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Homework Help: Volume of a surface of revolution

  1. Jul 14, 2014 #1
    1. The problem statement, all variables and given/known data

    The top of a rubber band bushing is designed to absorb vibrations in an automobile is the surface of a revolution generated by revolving the curve z = (0.5y^2) + 2
    for (0<= y <= 3) in the yz plane about the z axis.

    use the shell method to find its volume.

    2. Relevant equations

    V = 2pi ∫radius*height dy

    3. The attempt at a solution

    v = 2pi ∫y(4-0.5y^2) dy evaluated from 0 to 3.

    This is wrong, how to i go about setting up the integral for this? Have trouble with these.
  2. jcsd
  3. Jul 14, 2014 #2
    Imagine that the curve revolve around the z-axis, and you can divide this volume into as many cylinders as u like to approximate this volume,

    volume of cylinder,
    [itex]V_{cylinder} = \pi r^2l[/itex]

    but in this case, the [itex]r = y, l = dz[/itex]
    thus, your small cylinder volume, [itex]dV[/itex]
    [itex]dV = \pi y^2 dz[/itex]

    by integrating both side,
    [itex]V = \pi \int y^2 dz[/itex]
  4. Jul 15, 2014 #3


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    This is wrong. Yes, y is the radius but the height is the distance from the point (y, .5y^2+ 2) to the top of the figure which is at (y, .5(3)^2+ 2)= (y, 13/2). The height is 13/2- y^/2- 2= 9/2- y^2/2. 4.5- 0.5y^2, not 4- 0.5y^2.

  5. Jul 15, 2014 #4


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    I don't see how you deduce that. It says the curve is the top; doesn't say where the base is, but the XY plane seems a reasonable assumption.
  6. Jul 15, 2014 #5
    alright, so using advice from HallsofIvy, i set the integral up like this:

    V = 2pi ∫y(9/2 - 0.5y^2) dy , evaluated from y = 0 to 3.

    it comes out to 63.6174.

    This also appears to be wrong, and although the site does not give me the answer they want, in the guided practice they demonstrate solving the problem by switching to the x and z axis, and then
    setting up an integral for volume in terms of x.

    How do I switch variables?

    Attached is a photo of the guided solution to a similar problem.

    Attached Files:

    Last edited: Jul 15, 2014
  7. Jul 15, 2014 #6


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    So what if you take the curve to be the top (as stated in the OP) and the base to be z=0?
  8. Jul 15, 2014 #7
    Okay, so just looking at the lid of this figure, it is a circle, with radius of 3, and an area of 9pi.

    The rest of it looks sort of like a cup, that tapers down, the radius getting smaller and smaller, and the sides steeper and steeper until y = 0, at which point z = 2,[STRIKE] but for the sake of argument that is irrelevant [/STRIKE]... or is it?

    Since the cup is actually "floating" above the y axis in 3 space, then the calculations need to be adjusted accordingly.

    The the maximum height of a rod placed in the center, right where the axis of rotation is, would be about 9/2 units in length, before it started peeking out of the cup.

    Thus the figure is inside a cylinder of height 9/2, radius 3.

    So for tiny pipes starting at the z axis and radiating outward, the radius of each pipe would be y units, but the
    height of each pipe would be z - 2, to account for the gap by floating.

    According to the Shell Method, the volume of a solid of revolution can be calculated by
    V = 2pi ∫radius*height*width

    Since in this case y happens to be my independent variable, I want to write it like this:

    V = 2pi ∫ y * (0.5y^2 ) dy , evaluated from y = 0, to y = 3.
  9. Jul 16, 2014 #8


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    Now you're finding the volume between z=2 and z=2+0.5y2.
    I believe the question wants the volume between z=0 and z=2+0.5y2.
  10. Jul 17, 2014 #9
    Aw jeeze, still messing up. How about I treat the line z = 6 as the top curve, and z = 0.5y^2 as the bottom curve.

    To find the height of a tiny slice,

    Height = 6 -0.5y^2

    Radius = y

    Width = dy

    Volume = 2pi integral( y*(6-(0.5y^2)))dy

    Evaluated from 0 to 3
  11. Jul 17, 2014 #10


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    Why would you do that? Why not z = 2+0.5y2 as the top and z = 0 as the bottom, as I keep suggesting?
  12. Jul 18, 2014 #11
    Ah. Yes because, well attached is a photo.

    Part of the problem is that I have been thinking about a different equation for the problem, different coefficient, so the previous response was also a bit off.

    But the rest of it is as follows:

    As illustrated in the photo, forgive me for the poor scaling/artistry, the volume trapped between z = 0 and z = y^2 is a very large chunk in comparison to the orange portion which is between z = 4 and z = y^2 + 2.

    And to prove this: I did the integral both ways:

    First way, 2pi ∫(y*(4-(0.5y^2 + 2 )))dy
    From y = 0 to y = 3, the answer is -7.06858.

    Now this should be positive, yes, so took absolute value of it for 7.068.

    Second way, 2pi ∫(y*(0.5y^2 +2))dy
    From y = 0 to y = 3, the answer is 120.165.

    Now, which is correct? Well finally, for a cylinder of radius 3, and height 4, the total volume by v = pi r² h = 113.09 cm^3

    Weird, right? BUT, if and only if, you subtract out the second way, 113.09 - 120.165, equals -7.068.

    Take absolute value, you get 7.068. :D

    So therefore, it appears we were each finding a different piece of the whole solid of revolution.
    The first way is the little cup, the second way is the large base from which the cup was cut or milled out.

    And yes, I should flip my integral so the answer comes out positive.

    Either way, the difference is still the little cup that I am looking for.

    Attached Files:

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