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### Course: Calculus, all content (2017 edition)>Unit 6

Lesson 11: Shell method

# Calculating integral with shell method

Evaluating integral set up with shell method for two functions. Created by Sal Khan.

## Want to join the conversation?

• This is the same question he answered using the washer method. Sal arrive to the same answer as the washer method. So the method he used can be use to get the same answer as long as the same question does not change?
• He is trying to show with a worked example that the shell method really does work (granted that you believe in the disc method). That is the great thing about math: there are so many different ways to arrive at an answer, and they all work!
• How and when we use the shell method? I m confused how can I classify it to other methods like Disc and Washer.
• In BASIC questions, they will give you y = f(x) as the main "funny boundary" (maybe y = x² or y = sin(x) ), while the other sides of the region will be horizontal or vertical lines (x-axis, y-axis, x = 3, etc.). Or possibly y1 = f1(x), y2 = f2(x) for the "top" and the "bottom" of the region. In these cases, here is the idea:
1) IF the region is then rotated around a horizontal line (x-axis, or y = k), then you probably want to use discs or washers (depending on whether there is a hole in the middle). This is because slicing the shape with a straight vertical knife will give discs or washers, and the radius is determined by the "curvy" function y = f(x).
2) IF the region is rotated around a vertical line (y-axis, or x = k), then you probably want to use cylindrical shells. This is because slicing the shape into shells will give you shells whose height is determined by the "curvy" function y = f(x).
In both of these cases, you would end up doing a "dx" integral.

As always, there can be tricky exceptions to this general rule. For example, if they give you x = f(y), then everything is reversed, so that you end up with a "dy" integral: discs/washers if rotated around a vertical line and shells if rotated around a horizontal line.
• Why aren't the limits for the volume 0 and 2?
• The bounds are the places were the two functions intersect.
• why is the integral in terms of x? can it be in terms of y?
• The integral has to be taken in terms of what axis the shells or washers are stacked about.
• I had a general question on Solids of Revolution. When/How do we know when we are to use dy or dx and when/how do we get to know when to use the shell or the disk/washer method or does it not matter ?
(1 vote)
• I'm not aware of any general answer to this question. Many problems can be solved either way with about equal difficulty. Depending on the formulas involved, in some cases you may find that one way forces you to solve a difficult integral while the other way gives you an easy one, and it may not be obvious which way is easier until you work on it a while.
• Why can't you do surface area of the the top the the shell times the height. Ie. Hollowing out a cylinder?
• why is the constant not added while taking the integral of the expressions ??
(1 vote)
• When we have a definite integral, we don't add a constant. We are just using the first fundamental theorem of calculus:
b
∫ f(x) dx = F(b) - F(a)
a
However, if we have an indefinite integral, we have to write +C at the end.
• I know that when the equation lands beneath the x-axis, the area is negative, so when finding the volume around the x-axis, would it be required to manually flip the equation in order to get a positive volume? Because it seems to be that the problems I've done tend to require me to do that. Unless I'm doing something wrong.
(1 vote)
• For this sort of problem we shift our frame of reference from the x or y axis to the axis of rotation. In this example that is done when we define the radius as (2-x) rather than x, and the width as (x^1/2 - x^2). Were it rotated about the y-axis then the radius would just have been x, and the width would have been (x^2 - x^1/2). By deliberately shifting our frame of reference to match the axis of rotation we ensure the volume is positive regardless of where the curve falls relative to the x and y axes.
(1 vote)
• why not call this the "ring" method?
(1 vote)
• Isn't the equivalent of 1/4, 20/60? I think an error was overlooked and the answer simplifies to be 8(pi)/5.
(1 vote)
• I'm 4 years late, but 20/60 = 1/3 since 20/20 is 1 and 60/20 is 3 (or put it into a calculator). So, no, I don't think it's an error.
(1 vote)

## Video transcript

In the last video we were able to set up this definite integral using the shell or the hollow cylinder method in order to figure out the volume of this solid of revolution. And so now let's just evaluate this thing. And really the main thing we have to do here is just to multiply what we have here out. So multiply this expression out. So this is going to be equal to-- I'll take the 2 pi out of the integral. 2 pi times the integral from 0 to 1. Let's see, 2 times the square root of x is 2-- I'll write it as 2 square roots of x. But I'll write it as 2x to the 1/2. It'll make it a little bit easier to take the antiderivative conceptually, or at least in our brain. So two times the square root of x is 2x to the 1/2. 2 times negative x squared is negative 2 x squared. And then we have negative x times the square root of x. Well, that's x to the first times x to the 1/2. That's going to be negative x to the 3/2 power. And then we have negative x times negative x squared that's going to be positive x to the third power. And all of that dx. And so now we're ready to take the antiderivative. So this is going to be equal to 2 pi times the antiderivative of all of this business evaluated at 1 and at 0. So the antiderivative of 2 times x to the 1/2 is going to be 2-- it's going to be-- let's see. We're going to take x to the 3/2 times 2/3. So it's going to be 4/3 x to the 3/2. And then for this term right over here it's going to be negative 2/3 x to the third. And you could take the derivative here to verify that you actually do get this. And then right over here, let's see, if we incremented this, you get x to the 5/2. And so we're going to want to multiply by 2/5. So minus-- let me do this in another color. Let's see, so this one right over here, it's going to be minus 2/5 x to the 5/2 power. Yep, that works out. And then finally you're going to have x to the fourth over 4 plus-- let me do that in a different color-- plus x to the fourth over 4. That's this term right over here. And now we just have to evaluate at 1 and 0. And 0, luckily, all of these terms end up being a 0. So that's nice and cancels out. And so we are just left with-- we're just-- [INAUDIBLE] cancel out. It just evaluates to 0. So this is just 2 pi times when you evaluate all this business at 1. So that's going to be 4/3 minus 2/3 minus 2/5 plus 1/4. And the least common multiple right over here looks like 60, so we're going to want to put all this over a denominator of 60. So it's going to be 2 pi times all of this business over a denominator of 60. And 4/3 is same thing as 80/60. Negative 2/3 is the same thing as negative 40/60. Negative 2/5 is the same thing as negative 24/60. And then 1/4 is the same thing as 15/60. So this is equal to-- and actually this will cancel over here, and you'll just get a 30 in your denominator. So in your denominator, you get a 30. And up here 80 minus 40 is 40. 40 minus 24 gets us to 16. 16 plus 15 is 31. So we get 31 times pi over 30 for the volume of the figure right over there.