If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Disc method around y-axis

AP.CALC:
CHA‑5 (EU)
,
CHA‑5.C (LO)
,
CHA‑5.C.1 (EK)
Finding the volume of a figure that is rotated around the y-axis using the disc method. Created by Sal Khan.

Want to join the conversation?

  • blobby green style avatar for user Blu Jay
    How will I know if I'll use the disk, ring or shell method? And when do I use vertical or horizontal?
    (24 votes)
    Default Khan Academy avatar avatar for user
    • spunky sam blue style avatar for user Ethan Dlugie
      It really depends on the situation you have. If you have a function y=f(x) and you rotate it about the x axis, you should use disk (or ring, same thing in my mind). If you rotate y=f(x) about the y axis, you should use shell.
      Of course, you can always use both methods if you can find the inverse of the function. If I wanted to rotate y=x^2 about the y axis, that would be equivalent to rotating x=√(y) about the x axis. I prefer to not bother with finding the inverse of the function. I would just identify the situation and use the appropriate method.
      By the way, if you have a function that is reverse, that is to say x=f(y), everything I said would be reversed. You would use disk about the y axis and shell about the x axis.
      (44 votes)
  • aqualine seed style avatar for user Diyar Ghazi
    Anyone that might explain why Sal takes the principle root of y at ? For some reason it is completely lost on me..
    (11 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Daniel Cimino
      He did that because he is rotating around the y-axis and he is trying to find the volume. As Sal showed, you need to find the radius of each disk so as to apply it into A = (pi)r^2 and then V = A(dy). Notice that it is in terms of dy, not dx. Therefore, the equation y=x^2 needed to be changed into terms of x, otherwise you would be finding a radius and thus an area and thus a volume of a solid that is irrelevant to this problem.
      (7 votes)
  • female robot grace style avatar for user Ruth
    At , why did we use the interval from 1 to 4 instead of from 0 to 4?
    (7 votes)
    Default Khan Academy avatar avatar for user
  • leaf green style avatar for user Alexander Jacoby
    What are the practical uses of the disk method (or finding the volume of solids of revolution in general)? What are some physical applications of these problems?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • leaf green style avatar for user tanishq.aggarwal.11
    The integral used in the problem is in terms of y. However, since y=x^2 and thus dy=2x dx, can you redefine the integral so that it is from 1 to 2 and its expression is 2pix^3 dx?
    (6 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user jimstanley49
      That's a very creative approach. It seems to work, but involves substituting things around and updating the bounds of integration. You still have to solve for x in terms of y (the square root bit) to transform the bounds from y to x and there is a little more simplification required at the end. This adds points at which stupid algebra errors can creep in.
      If it helps you get the right answer and it does give you the right answer every time, go for it. Though you may have to argue with your teacher for a couple of "show your work" points because you do it differently. I leave it to someone more experienced to show proof.
      (7 votes)
  • blobby green style avatar for user sbu hlombe
    how do we calculate the volume if the question said roteted about the line y=3, x=5 on the same equation of y=x^2
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Haap
    how do you decide the limits of the integral?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • starky ultimate style avatar for user Alasta Firkins
    If we are integrating with respect to y, why don't we have to adjust the boundaries?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Thien D Ho
      Why do we? We integrate from y=1 to y=4 to find the volume, those are the boundaries regarding to y-axis. And Sal already manipulated the equation of the function into y: x=radical y or y^1/2 before carry out the integration.
      (2 votes)
  • leaf green style avatar for user Yu Aoi
    is the volume always positive?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user qudrat mommandi
    could you have done this by integrating in terms of x? instead of taking horizontal slices from top to bottom in terms of the y value with the depth being dy couldn't you take vertical slices in terms of x with the width being dx?

    I'm confused because I hear if you're rotating about the y axis you should always try to put your equations in the form of x=...something in terms of y, and do the differentiation in terms of y but it seems unnecessary.
    (1 vote)
    Default Khan Academy avatar avatar for user

Video transcript

Here we have the graph, or part of the graph, of y is equal to x squared again. And I want to find the volume of another solid of revolution. But instead of rotating around the x-axis this time, I want to rotate around the y-axis. And instead of going between 0 and some point, I'm going to go between y is equal to 1 and y is equal to 4. So what I'm going to do is I'm going to take this graph right over here. I'm going to take this curve. And instead of rotating it around the x-axis, like we did in the last few videos, I'm going to rotate it around the y-axis. So I'm going to rotate it around just like that. So what's the shape that we would get? So let me see if we can visualize this. So the base is going to look something like that if we could see through it. And then this up here, the top of it, would look something like that. And we care about the stuff in between. So we care about this part right over here, not the very bottom of it. And let me shade it in a little bit. So it would look something like that. So let me draw it separately, just so we can visualize it. So I'll draw it at different angles. So if I were to draw it with the y-axis kind of coming out the back, it would look something like this. It would look something like-- it gets a little bit smaller like that. And then it gets cut off right over here, right over here like that. So it looks-- I don't know what shape you could call it. But I think hopefully you're conceptualizing this. Let me do it in that same yellow color. The visual-- that's not yellow. The visualization here is probably the hardest part. But as we can see it's not too bad. So it looks something like this. It looks like maybe a truffle, an upside-down truffle. So this right here, let me draw the y-axis just to show how we're oriented. So the y-axis is popping out in this example like that. Then it goes down over here. And then the x-axis is going like this. So I just tilted this over. I tilted it over a little bit to be able to view it at a different angle. This top right over here is this top right over there. So that gives you an idea of what it looks like. But we still haven't thought about how do we actually find the volume of this thing? Well, what we can do, instead of creating discs where the depth is in little dx's, what if we created discs where the depth is in dy? So let's think about that a little bit. So let's create-- let's think about constructing a disc at a certain y-value. So let's think about a certain y-value, and we're going to construct a disc right over there that has the same radius of the shape at that point. So that's our disc. That's our disc right over here. And then it has a depth-- instead of saying it has a depth of dx, let's say it has a depth of dy. So this depth right over here is dy. So what is the volume of this disc in terms of y? And as you could imagine, we're going to do this definite integral, and it is a definite integral, with respect to y. So what's the volume of this thing? Well, like we did in the last video, we have to figure out the area of the top of each of these discs. Or I guess you could say the face of this coin. Well, to find that area it's pi r squared. If we can figure out this radius right over here, we know the area. So what's that radius? So to think about that radius in terms of y, we just have to solve this equation explicitly in terms of y. So instead of saying it's y is equal to x squared, we can take the principal root of both sides, and we could say that the square root of y is equal to x. And this right over here is only defined for non-negative y's, but that's OK, because we are in the positive x-axis right over here. So we can also call of this function right over here x is equal to the square root of y. And we're essentially looking at this side of it. We're not looking at this stuff right over here. So we're only looking at this side right over here. We've now expressed this graph, this curve, as x as a function of y. So if we do it that way, what's our radius right over here? Well, our radius right over here is going to be f of y. It's going to be the square root of y. It's going to be the square root of y is our radius. So it's going to be a function of y. I don't want to confuse you if you thought this was f of x, and actually this is f of y. No, it would be a function of y. We could call it g of y. It's going to be the square root of y. So area is equal to pi r squared, which means that the area of this thing is going to be pi times our radius, radius squared. Our radius is square root of y. So this thing is going to be equal to pi-- the square root of y squared is just pi times y. Now, if we want the volume, we just have to multiply the area of this surface times the depth, times dy. So the volume of each of those discs is going to be pi y times dy. This gives you the volume of a disc. Now, if we want the volume of this entire thing, we just have to sum all of these discs for all of the y-values between y is equal to 1 and y is equal to 4. So let's do that. So we just take the definite integral from y is equal to 1 and y equals 4. Just as a reminder, definite integral is a very special type of sum. We're summing up all of these things. But we're taking the limit of that sum as these dy's get shorter or get, I guess, squatter and squatter or smaller and smaller, and we have a larger and larger number of these discs. Really, as these become infinitely small and we have an infinite number of discs, so that our sum doesn't just approximate the volume, it actually is the volume at the limit. So to figure out the volume of this entire thing, we just have to evaluate this definite integral in terms of y. And so how do we do that? What's it going to be equal to? Well, we could take the pi outside. It's going to be pi times the antiderivative of y, which is just y squared over 2, y squared over 2 evaluated from 1 to 4, which is equal to pi times-- well, if you evaluate it at 4, you get 16 over 2. Let me just write it out like this. 4 squared over 2 minus 1 squared over 2, which is equal to pi times 16 over 2 is 8, minus 1/2. And so we could view this as 16/2 minus 1/2, which is equal to 15/2. So this is equal to 15/2 times pi. Or another way of thinking of it is 7 and 1/2 times pi. But this is a little bit clearer. So we're done. We found our volume not rotating around the x-axis, but rotating around the y-axis, which is kind of exciting.