 In this video we provide the solution to question number 12 for practice exam number 1 for math 12-20, in which case we have a work problem, specifically we have a water pumping problem of a tank. So let's first talk about the shape of the tank. The shape of the tank is in fact going to be a solid of revolution itself. A tank has the shape of the surface generated by revolving the parabolic segment y equals x squared on the interval 0 to 3 for x about the y-axis, and all of this is measured in feet. So before we even talk about the pumping problem, let's make sure we understand the shape of our tank. So we're going to have our x-y-axis as illustrated right here. I'm going to draw a parabola, like so, yeah that looks pretty good, in which case we might do something like we should label things. We're going to have x equals 0 right here, and then let's say that this point over here, and then not that far, we'll go like let's go I guess to the edge right here, this would be x equals 3. This thing does not have to be perfectly drawn to scale or anything like that. This point right here for future reference would be the point 3 comma 9, because after all this is the curve y equals x squared. We're going to take this curve from 0 to 3, and we're going to revolve it around the y-axis. So I'll draw the other side here for like completeness sake, and then I'll also draw a line at the very top of this thing. And so while this is not perfectly drawn to scale, this region over here is the thing that is our tank. But of course you have to think of that parabola and spin it around in three dimensions. If the tank is full, and it's full with a fluid weighing 100 pounds per cubic feet, then let's notice that's not water right, but I mean it is, it's something. But notice the units here, we're using feet, we're using pounds. We don't need to multiply our results by 9.8, because with scientific units, you often measure this with mass, mass is not a force, you have to multiply by the force due to gravity. With standard units, pounds is already a force, it's already a weight, it already has that built into it. So it's going to be 100 pounds per cubic foot in that situation. So that'll come into play later. So we need to set up an integral to find the work required to pump the contents of the tank five feet above the top of the tank. So I want to make mention, like we mentioned over here, the top of the tank is happening at y equals 9. We need to then go five more feet up the line here. And so in the end, the top of the tank, sorry, we have to pump the water to the point y equals 14. That'll be important for us in just a little bit, okay? So we need to set up the integral to calculate this work, but we are not going to evaluate it. So our strategy for doing this is to look at a typical cross-section. If we were to cut our tank into little slices, they would be circular slices, cylindrical slices, and then we have to move these to the top of the tank. Now let's do it in steps. The first thing we're going to do is we're going to calculate the volume of one of these slices. Because it's a solid of revolution, each slice is going to be a disk. And so its volume will be pi r squared. The thickness here is going to be a dy, because the thickness of one is a small change of the y-coordinate. So we get that the volume will be pi r squared dy. R is not a very good one if I have to integrate with respect to y. So how do we get rid of the r? Now the r is the radius of this cylinder, of course, but it's also the distance from the x-axis to this point on the very edge of the circle, whose coordinates are going to be x comma y. So in particular, the radius of this disk is going to be the x-coordinate there. So we can improve upon that, pi x squared dy. But remember, we want to integrate with respect to y. So I have to rewrite the x's in terms of the y's. But as this point lives on the parabola, the relationship y equals x squared comes into play. So since I have an x squared, I can rewrite that as a y. And we get the integral pi times y times dy. That gives us the volume of that slice. I didn't need to compute the weight. Weight is the force due to gravity. Although w makes sense, because w might also mean work. I'm going to actually write this as ff for force. To find the force that is the weight of the object, we're going to take its density, which is pounds per cubic foot. And then we're going to multiply it by its volume. So we have this constant rho times volume. That gives you the weight of the object. The rho is 100 pounds per cubic foot. The volume we just did, pi y dy. And so then we're ready to talk about the work here. The work is the integral of force times distance. I guess there is one other thing we haven't considered yet. What is the distance that has to be traveled here? Now if you take a typical slice like this, this has to move to five feet above. So the distance that a typical slice has to go is it has to go from y equals 14 to that has to go up to there from the location of the slice. This slice is happening at the address of y. So the distance we have to travel is going to be 14 minus y in that situation. So then if we apply that to the work that we were doing just a moment ago, we're going to end up with a 100 pi that came from the force integral. I just put those coefficients in the front. We have y for the force. I'm going to put the dy at the end. So then I'm going to put a 14 minus y right here, dy. That's the integrand. Then the next thing we have to consider is the locations of the disk. That is, what are the bounds of integration? Our disk can range anywhere from the very bottom of the parabola. That's going to happen at y equals zero. And they can go all the way to the top of my parabola, which actually was bounded above here at x equals three, aka y equals nine. So the upper bound is going to be nine. So even though the disk will be pumped up to the level y equals 14, the integral's bounds are the locations of the disk, not where they end up, but where they are. And these disks live anywhere from y equals zero to y equals nine. And so I'm going to put a nice little box around this one. You can see why I changed the font color. This is the very final answer. This gives us the integral to measure the volume of this thing. I confess it's not a very hard integral to compute, but the instructions did say do not evaluate it. All of the points here are for setting up the integral, not from its evaluation.