Differential form of Conservation of Momentum related to the Navier-stokes equation In the context of the Navier-Stokes equation the Conservation of Momentum is given in my textbook as $$\frac{d}{dt}\int_{W_t} \rho \mathbf{u}dV = -\int_{W_t}\rho \cdot \mathbf{n} - \mathbf{\sigma}(\mathbf{x},t)dA$$ where $\sigma(\mathbf{x},t)$ is called the $\textit{stress tensor}$ and is written as $$\sigma(\mathbf{x},t) = 2\mu[D - \frac{1}{3}(div(u))Id]+\xi (div(u))Id$$ where $\mu$ is called the first coefficient of viscosity and $\xi = \lambda + \frac{2}{3}\mu$ is called the second coefficient of viscosity, and $D$ called the deformation densor.
Plugging in the above $\sigma$ into the integral version of the conservation of momentum apparently yields the differential form of the conservation of momentum below:
$$\rho \frac{D \mathbf{u}}{Dt} = -\nabla \rho + (\lambda + \mu )\nabla (div(\mathbf{u}))+\mu\Delta \mathbf{u}$$
However, I cannot get the same result. Could someone please show me how this is done? Here $\frac{D}{Dt} = \partial_t + u\cdot \nabla$ denotes the material derivative. I can plug in $\sigma$, but how are the integrals removed?
 A: First, you need to recognize that $W_t$ represents a material control volume that is time-dependent as it moves and deforms with the fluid.  Then you pass the derivative under the integral on the right-hand side by applying Reynolds transport theorem to obtain,
$$\frac{d}{dt} \int_{W_t} \rho \mathbf{u} \, dV = \int_{W_t} \frac{\partial (\rho \mathbf{u})}{\partial t} \, dV + \int_{\partial W_t}\rho \mathbf{u} \mathbf{u} \cdot \mathbf{n} \, dS,$$
where $\mathbf{n}$ is the outer unit normal vector field at the surface $\partial W_t$ bounding the control volume and the integral on the far right-hand side is a surface integral.
By the divergence theorem, it follows that
$$\tag{1}\frac{d}{dt} \int_{W_t} \rho \mathbf{u} \, dV = \int_{W_t} \frac{\partial (\rho \mathbf{u})}{\partial t} \, dV + \int_{W_t}\nabla \cdot(\rho \mathbf{u} \mathbf{u}) \,dV = \int_{W_t} \left[\frac{\partial (\rho \mathbf{u})}{\partial t} +\nabla \cdot(\rho \mathbf{u} \mathbf{u})\right] \,dV $$
The right-hand side you have written is incorrect.  This should represent the net force acting on the control volume due to pressure and stress at the surface $\partial W_t$. Correctly written it is
$$\int_{\partial W_t}(-p\mathbf{n} + \mathbf{\sigma} \cdot \mathbf{n}) \, dS,$$
where $p$ is the pressure field and $\sigma $ is the (deviatoric) stress tensor.  Applying the divergence theorem, we get
$$\tag{2}\int_{\partial W_t}(-p\mathbf{n} + \mathbf{\sigma} \cdot \mathbf{n}) \, dS = \int_{W_t} (- \nabla p + \nabla \cdot \sigma) \, dV$$
Equating (1) and (2), we have
$$\int_{W_t} \left[\frac{\partial (\rho \mathbf{u})}{\partial t} +\nabla \cdot(\rho \mathbf{u} \mathbf{u})\right] \,dV = \int_{W_t} (- \nabla p + \nabla \cdot \sigma) \, dV$$
Since this is true for any control volume $W_t$ we can eqaute the integrands, to obtain
$$\tag{3}\frac{\partial (\rho \mathbf{u})}{\partial t} +\nabla \cdot(\rho \mathbf{u} \mathbf{u}) =- \nabla p + \nabla \cdot \sigma $$
Now it is just  a matter of simplifying both sides.
For the left-hand side we have
$$\tag{4}\frac{\partial (\rho \mathbf{u})}{\partial t} +\nabla \cdot(\rho \mathbf{u} \mathbf{u}) = \rho \left[\frac{\partial \mathbf{u}}{\partial t}+ \mathbf{u} \cdot \nabla \mathbf{u}\right] + \left[\frac{\partial \rho}{\partial t}+ \nabla \cdot (\rho\mathbf{u}) \right]\mathbf{u}$$
Conservation of mass is expressed in terms of the equation of continuity
$$\frac{\partial \rho}{\partial t}+ \nabla \cdot (\rho\mathbf{u}) =0,$$
which can be derived in a similar way. Substituting into (4) , we get
$$
\frac{\partial (\rho \mathbf{u})}{\partial t} +\nabla \cdot(\rho \mathbf{u} \mathbf{u}) = \rho \left[\frac{\partial \mathbf{u}}{\partial t}+ \mathbf{u} \cdot \nabla \mathbf{u}\right]= \rho \frac{D\mathbf{u}}{Dt}, $$
and substituting into (3) with this result,
$$ \rho \frac{D\mathbf{u}}{Dt} = -\nabla p + \nabla \cdot \sigma$$
It remains for you to expand $\nabla \cdot \sigma$ using the expression you gave for $\sigma$ in terms of the deformation tensor which should be $D = \nabla \mathbf{u} + [\nabla \mathbf{u}]^T.$
