proof involving divergence theorem for fluid dynamics I have been trying to derive from $(\vec{u}.\nabla)\vec{u}=-\frac{1}{\rho}\nabla P$ $*$
that $$\rho\iint_{S}(\vec{u}.\vec{n})\vec{u}dS=-\iint_Sp\vec{n}dS$$ where as usual $\vec{n}$ is a unit vector outward from $S$. 
I first took the volume intergrals of both sides of $*$ which gave the following 
$$\iiint_{V}(\vec{u}.\nabla)\vec{u}dV=\iiint_{V}-\frac{1}{\rho}\nabla P dV$$ now by the divergence theorem for scalars the RHS of the above equation becomes
$$\iiint_{V}(\vec{u}.\nabla)\vec{u}dV=-\frac{1}{\rho}\iint_{S}p\vec{n}dS$$, but this is where i'm stuck, i know i need to rewrite the LHS before i can arrive at the equtaion i'm aiming for i'm just unsure how to do this, i assume i will be needing to use at least two integral transforms to get my required results, any help with this would be highly appreciated. 
This gives a formal proof of the statement that, in a steady incompressible flow, the total force due to hydrostatic pressure exerted on a control surface is equal to the net outward rate of momentum outflow.
 A: This form of the momentum equation, where the density is treated as a constant, holds for an incompressible fluid with the added condition $\nabla \cdot \mathbb{u} = 0$.
Using Cartesian components and the Einstein convention for repeated indices ($a_ib_i = \sum_ia_ib_i$), the ith component of $\mathbb{u} \cdot \nabla \mathbb{u} $ is
$$(\mathbb{u} \cdot \nabla \mathbb{u})_i = u_j \frac{\partial u_i}{\partial x_j} = \frac{\partial}{\partial x_j}(u_iu_j) - u_i \frac{\partial u_j}{\partial x_j}.$$
For an incompressible fluid we have $\nabla \cdot \mathbb{u} = \frac{\partial u_j}{\partial x_j} = 0.$
Hence, the ith component of the integral is 
$$\tag{*}\int_V (\mathbb{u} \cdot \nabla \mathbb{u})_i \, dV = \int_V \frac{\partial}{\partial x_j}(u_iu_j) \, dV = \int_S u_i u_j n_j \, dS = \int_S (\mathbb{u} \cdot \mathbb{n})u_i \, dS$$
and in vector form
$$\int_V \mathbb{u} \cdot \nabla \mathbb{u} \, dV= \int_S (\mathbb{u} \cdot \mathbb{n})\mathbb{u} \, dS.$$
Note that in (*) we applied the divergence theorem in the form
$$\int_V \frac{\partial f}{\partial x_j} \, dV = \int_S f n_j \, dS.$$
