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Shell method with two functions of x

Using the shell method to rotate around a vertical line. Created by Sal Khan.

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  • piceratops ultimate style avatar for user Waleed Junaidy
    why is the radius, or x, 2 - x?
    (37 votes)
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    • spunky sam blue style avatar for user Ethan Dlugie
      The radius of each cylindrical shell is the horizontal distance from the current x value to the axis of rotation. So if we rotate about the line x=2, the distance between our current x position and the axis of rotation is 2-x.
      Likewise, if we rotate about the y axis (aka x=0) the radius is x-0=x.
      (45 votes)
  • blobby green style avatar for user Minh Anh Nguyen
    At why can't the radius be 2-x^(1/2). Isn't the distance from the axis of rotation to the furthest point on the shell (the point on the function x^(1/2)), the value of the radius?
    (8 votes)
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    • blobby green style avatar for user Mack Andrews
      For any given x-value, the radius of the shell will be the space between the x value and the axis of rotation, which is at x=2.

      If x=1, the radius is 1, if x=.1, the radius is 1.9.

      Therefore, the radius is always 2-x.

      The x^(1/2) and x^2 only come into play when determining the height of the cylinder.
      (3 votes)
  • leaf green style avatar for user mad4soccer1
    How do you decide whether to use disks or whether to use shells? Is there something to look for that will indicate what problems should be solved using one vs the other?
    (8 votes)
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  • blobby green style avatar for user keegan kimbrough
    What it we rotated about -2 instead of 2. Would it then be x+2? TY!!
    (7 votes)
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    • spunky sam blue style avatar for user Ethan Dlugie
      Indeed it would be x+2 which is really a simpler way of writing x–(-2), the distance from the current x coordinate to the line x=-2.
      Just make sure that your distance/radius factor is always positive when you take the shell method. By that I mean not to use -2–x. You will end up with a negative volume if you perform that integration.
      (3 votes)
  • leaf blue style avatar for user Matthew Daly
    How did Sal just throw in the factor of dx at ? The volume of a real shell is not the area of its outer surface times the depth of the shell, so why should that be the volume when the shell is infinitesimally thin?
    (4 votes)
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    • male robot hal style avatar for user Matthew Dai
      The volume of the shell, as stated in previous videos, is the circumference times the height of the shell times the width. Here, the width is dx. However, it does not really matter in the end, because you are just taking the definite integral to find the area.

      Hope that helps!
      (2 votes)
  • blobby green style avatar for user Eddy Zavala
    If rotation occurs about the Y-axis or X=0, the radius would simply be X correct?
    (2 votes)
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    • hopper cool style avatar for user Oliver Dahl
      correct!

      keep in mind that if we rotate around y=c in D we change the integral:

      v(K)=π a(integral)b |(g(x)-c)^2-(f(x)-c)^2| dx,

      for the volume of body K where D={(x,y):a<=x<=b,f(x)<=y<=g(x)}

      this means that we subtract the value of c if we move the axis of rotation up

      and that we add the value of c if we move the axis of rotation down

      i hope this was a little helpful!
      (2 votes)
  • duskpin ultimate style avatar for user Lezzlly Real
    At how did you find the interval from 0 to 1 when looking at volume?
    (2 votes)
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  • blobby green style avatar for user matt.claude.lafrance
    How is the radius 2-x. He always says that the radius is the function minus what it's being rotated around so wouldn't it be x-2? Because in every other video that is the case but in this one it's the value 2- x, why is that?
    (1 vote)
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    • blobby green style avatar for user Creeksider
      Many of these problems can be solved in multiple ways. For this one, you're rotating around the line x = 2, so you can find the radius by going from that point to the left, as Sal did, getting 2-x, or from that point to the right, getting x - 2, the version that seems more natural to you. Either one produces the same result, but Sal has chosen the easier way because when you measure on the left your integral is from 0 to 1 (the x-values on the left that are within range of the figure we're analyzing), whereas if you measure on the right your integral is from 3 to 4. It's almost always easier to evaluate expressions at 0 and 1 than at 3 and 4.

      It would be good practice for you to calculate the integral the way you suggest, using x - 2 as the radius and evaluating the integral from 3 to 4, and confirming that the result is the same.
      (3 votes)
  • blobby green style avatar for user kma1660
    at , why are we integrating from 0 to 1? That would mean our radius (2-x) goes from 2 to 1, but don't we really want the radius from x=0 to x=1? Sorry if I'm not explaining myself well, but I thought we should integrate from 1 to 2 so the radius goes from 1 to 0.
    (1 vote)
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    • piceratops ultimate style avatar for user wouter-muyres
      In this example the x values only resemble the distance from the origin of the axis, you will have to synthesize the axis of revolution(x=2) and the x values to make the radius of the cilinder.
      Because the graph is twisted around the x=2 axis, the radius = the difference between the axis we twist the graphs around(x=2) and the x values between x=0 and x=1(to include all the possible x values we integrade from x=0 to x=1).
      We are only interested in the solid of revolution of the shape on the left side. Graph the formulas between x=1 and x=2 for your visualization, the shape looks different.
      (1 vote)
  • blobby green style avatar for user nematwaseem
    These videos are inaccessible for me for some reason. I was only able to watch the videos previous to this one. What do I need to do to access it?
    (1 vote)
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Video transcript

This right here is a solid of revolution whose volume we were able to figure out in previous videos, actually in a different tutorial, using the disk method and integrating in terms of y. What we're going to do right now is we're going to find the same volume for the same solid of revolution, but we're going to do it using the shell method and integrating with respect to x. So what we do is we have the region between these two curves, y is equal to square root of x and y is equal to x squared. And we're going to rotate it around the vertical line x equals 2. And we're going to do that at the interval that we're going to rotate this space between these two curves is the interval when square root of x is greater than x squared. And so it's between 0 and 1. And so let's try to do it with the shell method. And so to do that, what we do is we want to construct a shell. Let me do this in a different color. Let's imagine a rectangle right over here. It has width dx. and its height is the difference of these two functions. And so if I were to draw it right over here, it would look something like this. It'd be there, and then it is a shell, it's kind of a hollowed-out cylinder. So it would look something like this. Just like that. And it has some depth, that's what the dx gives us. So we have the depth that looks something like that. And then let me shade it in a little bit, just so we can see a little bit of its depth. So when you rotate this rectangle around the line x equals 2, you get a shell like this. So let's think about how we can figure out the volume of this shell. Well, we've already done this several times. The first thing we might want to think about is the circumference of the top of the shell. We know circumference is 2 pi times radius. We just need to know what the radius of the shell is. What is that distance going to be? Well, it's the horizontal distance between x equals 2 and whatever the x value is right over here. So it's going to be 2 minus our x value. So this radius, this distance right over here, is going to be 2 minus x. And so the circumference is going to be that times 2 pi. 2 pi r gives us the circumference of that circle. So 2 pi times 2 minus x. And then if we want the surface area of the outside of our shell, so the area is going to be the circumference 2 pi times 2 minus x times the height of each shell. Now, what is the height of each shell? It's going to be the vertical distance expressed as functions of y. So it's going to be the top boundary is y is equal to square root of x, the bottom boundary is y is equal to x squared. So it's going to be square root of x minus x squared. Let me do this in the yellow. So it's going to be square root of x minus x squared. And so if you want the volume of a given shell-- I'll write all this in white-- it's going to be 2 pi times 2 minus x times square root of x minus x squared. So this whole expression, I just rewrote it, is the area, the outside surface area, of one of these shells. If we want the volume, we have to get a little bit of depth, multiply by how deep the shell is, so times dx. And if we want the volume of this whole thing, we just have to solve all the shells for all of the x's in this interval and take the limit as the dx's get smaller and smaller and we have more and more shells. And so, what's our interval? Well our x's are going to go between 0 and 1. So that right over there is the volume of this figure.