 Welcome back everyone to our lecture series, Math 1220 Calculus II for students at Southern Utah University. As usual, I will be your professor today, Dr. Angel Misseldine. Now, in this lecture, I want to start talking about average value of integration. But before we do that, I want to do one more example about work problems that we saw in section 6.4 of James Stewart's Calculus textbook. And so in this example, what we want to do is consider the work it takes to pump water out of a tank. And we're going to pick some tanks of some interesting shapes because the shape of the tank will determine how difficult it is to pump out and such. And these water pumping work problems are going to, in many ways, there certainly work problems, but they're going to also kind of feel a little like these volume problems we did earlier in this chapter. So consider, for example, a tank that has shaped like an inverted circular cone. That's just a fancy way of saying an ice cream cone. It looks like an ice cream cone. The height of this cylinder, sorry, not cylinder, the height of this cone is going to be 10 meters tall. Since it is a cone, there is a circle on the top of it. It'll have a base diameter of 8 meters or 4 meters if we talk about the base radius. And we fill this thing up with water, but it's only filled up to 8 meters tall. So the water has a height of 8 meters. And given that the tank itself has a height of 10 meters, there is a 2 meter gap from the top of the tank to the top of the water. And so what we want to do is find the work required to empty the tank by pumping all the water to the top of the tank. So all this water here needs to come up to the top. And so how does one begin trying to approach this thing right here? Well, using this accumulation technique, that is this technique of integration, we want to dissect the problem into small, small, simple problems, which on those simple problems, we can assume some type of constant or linear property, and then take the limit of the sum of all those approximate solutions to get the final answer, and that limit will be integral. And so how are we going to do that? Well, we should pick some type of coordinate system. And so in this exercise, we're going to choose the top of the tank to be our x equals 0 coordinate. And we're going to align the x-axis along the axis of the cone. It seems like a convenient choice there. And so the deeper we get into the cone, that'll be larger x-coordinates, right? So this is x equals 0. This right here would be x equals 2. And the very bottom of this would be x equals 10. There are other ways one could choose a coordinate system. It doesn't make sense to place x equals 0 at the apex of the cone. And you could do positive x along this way. You could also place x equals 0 at the top of the water. There's some benefit of doing those type of things. Again, for this exercise, I'm going to choose x equals 0 to be up here. And then we're going to integrate, these are going to be positive x-coordinates along the axis right here. It's also just a matter of flavor. I chose strawberry ice cream over chocolate ice cream to continue that ice cream cone analogy going on right here. And so once we have this axis chosen, and when you choose an axis, you should specify what does 0 mean, what does positive x mean. And so now we're going to start subdividing the problem into little pieces like you can see right here. And so a typical cross-section is labeled for you with this rectangle right there. And so what we're going to do is consider what is the work to take one cross-section up to the top of the tank at x equals 0. So how do we go about doing that? And this is where we have to be a little bit more creative with these work problems. So remember work fundamentally is forced by distance. So a distance has to be traveled and a force has to be exerted, right? The distance traveled is going to be fairly simple, right? The distance from the top of the cone to our cross-section is going to be this Xi star, which when we take the integral this will simply just be an x. So the distance that we have to travel is going to be x meters to take that cross-section to the top of the cone. That's great and wonderful. Well what force is being done here? Well when it comes to lifting, right? Pumping the water to the top of the tank is a lifting problem. When you come to lift something, the force associated to lifting is going to be its weight. And so we want to compute the weight of a single cross-section. Now how are we going to do that? Well the weight, so continue on with this discussion here, the weight is going to be its mass times the pole due to gravity. Now one has to be careful, if we're working with scientific units, you'll notice that in this diagram everything's labeled in terms of meters. So using scientific measurements we're going to use meters and kilograms and the like. So we're going to have to compensate for gravity, right? And so gravity in this situation oftentimes they just draw it as little g and this is going to be 9.8 meters per second squared. This is the acceleration due to gravity. And so if we call mass just little m, we're going to write this as a 9.8, 9.8 m. So we have to determine the mass of one slice here and one slice of water. This cross-section represents a slice of water and in particular if we look at from a different perspective the slice of water we're considering is going to be a disc of water, a.k.a. cylinder. And this is what I meant by it's going to kind of feel a little bit like these volume problems we looked in the past. So we have to consider, if we want to find the mass of this cylinder of water, we have to figure out how much water there is. To aid us in this direction we're going to use actually the density of water which is given as 1,000 kilograms per meters cubed. You don't necessarily need to memorize this number. I'll present it to you or I'll be given the homework of things. Likewise there's a similar one if you're working with pounds per cubic foot. In this situation the density of water is 1,000 kilograms per meter cubed. And so the way we're going to handle mass here is that density as it's given to you, it's often denoted as a row, density as 1,000 kilograms per meters cubed. Why do I keep on mentioning that? Well, when working with these scientific problems it's useful to look at the units, right? You have kilograms on top which is a measurement of mass. You have meters cubed on the bottom which is a measurement of volume, right? Density in general it's going to be a mass per volume, alright? So what we're interested in is finding the mass of the water here. So by manipulating this equation we see that density times volume is equal to the mass. And so that's the observation we want to use in this situation. The density of water we know, what's the volume of the water? We're talking about a typical cross-section which is going to be this cylinder right here. As we've seen in the past, volume of a cylinder is going to be pi r squared times the height of the cylinder, right? And so if we investigate this, the thickness of a single cross-section which will be the height of the cylinder, that's going to be a delta x or if you switch to the integral notation it'll be a dx. So we can record that down there, h equals dx. And so we're left with this radius that we have to determine. Because if we know the radius of a typical cross-section, we see here that the force is going to equal, like we saw before, 9.8 times the mass, which mass will be 1000 times the volume. So we get this 9800 times the volume which the volume is pi r squared dx. So what we're left with right now is we want to express the radius of a typical cross-section using writing as a function of x here, the variable x. It's just the distance, notice that's the distance measured to the top of the cone. How are we going to do that? In a situation like this, it's very, very useful to utilize the technique of similar triangles. Because when you look at these, when you look at this cone at every cross-section, at every cross-section you get an identical triangle right here. So there's a triangle that goes from the apex to the cross-section right here. It's a right triangle. But then there's also the whole tanks triangle as well, like so. And so if I redraw those triangles below side by side, we see that the big one is a right triangle. And we know that the height of this triangle is 10 meters because it's the height of the whole cone. The side, because it is a right triangle, the other leg is going to be 4 meters because that was the base radius of that cone. So this is like the whole cone. But then if we look at a small cross-section like so, what are the sides of this? These two triangles will be similar because these angles are congruent, these angles are congruent, and of course right angles are congruent. And so as these triangles are similar, their sides will be proportional to each other. We're going to use that proportionality argument to pass this thing forward. And so putting the blue triangle back inside the total triangle we get right here. We're looking for the radius r, which is right here. And we do know that this distance, the distance from the top of the cone to this cross-section is a distance of x. And so as the total distance is 10 meters, we can then translate that to tell us that the other side of the blue triangle is going to be 10 minus x, because the sum should add up to be 10 meters. And so using this relationship, the 10 coincides with the 10 minus x, the r corresponds with a 4, and we see that r over 4. So those combinations to the top two sides, r and 4, correspond with 10 minus x over 10, like so. In which case if we times both sides of the equations by 4, 4 cancels on the left-hand side, 4 and 10 both have a common factor of 2, so we get a 2 fifths, and multiplying that through, we end up with r equaling 2 fifths times 10 minus x. And if you want to, we can distribute that 2 fifths through the 10 minus x if you wanted to, because 5 does go into 10 two times, so you can get 4 minus 2 fifths x, but we'll leave this factored the way it is. And so now notice that we have the missing function r, and we're ready to put together our integral in the manner that we had. And so again, I'm taking this slow to illustrate everything we've done here. So the work required here, we're going to integrate. Remember, work is force by distance. So what have we seen? The force function, which was the weight, that remember, sorry, yeah, the force was the weight, and so as we saw above, right, our force was 9.8 times the mass, and let me write this down. So we had 9.8 times the mass, but the mass was the density times volume. So 1,000, so we kind of keep track of what we have so far. This is the acceleration due to gravity. This was the linear density, or this is the density of water, excuse me. Next, we have the volume of the water, which was pi r squared r. We got to be 2 fifths, 10 minus x. Make sure to square that entire quantity. The thickness was a dx. So be aware that this entire chunk right here is the volume. Volume times rho gives us mass, g times mass gives us the weight, the force, and then we're left with the distance that has to move, and as we mentioned earlier, the distance was x. So this is our value, d. It's good to do this analysis to see that you have all the ingredients. Story problems are sometimes difficult because you have all this interpretation that has to go through it, but also, story problems are nice because with all these units like mass and length, you can show that this thing actually does add up to be some type of work type thing. So we have our integral set up. The last thing we have to mention are the bounds, like how far does x travel here? Going back up to the picture, you can use any of these pictures if you want to, but the x-cordment, it can go from here, the top of the tank, which is at x equals 2, to the bottom of the tank, which is x equals 10. So for our integral here, we're going to integrate it from 2 to 10. I do want to clean this thing up a little bit because in its present form, it's not the most ideal for integration. Notice that 9.8 times 1,000, that'll be 9,800. So that's a constant multiple we can bring out. We can also take out the pi, we're integrating from 2 to 10. So let's left behind. We have a two-fifths that'll get squared. That's a four-twenty-fifths. We have an x. That doesn't look right. We have an x. We have a 10 minus x squared, like so, and then a dx right here. We could also take out the four-twenty-fifths factor, bring that out as well. If I want to integrate this 10 minus x squared times x, probably the best way is just to foil it out and just get a degree 3 polynomial. Let's do that. If we take 9,800 times the 4 over 25, 25 does go into this figure after all. Notice that 9,800, right? That's 9,800 times 100, right? And 25 does go into 104 times. And then we have another 4 to times that by. So 16 times 98, we get a coefficient out front of 1,568 pi. Integrate 2 to 10. If we foil out the 10 minus x squared, we get 100 minus 20x plus x squared dx. Distribute the x. That'll give us 100x minus 20x squared plus x cubed. Here's the degree 3 polynomial that I had promised, for which then using the power rule for anti-derivatives, we can get its anti-derivative quite quickly. We end up with 50x squared minus 20 thirds x cubed plus x to the 4th over 4, plug it in 2, plug it in 10. There is some arithmetic that has to be accomplished at this point. I won't bore you with the details of the arithmetic. It's really just plug and chug at this moment. If you plug in the 10, you should get 2,500 over 3. If you plug in the 2, you should end up with 452 over 3. Adding those together, you get 2048 over 3, which then that gives us an exact answer. Let's see, I don't know. The 1568 is not divisible by 3, and 2048 is actually a power of 2, but it's not divisible by 3. If we want an exact answer, we're done. But as this is a physics problem, an approximate solution is probably appropriate. We approximate it to however many significant digits we are looking for. I'm not going to worry about that in this situation. This number will be approximately 3.4 times 10 to the 6th mini-joules. We could write this as kilojoules or megajoules if we prefer, but we'll just leave it at 3.4 times 10 to the 6th joules. So this one took a while to explain all the pieces, and be aware that the computation of the integral wasn't really what I cared about in this problem. I mean, we had to do it to get the answer, yes, but the actual computation of the integral is fairly simple. The difficult part was setting this thing up, and as you're working through these applications and integrals, that's the part of the problem that I want you to focus on, focus on setting these puppies up.