 Welcome back everyone to the lecture series Math 1220, Calculus II for students at Southern Utah University. As usual, I'm your professor today, Dr. Andrew Missildine. It's good to have you back. We're going to start lecture six in our course here. It's going to be based upon section 6.3 from James Stewart's calculus textbook volumes using cylindrical shells or what we often call the shell method. The shell method is an alternative way of trying to find volumes of solids of revolution, you know, an alternative to the disk and washer method we had used before. The washer method, for which the disk method is just a special case, it tries to find the volume of these solids by slicing it into these very thin little pieces. Like if we took a carrot or a banana and sliced it into little pieces, we'd have a little disk along the way. The shell method tries to change things a little bit in that instead of having these slices that stand next to each other in a string, the shell method we take very thin shells, so to speak, and we put them inside of each other, much like a nesting toy of some kind. They stuck these cups inside of each other. And so this picture you see in front of you, this is courtesy of Stewart's textbook from section 6.3 here. I'm going to try to draw a picture that is similar in nature to it. Imagine we have our function. We have our function, here's the x-axis, and we have our function f of x, and so maybe it looks something like the following y equals f of x. And in this situation, we're going to rotate this thing around the y-axis as opposed to the x-axis. This doesn't make much of a difference, but it'll make it a little bit easier for this shell method consideration. So what we have here is we have a domain. So we're going to go from x equals a to x equals b. And so as usual, this defines some region right here. We've seen how to find the area. We want to calculate the volume of this thing if we were rotated around the y-axis. Well, what we're going to do is we're going to take the domain a to b and subdivide it the same way we've done before. So slice into pieces x1, x2, all the way up to some xi, xi minus one. You get the idea going off to b, which is xn, and a of course is just x0. This subdivides the area under the curve into these subintervals for which we can represent the area under the curve using these rectangles. So again, if we look at the rectangle, there's some f of xi star, which is a representative of the height of the rectangle. It's thickness. It's thickness is going to be a delta x. What happens if we take this rectangle here and we rotate it around the y-axis? What we end up with is what we mean by this shell, which the illustrations up here kind of give you some idea of what we mean by these shells. But what we're going to get is we're going to get a cylinder, a cylinder which has a noticeable height to it. And then taken from the middle of this cylinder is going to be even another cylinder, almost the same thickness. Almost the same thickness. It's meant to be very, very thin, right? And so for the sake of calculation, I'm going to draw it with a very noticeable thickness. So we have this cylinder which has been removed from the inside of a cylinder. So in some respect, it's kind of like a washer because we have a cylinder removed from a cylinder and they have the same center. The difference here though is that we're going to make the distance between the two get arbitrarily small. This is what we mean by a shell because we stack inside of this another shell and inside of that one, another shell, inside of that another shell. These shells are nested inside of each other like this illustration at the top gives us this right here. So what would the volume of a cylinder of a shell look like? Well, because they are just like washers, right? It's a cylinder with a cylinder removed from it. So we're going to take the volume of this thing. They have a height associated to them. You're going to notice that both of the two cylinders have the same height. We're going to call it H. What's different about them is they'll have the same center, but their radii are going to be different. There's one inner radius and there's this outer radius. So let's call the outer radius R2. Let's call the inner radius R1. And so the volume of this shell is going to be volume 2 minus volume 1, where volume 1 is the volume of the inner cylinder and volume 2 is the volume of the outer cylinder. Well, by the usual formula for the volume of a cylinder, you're going to get pi R2 squared H. That's the first one minus pi R1 squared H. Just the volume of the two cylinders attracted there. There are some common divisors. Let's factor out the pi and let's factor out the H. And that leaves this difference of squares R2 squared minus D2 squared. Oh, I'm sorry. R1 squared is what I meant to say. Pi R squared minus R1 squared times that by H. This is the volume here. And so what we're going to do is after all this R2 squared minus R1 squared is a difference of squares. We are going to factor it as a difference of squares. So those R2 squared minus R1 squared, it's going to look like R2 minus R1 times R2 plus R1 H. Now what good does that give us right here? Well, we're going to make a slight modification to this to kind of illustrate why we care about this thing. Let me back up the equal sign a little bit. We're going to introduce a 2 over 2 inside of the equation. Of course, 2 over 2 is just 1, but what's the good of this thing here? Well, the reason we wanted this 2 here is mostly so that we could use this 2 right here. So let me rewrite this and kind of illustrate what's going on here. We're going to have this 2 pi that sits in front. Then we're going to have this R2 plus R1 over 2 right here. We're going to get this R2 minus R1 right there and we're going to get this H. And so what parts do we have here? We have this 2 pi which is a constant which is going to sit around there. You're going to see this R2 plus R1 over 2. What is this thing? This right here would be what we call the average radius. The average radius of the two cylinders. The average radius. So the idea here is as this delta X gets smaller and smaller and smaller, we get these thinner and thinner cross sections. A teeny tiny little cross section right there. Let's draw one super thin. As these things get thinner and thinner and thinner, R1 and R2 are going to converge towards the same location. And that's going to be this average radius that kind of sits in between the two. We're going to call it R bar. This statistics putting the line over the number often refers to finding the mean. And so that's exactly what we're doing here. This thing here is going to become our average radius R bar. Well what about the R2 minus R1? R2 minus R1 is going to be the distance between this inner radius and this outer radius. And we want this thing to get smaller and smaller and smaller and smaller. And so this difference of the two radii we're going to call this delta R. And so what we want to do is summarize what we have going on so far. The volume of this shell is going to look like 2 pi R bar H. I'm moving that in front and then a delta R. We need a delta R because as one is integrating, this delta R is going to become a dr, which is going to be necessary for integration. And so the way you want to think of this is as the number of subdivisions goes to infinity, this thing is going to converge to be this 2 pi R HDR. So we integrate with respect to the radius. We're going to have a 2 pi of the radius and then R, H here, the height. We have to treat H here as a function of the radius. H equals f of R there. Now if we come back to the original picture we had up here. Well, what was the radius? The radius would be the distance, the distance of one of these cross sections to our axis of revolution, right here. This is our radius and since we're rotating around the y axis, this radius would be none other than just an x. And so then in this situation our volume would look like the integral from A to B of 2 pi, the radius which is x times the height which would be f of x. And then the thickness of the shell which is a dx. And this right here gives us our so-called shell method. Which again, the volume of a shell you want to keep track of, you get this 2 pi R, the circumference of the shell, the height of the shell, then the thickness of the shell, this 2 pi R H delta R. Remember that, but when you plug in specifics like x it might look something like this. Let's look at a specific example here. So let's take the, let's find the volume generated by rotating the region bounded by the given curves about the y axis. We're going to take the curve y equals e to the negative x squared. That's this yellow curve you see right here. Then y equals x, that's the x, sorry, y equals 0, that's the x axis. x equals 0, that's the y axis, and x equals 1. Well, which would be about right here. You see the graph just kind of terminates at that moment right here. So we're going to take this blue region and rotate it around the y axis. Well, you can see what a typical cross section would look like with this guy right here. And so using the ingredients for the shell method, this one's perfectly set up for the shell method, the volume will equal the integral. Well, we want to go from 0 to 1. So we get x equals 0 to 1. Now we get 2 pi, the radius. The radius is the distance from the axis of revolution to a typical cross section. That's going to be an x right there. So we get 2 pi x. The next thing that shows up is the height. The height is the height of a typical cross section. This thing right here, how tall that is that? Well, the height is going to be the y-coordinate, but given that this y-coordinate, x comma y, is associated to the function, the y-coordinate is going to be the f of x, in this case, e to the negative x squared. And so we're going to plug that in for the y-coordinate. We get e to the negative x squared and then a dx. So this right here sets up the integral that we want to calculate. And then let's go about computing this. In this situation, e to the negative x squared, that might be a difficult thing to compute. But remember, these integrals, we can use any of the techniques of integration we already know. For example, u-substitution works out excellently here. Because when you have this exponential expression, e to the negative x squared, it's very nice to set the x on an equal to u. So we get u equals negative x squared. Then when we calculate the derivative, we get a negative 2x dx. When no one's looking, you put in a double negative. So you have now the negative 2x dx right there. And so that all squashes together to be our du. So if we record that down below, we're going to have a negative integral. We didn't suck up the pi in terms of the du. That's going to stick around. But the 2, the negative 2, the x, and the dx all come together. Their powers combined make Captain du here. And so this would then simplify. You're going to have a negative pi. You're going to have an e to the u and then a du. So that's how we can change the variables. But as we change the variables, we can also change the bounds here. Because these bounds are x-coordinates. x equals 0, x equals 1. So using this identification here, when x equals 0, well, let's actually do x equals 1 first, the top one. When x equals 1, u is going to equal negative 1. Notice we just took this equation and plugged it 1 into the equation. So u equals negative x squared, plug it in 1, we get negative 1. When x equals 0, we end up with u equals, in that case, a 0 as well. So then the bounds are going to be u equals 0 on the bottom and u equals negative 1 on top. If you don't like the smaller number being on top, that's not a big deal. You can switch the order. Switching the order of the bounds does negate the integral. But as we already have a negative sign in there, that actually works to our benefit. So our integral then becomes the integral from negative 1 to 0 of pi e to the u du. We now need an antiderivative of the function e to u and it's its own antiderivative. So we end up with pi e to the u as we integrate from negative 1 to 0. And do plug those both in there. We're going to have pi times e to the 0 minus pi e to the negative 1. e to the 0 is the same thing as 1, so you're just going to get a pi. And then e to the negative 1 is actually just 1 over e, so you end up with pi over e. If you want to factor, you could write pi times 1 minus 1 over e or something similar to that. But this would be the final expression, the final answer, the volume of the associated solid of revolution here. Now I do want to make a comment that we did this exercise using the shell method. Could we have used the washer method as well? If we had used the washer method, the cross section would have changed. Our typical cross section wouldn't be this rectangle we see right here. If we were doing this with the washer method or this disc method, our cross sections would have to look something like this, right? Because we're rotating around this thing right here. And that's an important thing to keep track of here. When 1 is using the shell method, the shell method, you look at cross sections which are parallel to your axis of revolution. But when you're using the washer method, your cross sections actually should be perpendicular to the axis of revolution. And so the thickness of this cross section would be a dy. And so this is sort of like the choice one has to make. Do I want to integrate this with respect to x or do I want to integrate with respect to y? Well, oftentimes our functions are given as y equals f of x. That's a situation we prefer dx. We could switch to dy. That's not always a painless process here, but for this one right here, this would be a very challenging one to do with the washer method, but it's a very nice one using the shell method. So we present this as an alternative because there are some regions for which the solids volume is easier to compute using the shell method as opposed to the washer method. And we'll see some more examples of that in the next video.