 So we're now going to take a look at the hydrostatic pressure distribution in a liquid and if you recall we derived the equation from the force within a fluid that we did some simplification and we came up with this integral where we said that we would assume the gravity scalar value is a constant. That was the equation that we came up with. Now g we will pull out of the integral so we'll be okay there. Now with the liquid density is usually a more or less constant incompressible. So what we'll do is we will pull density out as well. So when we're dealing with the liquid we can treat the liquid as being incompressible and with that we assume a constant density and what we're left with is p2 minus p1 is equal to minus rho g integral from 1 to 2 dz. That then simplifies just down to z2 minus c1. So that becomes the equation. Now we can rearrange that. z1 minus z2 is p2 over rho g minus p1 over rho g. So with this equation what does that tell us? If you know p1 at z1 you can determine or can solve for p2 at z2. So that is something that we'll use quite often when we look at things such as manometers. A couple of other things we can see here. Rho g appears quite often in the equations. Rho g is specific weight and in fluid mechanics we sometimes will give that gamma. That's what we show it as. Don't confuse that. It's not gamma cp over cv which you find in thermodynamics. Quite often the cp over cv in fluid mechanics we use a little k for that. The ratio of specific heats. Another thing that we see. We see the p over rho g. Let me write that on the other side. We see p over rho g. That is sometimes referred to as being head or pressure head. And so you'll hear people talk about head loss and head and stuff like that. That comes from hydraulics. The field of hydraulics. But p over rho g is the term that we use there. So that is the equation that we can use to solve for the pressure distribution in a liquid.