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Volume vs. Area of a Surface of Revolution

  1. Mar 19, 2008 #1
    f(x) is a continuous function of x, whose domain is [a, b]. Revolve the graph around the x axis. In doing so you will create a solid.

    Apparently, to find the volume of this solid, partition the solid into n cylinders along the x axis from a to b, each partition of the x axis containing some x*. A cylinder's radius is f(x*), so its volume is pi(f(x*)^2). Add the cylinders' volumes together. Take the limit as n approaches infinity; that is, take the integral of pi(f(x)^2)dx from x=a to x=b.

    So that's volume. To find the area of the surface of revolution, instead of using cylinders, partition the solid into n frustums of cones along the x axis from a to b, each frustum having two different circular sides, one with radius f(xi-1) and the other with radius f(xi). The surface area of a frustum is 2pi times the average of the radii times the arc length of the curve cut by the two circular sides of the frustum, this arc length being sqrt[1+(dy/dx)^2]. Add the frustums' surface areas together. Take the limit as n approaches infinity; that is, take the integral of 2pif(x)sqrt[1+(dy/dx)^2]dx from x=a to x=b.

    My question: Why the inconsistency? Why can't we take the area of the surface of revolution just by taking cylinders, like we did with volume (except of course taking the surface area of the cylinders, not the volume), and not have to worry about the arc length (that is, not have to worry about the fact that, for a given partition, one circular side will be different from the other, since you'd think it wouldn't matter when the partition becomes infinitely thin)? And if we're not allowed to do that, why are we allowed to use cylinders instead of frustums and ignore the arc length when taking volume?

    In my calculus II class we're using Stewart's Calculus as our textbook. The book doesn't really explain the inconsistency. I also own Spivak's Calculus, which doesn't seem to explain it either (although Spivak at least admits there's some "fudging" in what he's doing in that chapter, although I'm not sure if this is what he was referring to or not).

    I might not have explained my question well in the post. If you're confused about what I'm asking, I can try clearing it up.
    Last edited: Mar 19, 2008
  2. jcsd
  3. Mar 19, 2008 #2
    Off the top of my head (that is, take this with a huge grain of salt), I think the approximating surface doesn't really matter in the answer, but the cone/cylinder might give the simplest (or maybe easiest to visualize) way to get to the answer.
  4. Mar 19, 2008 #3
    No, it definitely changes the answer.

    The formula for the surface area of revolution is the integral of 2pif(x)sqrt[1+(dy/dx)^2]dx from x=a to x=b.

    What I'm asking is why the sqrt[1+(dy/dx)^2]dx is there for surface area, but for volume it's just dx. It certainly affects the answer.
  5. Mar 19, 2008 #4
    Here's my shot at helping you:

    To find surface area you are essentially rotating an (arc length*circumference) around a given axis, in this case the x-axis. Now for some continuous function f(x) from x=a to x=b in order to approximate an arc length you will utilize the distance formula such that

    Arc Length ~ [tex]\sqrt{[(\Delta x)^2] + [(\Delta y)^2]}[/tex]

    You can then multiply by:
    [tex]\sqrt{(\Delta x)^2}[/tex]/[tex]\sqrt{(\Delta x)^2}[/tex]

    Which gives:
    [tex][\sqrt{1+[(\Delta y)^2/(\Delta x)^2]}]\Delta x[/tex]

    Then take the lim as n approaches infinity, aka the integral to end up with your

    [tex]\int \sqrt{1+(dy/dx)^2} dx[/tex]

    And that would be where it comes from, I hope it helped some
  6. Mar 20, 2008 #5


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    As Will J said, cylinders will not give you the surface area for the same reason that horizontal lines will not approximate a line at an angle to the x-axis.

    That is, you cannot approximate the the length of, say, the line from (0,0) to (1, 1) by using segments parallel to the x-axis. Even if you take a large number of short horizontal segments, their sum will be 1, not [itex]\sqrt{2}[/itex]. Even in the limit, you get 1, not [itex]\sqrt{2}[/itex].
  7. Apr 8, 2008 #6
    Re: Volume-Surface area (Solids of Revolution)

    One can use these online applets to find Volume and Surface Area for Solids of Revolution

    Volume - Disks

    http://britishcomputercolleges.com/vu/Disks" [Broken]

    Volume - Washers

    http://britishcomputercolleges.com/vu/Washers.html" [Broken]

    Surface Area (Disks - Rectangular functions)


    http://britishcomputercolleges.com/vu/SurfaceArea.html" [Broken]
    Last edited by a moderator: May 3, 2017
  8. Apr 8, 2008 #7


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    Science Advisor

    That's enough. DexteronLine, if you do not stop posting links to your own applets, I will start deleting your posts that contain links.

    I am very impressed with them but they do not help the posters understand their specific problems and do not belong here.
  9. Apr 8, 2008 #8

    I apologize if the links I added were not suitable for the forum. I will not add links to my applets anymore.

    Thanks for the remarks about the program.
    I will learn alot from the postings on this forum

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