 Welcome back everyone to our lecture series Math 12-20, Calculus II for students at Southern Utah University. As usual, I'm your professor today, Dr. Angie Misselein. This video represents the first video for lecture 22 in our series. And so what we're going to do over the next couple of lectures is look at some, we'll continue looking at some applications of integration. We've learned recently about arc length and surface area, which are inherently geometric applications. We want to kind of step out of the realm of just pure mathematics and look at some applications from physics and engineering, some of which are selected from a section 8.3 of James Stewart's Calculus textbook. Now in this lecture and the subsequent videos, what I want to do is introduce the notion of hydrostatic force and hydrostatic pressure and see what this has to do with how we can use integration, I should say, to help us calculate these type of problems. Now before anyone panics, because some of the things we're going to talk about, because again they're very physical in nature, some of these questions and some of these problems are going to feel a lot like what we saw previously when we talked about physical work. And honestly speaking, those work integrals that we had to set up, some of them were quite challenging. And many of us were overwhelmed at the time probably when we learned about it. And so before anyone panics and think this is giving the same thing, there's going to be some stark differences going on with hydrostatic force problems. And those differences are actually going to make these things feel a lot easier. Now one reason it's going to feel easier is first of all, there's a lot less variability when it comes to a hydrostatic problem as opposed to the work problems we've seen before. Basically the variability will come to the shape of our plate, but that's about it, right? And also another thing I want to point out is one of the reasons these problems are going to feel easier is because as a learner of calculus, we are much better at setting up integrals than we were when we first learned about work, at least with respect to this lecture series here. And so have some faith in yourself and I think you're actually going to find this to be quite satisfying, these type of problems here. So to introduce this idea of what do we mean by hydrostatic force and hydrostatic pressure and such. Imagine you're a deep sea diver, maybe you have gone scuba diving before. Well, one thing that deep sea divers realize is that water pressure is going to increase as they dive deeper and deeper and deeper. You feel more pressure on you the deeper you go in the water. And this really is just a result of the weight of the water above you is going to increase the farther down you go. So imagine for example that we have some type of horizontal plate whose area is given as A and this plate can just be any oddly shaped thing. It sits here at the bottom of say our pond or ocean or whatever. And it has some fixed area A and it's submerged into some fluid and we'll assume this is a liquid and it'll be water for nearly all situations. So it's submerged in some fluid and it's a depth of D feet from the surface of the water. So we can see kind of like this continuum this conduit of water that just kind of sits on top of this thing. And so it is D units deep. All right. Well, if this fluid has a density of row. So we have this density. Often we denote this with the Greek letter row. Then the force exerted by the fluid onto the plate is that is to say the hydrostatic force. This is going to be the weight of the fluid above the plate because as this water sits above the plate, right? This water has a weight. Gravity is pulling down on that water and that water is applying a weight on the plate based upon its force, the so-called hydrostatic force. Now let's say that the volume of the water above the plate is volume A. Now there's some things that you can very easily say about this volume. The volume or the water that sits above the plate, we can think of as a geometric prism, right? It's a two-dimensional shape that's been stretched throughout the third dimension. So the volume of that plate can be computed as the area of the plate. So the volume of the water above the plate, I should say, it will be the area of the plate times the depth. So we have one face of the prism, which is the plate itself and then how long is the prism? So the volume of this water tower is going to be just the area times the depth, all right? And so if we were to multiply the volume, if we take the volume of the water and multiply by its density here, this density, we're talking about its mass per cubic unit, its mass per volume. That's what this density is measuring here. So in fact, the mass of the water is going to be rho times V. That's kind of a poorly drawn row there. Let's try that again. There you go. So the mass of this thing is going to be rho times V. And so applying Newton's second law of motion here, the force is going to be equal to the mass times acceleration, for which acceleration in this situation, since we're talking about weight, the acceleration will be, that is, the acceleration due to gravity. So force is gravity times mass. Well, like we mentioned, mass is going to be rho, the density times volume and volume is area times depth. And so be aware that gravity is going to be a constant, right? It depends on the planetary body we reside on. And so this will be Earth for nearly most examples we never consider, right? And also this row here is also going to be a constant. It's a constant with respect to the fluid. So with respect to water, this would be 1,000 kilograms per cubic meter or 62 and a half pounds per cubic foot, things like that. And so because these are constants, we're going to bring the row and G together. You're going to see this number right here. Oftentimes, it'll be like 9,800. That is 9.8 meters per second squared times 1,000 kilograms per meters cubed. And so then our variables will be consisting of the area of the plate which could change and then the depth. So that's an idea that gives us the basic idea behind hydrostatic force, the weight of the water crushing the plate from above. And so the deeper you get, the more forced there is because the depth is increasing. Now another concept we need to talk about is the physical interpretation of pressure. Now pressure in terms of physics is the ratio of force to the area which the force is distributed. Or mathematically speaking, pressure is equal to force divided by area. So this is like when you step on the ground with your foot versus when you like regular ground versus when you step on a nail with your same foot. The force is essentially your weight, right? So the force you feel in your foot is the weight of your body pressing down on your foot. And when you spread that force across the entire size of your foot, it's quite comfortable, not a big deal. But if you apply the weight of your body over a very small area, like just the point of a nail, that pressure dramatically increases because you have a much smaller area. And that's the difference we feel when we step on something sharp. So pressure is force per area. Or if you multiply both sides of this equation because it is an equation after all, you're going to get that force is equal to pressure times area. And so this is an observation we're going to use a lot as we talk about hydrostatic pressure here. Force will be equal to pressure times area. If you're not familiar with pressure here, I should mention that in the psi unit, with scientific units here, the basic unit of pressure is called one Pascal denoted PA named after the scientist Pascal, right? And one Pascal is a Newton per meter squared right here. So we take the scientific unit for force, which is typically a Newton and again for scientific measurements for for sort of traditional British measurements, this would be a pound. And then for area, we just take a length squared so meter squared. So Pascal is one Newton per meter squared. If we do use sort of American British units, then we would take pounds per pounds per square foot. Sometimes pounds per square inches use psi for pressure. Like if you look at your tire pressure, that's typically a PSI, but we might see something like that. So continuing on with this, we see that the pressure on the plate exerted by the fluid, the so-called hydrostatic pressure. Using this formula, we get the following. The pressure is equal to force over area. Now, if we take the hydrostatic force observation we had before, we're going to get the density times the acceleration due to gravity times the area divided or times or the area times by the depth and then divide that by area. Well, since area, of course, is the same thing, you cancel those things out and you'll just end up with rho gd. So one thing I want to mention about this is that the hydrostatic pressure is completely independent of the shape of the object. It doesn't matter how our plate is shaped. The pressure is determined entirely by the depth, right? D is the variable here, rho and g are constants that we'll see throughout. So a very simple calculation of this. Consider if we have a fluid, which is, let's say we submerge something into water. Like I said before, the pressure will be a thousand kilograms per meters cubed, not the pressure. Excuse me, the density is a thousand kilograms per meters cubed. And if the plate is submerged two meters below the surface, then the hydrostatic pressure will be 1,000 kilograms per meters cubed. Then you get 9.8 meters per second squared. That is the acceleration due to gravity. And then you times that by two meters, which is the depth in which case that then gives us when you multiply that together, you'll get 19,600 pascals. One Pascal is actually not a large amount of pressure. So oftentimes we'll write this as kilo pascals. So you see that this object, this plate when it's emerged two meters, will experience approximately 20 kilo pascals of pressure. And so calculate hydrostatic pressure is not too much of a challenge here. We basically just need to know the depth and the density and such. If we wanted to figure out the force exerted on this plate, we'd have to know the area of the plate. If we times pressure by area, we would then get the hydrostatic force. So this is sort of like the simple linear problem. As we will see in the next video is that in many situations the problem is not linear, but we'll use this linear problem to approximate using integration. So stay tuned for that coming up in just a second.