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Generalizing disc method around x-axis

AP.CALC:
CHA‑5 (EU)
,
CHA‑5.C (LO)
,
CHA‑5.C.1 (EK)
Generalizing what we did in the last video for f(x) to get the "formula" for using the disc method around the x-axis. Created by Sal Khan.

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  • female robot grace style avatar for user Strings to Eternity
    i do not understand how does the definite integral comes to know of the exact variation of radius and how to add em up ,
    (6 votes)
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    • blobby green style avatar for user Rodrigo Campos
      in technical terms, the definite integral is the Summation of the terms of a function with the first term equal to f(a), the second to f(a+dx), third to f(a+2dx) ..., term j equal to f(a+n.dx)=f(b), when the dx approaches 0, or you could say, when n approaches plus infinity. Each term is multiplied by dx, if you draw that you'll see that this Sum will give you the sum of many areas of rectangles. When the number of rectangles approaches inf. you will have a definite integral. Fortunately, someone proved that all of that I said equals to F(b) - F(a), when F(x) is the antiderivative of f(x)
      (8 votes)
  • male robot hal style avatar for user The Last Guy
    When finding the volume of the "disk", why do you use the formula for the area of a circle, and not the volume of a cylinder?
    (7 votes)
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    • old spice man green style avatar for user Dan Surerus
      Take a look at the Riemann sum that is a step before the definite integral. You can see there are physical cylinders that make up the volume for n= positive real number. You can use this formula as a better and better approximation for the volume as n becomes larger and larger.
      I think your question involves the limiting case that defines the definite integral as you let n go to infinity. The width of each cross sectional area "shell, washer...any shape" becomes infinitely small (dx). Limit as delta x approaches zero and the number of cross sectional areas you are adding up becomes infinite.
      (4 votes)
  • piceratops sapling style avatar for user ∑⨜|
    Why can't you just measure out each disc?
    (0 votes)
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  • leaf grey style avatar for user TQ
    Can you use this for the y-axis as well?
    (dy)
    (2 votes)
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    • leafers seed style avatar for user Travis Bartholome
      Yes, you can use any of the methods around either axis. (That is, disks, shells, and washers.) You just have to keep in mind that your variable of integration has changed; so for example, if you're integrating dy and rotating around the x-axis, you'd have to use shells rather than using disks like you would if you were integrating dx. Hope that makes sense.
      (3 votes)
  • old spice man green style avatar for user Ionuţ Ştefan
    So at is the bottom radius of the taken disk not bigger than the top radius? Is it not more something like a section of a cone rather than a cylinder? Or is it taken like a cylinder because of the limit?
    (3 votes)
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    • piceratops ultimate style avatar for user jimstanley49
      The radii are different, but as the height becomes infinitely small, so does the difference in radius. It's much like the "problem" with Riemann sums, where the edges of the rectangles stick out to one side or the other of the function graph. As the rectangles get narrower and narrower, the error disappears.
      So yes, the limit makes it a cylinder.
      (2 votes)
  • boggle yellow style avatar for user Gustavo Sáez
    In the examples shown thus far f(x) is greater or equal to 0, but what of we had a cubic that dipped below the x-axis, would we still be able to use that formula?
    (3 votes)
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  • orange juice squid orange style avatar for user Rehuel J. Galzote
    What if the x-axis of revolution is moved? How would you deal with it? Any example videos please. Thank you.
    (1 vote)
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    • leaf green style avatar for user nick.embrey
      You would still need to figure out the radius of the disks. The radius is the distance between TWO lines: the function f(x) and the axis of rotation. So if you were rotating around the line y=-2 (which is just shifting down two units from how it is now), then the radius of each disk would be x^2 + 2.
      (5 votes)
  • aqualine ultimate style avatar for user Kaleb Sagehorn
    Say that the axis that we revolve our function around is also revolving around another axis. How would one calculate that?
    (2 votes)
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  • blobby green style avatar for user Soumya Sambeet Mohapatra
    Hi Sal. A small question. Is there way to calculate the equation of the figure when y=x^2 is rotated about the x-axis? If there is a video could you please share it?
    (2 votes)
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  • stelly blue style avatar for user Anonymous
    why when calculating volume of disk with the y-axis, you are using x with respect to y
    (1 vote)
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    • duskpin ultimate style avatar for user ariya8.narayan7
      To find volume, first we have to calculate the area of a single 2D slice and rotate it around an axis. The integral accomplishes this by finding the area under a section of a curve.

      However, when using the y-axis as the axis of rotation, you have to find the area to the left of an equation.

      An easier method is to just rewrite the equation in terms of y. Effectively, this flips the way you interpret the graph and the area to the left of the graph is now the area "under" the graph.
      (1 vote)

Video transcript

What we're going to do in this video is generalize what we did in the last video. And, essentially, end up with the formula for rotating something around the X-axis like this using, what we call, the "disk method". And the point is to show you where that formula inside a calculus textbook actually comes from. But it just comes from the same exact principles we did in the last video! It's not advised to memorize the formula; I highly recommend against that because you really need to know what's going on. It's really better to do it in first principles where you find the volume of each of these disks. But let's just generalize what we saw in the last video. So instead of saying that "y = x^2", let's just say that this is the graph, the function that's right over here. Let's just generalize it and call it "y=f(x)". And instead of saying x is going from 0 to 2 let's say that we're going between a and b, so these are just two endpoints along the X-axis. So how would we find the volume of this? Well, just like the last video, we still take a disk just like this. But what is the height of the disk? The height of the disk is not just x^2 since we've generalized it. So the height is simply going to be whatever the height of the function is at that point. So the height of the disk is going to be f(x)! The area of the space of this disk is going to be πR^2. So our radius is f(x), and we're just going to square it. That's the area, that's the area of this face right over here. What is the volume of the disk? We're just going to multiply that by our depth, which going to be dx. And we want to take the sum of all of these disks from a to b, and we're going to take the sum of them, and we're going to take the limit as the "dx"s get smaller and smaller. And we have an infinite number of these disks. Thus, we are going to the integral of this from a to b! And this right here is the formula that you will see often in a calculus textbook for using the disk method as you're rotating around the X-axis. So I just wanted to show you that it comes out of the common sense of finding the volume of this disk. The f(x) right over here is just the radius of the disk, so this part over here is really just πR^2; we multiply it [area of the circle] times the depth; then we take the sum from a to b of all of the disks. And essentially since this is integral is the limit of all the disks getting narrower and narrower. And thus, we have an infinite number of these disks.