 In trying to express our conservation of momentum integral relation as a differential form, it is useful again to consider a unit cell of fluid. The volume of this unit cell is infinitesimally small, and for Cartesian coordinates we can consider it as having x, y, and z dimensions of dx, dy, dz, respectively. Its position is entirely arbitrary so long as it's not on a boundary, and if we define our axes as going up or positive in the right, up, and forward direction, then we can express any momentum occurring with respect to those axes as three ingress momentums and three egress momentums. All of the ingress momentums will have a momentum of density times the velocity in that direction times the velocity vector, and the egress momentums will include that ingress momentum plus however much the momentum changes over the course of the cube itself. So for example in the x direction here, we can see that we're taking the density times the x component of velocity times the velocity vector, and we are adding to that the rate of change of that momentum with respect to x multiplied by the x displacement. When I take these six ingress and egress points and apply them to our conservation of momentum integral relation, they will all appear in the control surface term. My three egress points will have a velocity that is in the same direction as the area vector, and as a result will appear as a positive term. For the ingress points, their velocity is in the opposite direction as the area, and as a result they will appear as negative terms. For the control volume, I can replace dv with dx, dy, dz, and I'm left with this relationship, and I can simplify that by recognizing that this term cancels this term because it's still multiplied by dy, dz. This term cancels this term, and this term cancels this term, leaving me with this relationship. When I bring dy, dz into the x component term, dx, dz into the y component term, and dx, dy into the z component term, I recognize that dx, dy, dz appears in all four terms, so it can be factored out. My next simplification comes from recognizing that if I split these products up, I can express them separately like this, and I can simplify that first term by recognizing that this is zero as per our conservation of mass. That means all I'm left with is this right term, which can be rewritten as the material derivative of velocity because these terms all look familiar, and the right-hand side of my conservation momentum equation simplifies all the way down to density times the material derivative of velocity with respect to time times the volume of the cube. Now we can look at the left-hand side of that equation. The sum of forces acting on the cell will be a result of body forces and the surface forces. For the body force, we're only going to be considering gravitational acceleration, which means that we can include that in our calculation by taking mass times acceleration. For the mass term, it's more convenient if we express that as density times volume. The volume of the unit cell again is dx dy dz. Therefore, the force acting as a result of gravitational acceleration as a vector would be density times the gravitational acceleration vector times dx dy dz. For the surface forces, we have to look back at our unit cell. To make it a little bit easier to follow, let's only look at the x-direction for now. In the x-direction, we have six possible surface forces. We have normal stresses on the left-hand right side of the cube and we have viscous stresses on the top, bottom, front, and back of the cube. Since the stresses are going to be acting in the outward direction for normal stress, in the rightward direction for the viscous stress on the top and front, and the leftward direction for the back and bottom, I recognize that I am going to be cancelling the regular stress terms in all three cases. For example, if you consider this term minus this term, all that we're left with is the rate of change of that stress across the cell. If it's easier to see written out in terms of equations, the sum of forces acting on the surface in the x-direction simplifies down to these three rate of change terms multiplied by their area of effect because the force exerted by that stress includes the area of effect. I recognize that each of those three terms includes dx dy dz, so again for convenience, I can factor that out. Furthermore, I can express the normal stress as a viscous stress plus the pressure exerted on that body and I can write the forces acting on the surface in the x-direction as the sum of the rate of change of pressure with respect to x plus the viscous stress in the x-x direction with respect to x plus the viscous stress in the y-x direction with respect to y plus the viscous stress in the z-x direction with respect to z times the volume of the cell. If I do that for the other two directions as well, I end up with these terms. For convenience, I can express them as gradients at which point when substituting back into our conservation of momentum equation I have a much more manageable equation. It's even more manageable when I recognize that dx dy dz cancels from all four terms. That leaves me with this equation which is my differential equation for the conservation of momentum. The hardest aspect of this equation is this term here. That term is a complicated dot product of the gradient operator and a three by three tensor and as a result, we typically try to avoid it where we can. There are two common ways to simplify that equation. The first is for inviscid flow where that term disappears. This is called Euler's equation named after the mathematician. The other common simplification is for constant density, constant viscosity which looks like this. When expanded for the three directions in Cartesian coordinates we are left with these three equations. These equations are tremendously useful and are named after the researchers that created the equations for us to use for the first time. Their names were Navierre and Stokes and as a result, these equations are called the Navierre-Stokes equations.