 Let's do some more practice of finding volumes of solids of evolution using the shell method that we've learned about previously, right? So in this example, we want to find the volume generated by rotating the region bounded by the two given curves you can see around the y-axis. We're going to take the curve y equals x squared, which is the lower curve in our picture right here. And then the other curve is y equals the quadratic x, 6x minus 2x squared, which is this upper curve you see right here. And again, we want to rotate this around the y-axis. And so we can see that there's these two natural, it doesn't tell us the bounds, x ranks from what to what, and we can see from the picture there's two natural bounds we can use here. One of them seems like the origin, 00. We should always be cautious when we trust too much in a picture because pictures can be sometimes deceptive. But it sure seems like 00 works, and you could check that if you plug in x equals 0, y equals 0, that's a solution. If you plug it in here as well, that is a point of intersection. We do need to figure out this other point right here. What is it? And whenever you want to find intersections between these functions, you set the two functions equal to each other. So the first function is y equals x squared. The second one is y equals 6x minus 2x. Set them equal to each other and solve. We could add 2x squared to both sides. That's going to give a 3x squared is equal to 6x. Actually, I mean, we might as well just set everything equal to 0, 3x squared minus 6x equals 0. Factor, you can pull out a common divisor of 3x that leaves behind x minus 2. And so this is going to give us the two intersections we're looking for, x equals 0 and x equals 2. So that's going to be the x-coordinate we want over here, x equals 2. To figure out the y-coordinate, plug it into either of the function, into y equals x squared, x2 would give us y is to the fourth, sorry, y is equal to 4. And you can double check with the other one as well. 6 times 2 is 12, minus 2 times 2 squared, that's an 8, 12 minus 8 is a 4 as well. That's the intersection we want. So that's important to know. So we know the points for which we're going to vary. If we want to rotate around the y-axis, we ask ourselves, do our cross sections, are they parallel to the axis of revolution or perpendicular? Well, if these, as these are both functions, f, y equals f of x and y equals g of x right here, a typical cross section you would see here in orange, the thickness of that thing that's going to be a delta x. That delta x is going to converge to be a dx, as these things get thinner and thinner and thinner. So we want to integrate this thing with respect to x. So the shell method is going to be the more preferred method to use in this one, as opposed to the washer method here. And so as we've seen with the shell method before, where the volume is going to equal the integral, we'll come back to that. But in fact, the shell method always has a 2 pi sitting in front, like that. Then you're going to take the radius. How far away from the axis are we? And that's what we can see right here. The distance between the axis and the cross section is the x coordinate, which is in that case, this means it's going to be x. So we get 2 pi x. Then we need the height of the function, the height of the rectangle, I should say. The height is going to be this length right here. How far apart are these? Well, the top of the rectangle is 6x minus 2x squared. The bottom is an x squared. So the length is going to be the difference of those things. 6x minus 2x squared minus x squared. Do put some parenthesis around that. 6x minus 2x squared minus x squared. We can simplify that in a bit, but I'm putting it this way so we emphasize where it came from. And then what's the thickness of this rectangle? Like we mentioned before, that's exactly a dx right there. And so since we know that we're integrating with respect to x, we need to find bounds for which x varies from x equals what? x equals what? Well, the smallest we get is over here at x equals 0. The largest we get is over here at x equals 2. And so now we're ready. We've set up this integral. We're now ready to simplify it and to compute the volume. So mostly it's the negative 2x squared, combining with the negative x there that we want to do. So we're going to get 2 pi, the integral from 0 to 2. We get x times 6x minus 3x squared like we did before. That is, this is kind of similar to what we had before. Distribute the x, I would say. I mean, we could do, we could try some type of u substitution, but I think distribution is probably a little bit simpler here, just an algebraic maneuver that makes life a little bit easier. We get 6x squared minus 3x cubed. Integrate with respect to x. Our function is now prepped for surgery. Let us then integrate. So by the usual power rule of integration, we want to find some anti-derivatives. We'll start with 6x. That'll become a 6x cubed over 3. And then we're going to get 3x to the fourth over 4. Going from 0 to 2. Admittedly, 3 does go into 6 two times. If we distribute this 2 onto the two pieces, 2 does go into 4. And so this would look like, I'm going to leave the pi actually factored out for now. We're going to end up with a 4x cubed minus 3 halves x to the fourth. I can't see any other way of simplifying that until we plug in the 0 and the 2. Now the good news is we plug in 0. Everything is a multiple of x, so it'll just vanish at 0. So plug it in 2 is where the interesting thing is going to happen. We end up with a pi times 4 times 2 cubed minus 3 over 2 times 2 to the fourth. Like so. You'll notice that one of the 2's here cancels with this 2 right here. So we're left with 2 cubed, which is just an 8. I'm going to factor that thing out, because both of these things are divisible by 8. So now you get an 8 pi and you're left with just a 4 minus a 3. 4 minus 3 is just a 1, so we end up with the volume of this being 8 pi, like so. And so we calculate this using the shell method. If one was using the washer method to try to calculate these things, notice a typical cross section would look like this rectangle right here. And so as you rotate this thing around and around and around the y-axis, there's going to be a hole in it. And with the washer method, you have to pay attention to what's the outer hole, what's the inner radius, what's the inner radius to compensate for the hole. One last thing I really like about the shell method is the shell method automatically keeps track of holes. You don't have to worry about a longer radius and a shorter radius. There's only one radius to work with, because we're really just focusing on the average radius. And so honestly, if I had to pick between the method, I think the shell method actually does turn out to be a little bit cleaner in terms of calculations typically, not always. There are times where the washer method is preferable. And honestly, I would tell you, you pick the shell method when your cross section is parallel to the axis, you pick the washer method or do this method when your cross sections are perpendicular to the axis. Let's take another example here. Let's use the shell method to find the volume of this solid revolution that's obtained by rotating around the x-axis, the curve y equals the square root of x when you go from 0 to 1, as you see the picture right here. Now if we're going to use the shell method to calculate this one, we've done this with the washer method before, the disc method I should say. But let's see how it would look like with the shell method here. If you're going to do the shell method and you're rotating on the x-axis, your cross sections have to be parallel to your axis, which would tell us that the thickness is going to be a delta x or a dy, sorry, a delta y, which becomes a dy when you take the integral here. And so then the radius is going to be this right here, by, which if you're going to restrict y, that's great. And then you have the height here. What's this distance right here? How long is this thing? Well, you have this x-coordinate one, subtract from it the x-coordinate of this point right here. So this x comma y. And we know this is the function x equals y squared. Where did that come from? Well, that's because we started off with y equals the square of x. But if we're going to integrate with respect to y, it turns out we don't want a function in terms of x, we want a function in terms of y. So if you solve for x, you get x equals y squared. And so that's the y-coordinate that shows up right here. So you're going to get 1 minus y squared. This is the height of your rectangle. The radius that you spin in will be y. And the thickness is always a delta y, in this case dy. And so I think we're now good to set up our integral. The volume would equal the integral 2 pi. Your radius, which here is a y. Your height, which is going to be a 1 minus y squared. And then your thickness, which is going to be a dy, like so. We didn't have to pay attention to the bounds. Since we're integrating with respect to y, these are going to be y-coordinates, y equals whatever, 2 y equals whatever. What are the smallest value of y you can get? That's going to be down here at y equals 0. What's the biggest y-coordinate you get? It's going to be this point right here. So when x equals 1, you can plug that into this equation right here. When x equals 1, y equals 1. So that's this point here, 1, 1. So y equals 1. And so now we're ready to calculate this integral. Only recommendation, I would say, is distribute the y here. And so the volume would equal 2 pi, the integral from 0 to 1. We get y minus y cubed dy. So finding the anti-derivative, we're going to get a y squared over 2 minus a y to the fourth over 4 from 0 to 1. Let's like the last example. When we plug in the 0, everything will vanish. You plug in the 1, we'll just get a bunch of fractions. So we end up with this 2 pi times 1 half minus 1 fourth. Distributing the 2 through helps a little bit. That would give us a 1 minus a half. And then 1 take away a half is itself a half. We get the volume to be pi halves. And so 1 could have done this problem using the washer method as well. Or in this case, the disk method. Because with the disk method, your cross sections will be looking something like this. Right? And so if we were to use the disk method here, we would integrate with respect to x, right? So because our thing looks like pi r squared dr, the radius, the radius in that case would be your y-coordinate, which would be the square root of x. You square it. The thickness was a dx. You integrate from 0 to 1. You go like this. And then if you simplify it, of course, notice the square of the square root just gives you an x. So you end up with 0 to 1 pi x dx. And so integrating that, you're going to get pi x squared over 2 from 0 to 1. Plug it in, 0 makes it disappear. Plug it in, 1 ends up with a pi over 2. So when you compare this one here, the washer method is a much simpler approach to this exercise, what we see here in red. The shell method took a little bit more involved here. There are many situations where both will work, and so one might be simpler than the other, and it really comes down to what do you want your cross sections to look like? Are they parallel to the axis or perpendicular to the axis? It makes a difference. And so we learn these two different methods, the washer method and the shell method, because like I said, even though sometimes they both work, one typically works a lot better than the other, and it's good for us to use sort of a judgment call. What is the right thing to do right here? This one first told us to use the shell method, but honestly in practice, as we see right here, the disk method turned out to be a much cleaner way of calculating this solid or revolution. So when you have a choice, make the best choice you can.