AP®︎ Calculus AB (2017 edition)
- Shell method for rotating around vertical line
- Evaluating integral for shell method example
- Shell method for rotating around horizontal line
- Shell method with two functions of x
- Calculating integral with shell method
- Shell method with two functions of y
- Part 2 of shell method with 2 functions of y
- Shell method worksheet
- Shell method
Introducing the shell method for rotation around a vertical line. Created by Sal Khan.
Want to join the conversation?
- I need some help with the conceptual part. Conceptually, these shells stack INTO each other, kind of like a Russian Porcelian doll set right? Whereas the disc method you are simply laying discus on top of each other, kind of like laying pizzas on top of each other.(44 votes)
- I'm having a confusion that outer surface area of a cylinder is 2pi*r*h........from where that dx came from...........plz can you explain in terms of cylinder(9 votes)
- The surface area of a cylinder has zero thickness, so it can't be used to create something that has any volume. For a volume calculation, we need something with at least a little thickness, and in this case the small increment of thickness is in the horizontal direction, so we call it dx. It's as if we made a cylindrical shell by rolling up a piece of paper. The volume of that shell would be the surface area of the paper (2πrh) times the thickness of the paper (dx).(33 votes)
- So when do you use the shells method as opposed to the washers method?(15 votes)
- I get that multiplying the surface area of a cylinder by a very small thickness would approximate the volume of that very thin cylinder. But it seems that that remains always an approximation since the inner surface of a cylinder will always have a smaller circumference than the outer surface. So the result would have to be an over or under approximation, especially if the calculation is repeated. The discrepancy get smaller but the repetitions increase. Am I missing something?(10 votes)
- That comes down to the fact that the integral exists if and only if the supremum of all lower approximations equals the infimum of all upper approximations. If this is the case, the integral is equal to their common value, and furthermore, this value is unique.
- How do I know when to use the shell method over the disk/washer method??(5 votes)
- if your function is hard to define explicitly in terms of y, so like this function (x-1)(x-3)^2 is hard to write in terms of y(1 vote)
- I'm still confused on how to find the radius whenever it isn't on one of the axes, can someone help me?(8 votes)
- The shell looks like the washer method-How are they different?(2 votes)
- It's basically the same as the washer method, but you want to use this method when it's difficult to express an equation as a function of y or x, as Sal said at0:51.(6 votes)
- I am not sure, but is the volume of the shell equal to the volume of the parallelepiped?? I think the volume of the shell must be equal to the volume of the bigger cylinder mines the volume of the smaller cylinder.(4 votes)
- The volume of the shell must be equal to the volume of the outer cylinder minus the volume of the inner cylinder!!! In the formula V=2Пrh*thickness r is the average radius of tte shell (the radius of the outer circle minus the radius of the inner circle? and all that must be divided by 2)!!!!(2 votes)
- What if your function is rotating around a vertical line other than x=0. For example what if the function rotated around x=2. How does the affect the shell method function (S 2pixf(x)dx)?(3 votes)
- You can't actually revolve this function around x = 2 because that line passes through the function and so rotating f(x) would result in an overlap.
However, we an try revolving it around x = 1. Conceptually, the radius of the shell was x. Now we have moved the vertical line 1 unit closer to f(x). Because of this, our radius has decreased by 1, so our new radius is (x-1).
Therefore, the new integral is (S 2pi*(x-1)*f(x) dx)(2 votes)
I've got the function y is equal to x minus 3 squared times x minus 1. And what I want to do is think about rotating the part of this function that sits right over here between x is equal to 1 and x equals 3. And x equals 3 and x equals 1 are clearly the zeroes of this function right over here. And I want to take this region and rotate it around the y-axis. And if I did that, I'd get a shape that looks something like that. And I want to figure out the volume of that shape. And what we're going to do is a new method called the shell method. And the reason we're going to use the shell method-- you might say, hey, in the past, we've rotated things around a vertical line before. We used the disk method. We wrote everything as a function of y, et cetera, et cetera. We created all of these disks. We figured out the volume of each of those disks. But the problem here is this is hard to express as a function of y. How do you solve explicitly for y right over here? So instead, we're going to keep things in terms of x and have a different geometric visualization for how we can come up with the volume. What we're going to imagine instead-- instead of constructing disks, we're going to construct shells. And what do I mean by a shell? So for each x at the interval, on this kind of cut of it, we can construct a rectangle. And what happens if we were to rotate this rectangle? So this is the rectangle right over here. What happens if we rotate this rectangle around the y-axis along with everything else? I'll try my best attempt to draw it. It's going to look something like this. This is challenging my art skills, but I think I can handle it. So it's going to look something not too dissimilar to that right over there. So it looks like a hollowed-out cylinder. I guess that's why we call it a shell. And it's going to have some depth. The depth is going to be dx. And the height right over here is going to be the value of my function. The height is f of x. In this case, f of x is x minus 3 squared times x minus 1. How do we figure out the volume of a cylinder like this? Well, if we can figure out the circumference of the cylinder, and then multiply that circumference times the height of the cylinder, we'd essentially figure out the area of the outside surface of our cylinder. And then if we multiply the area of the outside surface of our cylinder by that infinitesimally small depth, then that'll give us the volume-- I shouldn't say cylinder-- of our shell. So let's try to do it. What is the circumference of a shell? What is the circumference of one shell going to be? Well, it's going to be 2 pi times the radius of that shell. We need to express this as a function of x. And so what is that going to be? It's going to be 2 pi. So for a given x, what is the radius? Well, the radius right over here is just the horizontal distance between the y-axis and that x. So that's just x. So the circumference, in this case, is just going to be 2 pi times x. Now, what is the height going to be for any one of those shells? The height is going to be f of x. This is right over here. And so what is going to be the surface area of the outside? So let me put this in quotes-- "outside" surface area. I'm not worried about the depth right now, the dx. I'm not worried about this top part and the bottom part. I'm just worried about the outside surface area. Well, the outside surface area is just going to be the circumference times the height. It's going to be 2 pi x times f of x. And in this situation, in the situation we're looking at right over here, that's going to be 2 pi x times x minus 3 squared times x minus 1. Now, what's going to be the volume? So the volume of the shell-- shell volume-- is just going to be all this business times dx. So it's going to be 2 pi x times f of x dx. And so now we're ready to integrate over the interval. So the volume of our entire shape is going to be the definite integral. We're going to integrate over all the x's in the interval, from x is equal to 1 to x is equal to 3 of this thing. And we could take the 2 pi out front. So we'll put 2 pi out front. And on the inside, we have x times f of x, which in our situation, is this business. So it's going to be x times x minus 3 squared times x minus 1. And then, of course, we have our dx. So there you have it. Using the shell method, we have set up our definite integral for the volume of this strange-looking shape right over there.