volume discretization, I want to integrate that over a finite volume. The first part $\dfrac{\partial \vec{u}}{\partial t}$ as well as the right-hand side $f$ are somewhat easy
to integrate. I was struggling to apply the Gauss-Theorem to the middle part $\left(\vec{v}\cdot \nabla\right)\vec{v}$.

From the basics I know about the Gauss-theorem, I should find a function $F$ which holds: $div(F) = \left(\vec{v}\cdot \nabla\right)\vec{v}$ so I can easily apply the Gauss-Theorem but I wasnt able to find such a function.
From my fluid dynamics-course I remember the momentum-equations derived in a different manner resulting in:
$$\int_B \vec{v} (\vec{v}\cdot \vec{n}) dS$$

Now my question is how one could derive that equation from the navier-stokes differential equation using the Gauss-theorem (or any other method of integration).
Basically I want an understanding / simple way to show that:
$$\int_V{\left(\vec{v}\cdot \nabla\right)\vec{v}} dV = \int_B{\vec{v}\cdot (\vec{v} \cdot \vec{n}) dS}$$
I am very happy for any help!
 A: You won't be able to find a vector field $\vec{F}$ with $\nabla\cdot \vec{F} =(\vec{v}\cdot\nabla)\vec{v}$, since the left side is a scalar and the right side is a vector. Really what you need is to integrate by parts* and use the divergence-free condition on $\vec{v}$.
$$\begin{align}\int_V (\vec{v}\cdot\nabla)\vec{v} \ dV &= \sum_i\int_V v_i(\partial_i\vec{v}) \ dV \\ 
&= -\sum_i \int (\partial_i v_i) \vec{v}\ dV+\sum_i\int_{\partial V} (v_i n_i)\vec{v}  \ dS\\
&= -\int_V (\nabla\cdot\vec{v})\vec{v} \ dV + \int_{\partial V} (\vec{v}\cdot\hat{n})\vec{v} \ dS\end{align}$$
Typically we assume the fluid is incompressible, so $\nabla\cdot\vec{v}=0$, which gives the desired identity.
*I'm using the following integration by parts formula, for $f,g$ smooth scalar functions, applied to each component of $\vec{v}$ (since the identity above is a vector identity).
$$\int_V f\partial_i g \ dV = -\int_V (\partial_if) g\ dV + \int_{\partial V} (fg) n_i dS$$
Where $\hat{n} = (n_1, \dots, n_d)$ is the unit normal to the region $V\subset\mathbb{R}^d$. You can prove it using the divergence theorem applied to the vector field $F$ with $i$th component equal to $fg$ and all other components equal to zero. Indeed, $\nabla\cdot F=\partial_i(fg) = (\partial_if)g+f\partial_ig$ and then just integrate this identity and use the divergence theorem.
