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Shell method for rotating around horizontal line

Find the volume of a solid of revolution by rotating around the x-axis using the shell method. Created by Sal Khan.

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  • mr pants teal style avatar for user Patrik Montgomery
    What does "solid of revolution" mean?
    (5 votes)
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    • blobby green style avatar for user Creeksider
      A solid of revolution is a solid figure formed by rotating a two-dimensional figure around an axis. For example, you can create a sphere by rotating a half-circle around the line formed by its endpoints, or a cone by rotating a slanted line segment around a vertical line dropped from its higher endpoint to a point level with the line segment's lower endpoint. Many other figures can be rotated to produce other types of shapes. Donuts, anyone?
      (29 votes)
  • blobby green style avatar for user bcbrittanycong
    when would you have to solve for y?
    (4 votes)
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  • aqualine tree style avatar for user Sarthak
    Wouldn't this problem just be simpler using disks in terms of x, from 0 to 8?
    (7 votes)
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    • piceratops ultimate style avatar for user Zach Gansworth
      In this case, you really could have done it easily either way. However, in some cases using the disk method is not always easy. For example, if we were rotating part of the graph y=(x-3)^2*(x-1) around the y-axis (Sal actually does this in the video titled Shell method for rotating around vertical line), it would require writing x as a function of y, which is not very easy to do in this case. Using the shell method allows us to use the function as it is in terms of x, making it a lot easier to find the volume.
      (3 votes)
  • blobby green style avatar for user WadePearrell
    Why did you use y for the radius instead of y^3?
    (7 votes)
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  • blobby green style avatar for user milanbaudelaire
    I don't understand why the height is equal to the function when revolving about the y-axis, but it's equal to some x-value minus the function when revolving about the x-axis.
    (5 votes)
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    • leaf blue style avatar for user Stefen
      It only seems that way because of the examples used so far, which were bounded by the x axis, or, y=0.
      What you are doing is making sure you are only integrating within the correct region.
      Here is a similar solid as in the video, but rotated about the y axis, whose region of integration is bounded by y=8 above and y=x^3 below - which means the height is given by h = 8 - x^3. http://bajasound.com/khan/khan0010.jpg

      Identifying the limits of integration is a very important skill to develop.
      (3 votes)
  • blobby green style avatar for user BrianMBrannon
    Sal, when you evaluated the integral from 0-2 you only found the volume of the shape above the x-axis. To get the volume of the entire shape you should have multiplied that by two or took the integral from -2 - 2. So the actual volume should be 192pi/5. Right?
    (3 votes)
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    • leaf red style avatar for user Bob Fred
      nope the idea of shells is that you integrate circumferences*heights which covers both sides (since you get a full circle) from integrating just 1 side (you can see it from the picture that as you integrate outwards you integrate the top and bottom areas at the same time)
      (6 votes)
  • male robot donald style avatar for user Parviz Latipov
    Do we get the same answer if we solved with Disk method?
    (2 votes)
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  • orange juice squid orange style avatar for user Abhisek Nandy
    I have a question so I know the volume is supposed to be 2pi *r*h so is he referring to the depth as height to dy or did he simply incorporate the height into 2pi*y
    (3 votes)
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  • marcimus red style avatar for user John Matthew Redman
    Again randomly inserting a y-value, is it the small radius? I see no x=y function anywhere in this problem at all.
    (3 votes)
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  • leaf green style avatar for user Korab Cocaj
    I didn't get how did he sum up all the cylinders inside. By doing so, you surpass the volume of the the intended shell.
    (2 votes)
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    • leaf blue style avatar for user Stefen
      If you understand how summing the "infinitely narrow" rectangles under a graph yields the area under the graph, it is not a giant leap to understand how the summing of "infinitely narrow" shells within the bounds of a rotated graph yields the volume of the solid bounded by the rotated graph.
      I suggest watching the video again with this in mind. Remember, we are summing an infinite number of infinitely thin shells.
      (2 votes)

Video transcript

What we're going to do in this video is take the function y is equal to the cube root of x and then rotate this around the x-axis. And if we do that, we get a solid of revolution that looks like that. And we're doing it between x is equal to 0 and x is equal to 8. And you get something that looks like this. And you could find the volume of this actually quite easily using the disk method. But just to show you that you could do it an alternate way, we're going to use the shell method. But we're going to use the shell method now to rotate around a horizontal line, and specifically the x-axis. So how would we do this? Well, what we want to do, what you could imagine, is constructing a rectangle that looks like this. Let me do it in this salmon color. So you have a rectangle that looks something like that. Its depth or its height, you could say, is dy. And then its length right over here is going to be 8 minus whatever x value this is. Let me make this clear. The width is going to be 8 minus whatever x value this is right over here. And you might already realize that if this is going to be dy, we're going to be integrating with respect to y over an interval of y. And so we really want to have everything in terms of y. So this x value, whatever it is, as a function of y. So if y is equal to the cube root of x, we can cube both sides, and we can get x is equal to y to the third power. These are equivalent statements right over here. And so, this distance right over here is going to be 8 minus y to the third, if we were to express it in terms of y. And then when you rotate that thing around the x-axis, it's going to construct the outside of the cylinder or shell, as we like to call it. And I'll try my best to draw that shell. So it will look something like this. So you're going to have a shell that looks like this. And there you go. And actually, this shares a common boundary right over here. So you're going to have a shell that looks something like that. So that hopefully helps a little bit. Let me give it some depth. You'll have a shell that looks something like that. If we could figure out the volume of that shell, which is really going to be the area of the outer surface area times the depth. And then if we were to sum up all of those shells, this shell is for a particular y in our interval. If we were to sum up over all of the y's in our interval, all the volumes of the shells, then we have the volume of this figure. So once again, how do we figure out the volume of a shell? Well, we can figure out the circumference of I guess the left or the right, in this case-- the left or the right of our cylinder. If we can figure out that circumference, that circumference is equal to 2 pi times the radius. And what is the radius? The radius of these things are just going to be your y value. That's this distance right over here. That's just going to be the y value. So it's going to be equal to 2 pi times y. And then if we want the area of the outer surface, we just multiply the circumference times the width of our cylinder. So let me write it over here. So outer surface area is going to be equal to our 2 pi y times our 8 minus y to the third. 8 minus y to the third is just this length. You multiply that times circumference. You get the outer surface area. Now, if you want to find the volume of this one shell, it's going to be the outer surface area. 2 pi y times 8 minus y to the third times the depth, times this dy right over here. I'll do the dy in purple. So that's the volume of one shell. If we want to find the volume of the entire solid of revolution, we have to sum all these up and then take the limit as they become infinitely thin, and we have an infinite number of these shells. So we're going to take the sum from, and remember, we're dealing in y. So the volume is going to be equal to, so what's our interval in terms of y? So y definitely starts off at 0, and when x is equal to 8, what is y? Well, 8 to the 1/3 power is just 2. So y is 2, this value right over here. Let me make it a little bit clearer. This value right over here is 2. So y goes from 0 to 2, and we've set up our integral. And this one looks pretty straightforward. So I think we can crank through it in this video. So this is going to be equal to, we can take out the 2 pi, 2 pi times the definite integral from 0 to 2. Let's multiply the y of 8y minus y to the fourth. All of that dy. This is going to be equal to 2 pi times the anti-derivative of this business. Anti-derivative of 8y is 4y squared. Anti-derivative of 1 negative y to the fourth is negative y to the fifth over 5. And we're going to evaluate it at 0 and 2. So this is going to be equal to 2 pi times, we evaluate this business at 2, 2 squared is 4 times 4 is 16. 16 minus 2 to the fifth is 32. So minus 32/5. And then, you evaluate this stuff at 0, you just get the 0. So that's what we're left with. And now, we just have to simplify this thing a little bit. So let's see. 16 over 5, this part right over here. 16 is the same thing, I should say, as 80/5. And from that, we are subtracting 32/5. And so, that is equal to 48/5. So all of this business is equal to 48/5. Did I do that right? 80 minus 30 is 50, and then minus another 2 gets us to 48. So this is 48/5 times 2 pi. And now we deserve our drum roll. 48 times 2 is 96. Let me do this in a new color, just to emphasize that we're at the end. 48 times 2 is 96 times pi over 5. So once again, this is something that you could have solved using the disk method in terms of x. And we're just showing that you could also solve it in the shell method in terms of y. The shell or the hollowed out cylinder method, whatever you want to call it, in terms of y.