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Surface Area of Revolution

  1. Mar 24, 2008 #1
    The Volume of a revolved function can be given by the integral of pi*f(x)^2*dx. For the arc length of a graph, a different integral is available. I understood the proof of these two and their integration is understandable. From such, I was actually expecting that the surface area of revolution would be the integral of 2*pi*f(x)*dx, in which it would be a summation of infinitesmally thin hoops over an interval deltax. This actually seems logical to me...but why doesn't it work. Apparently, the surface area of revolution is given by the integral of 2*pi*f(x)*dL, in which dL is infinitesmally small arc length and can be replaced by a function in respect to dx...which makes things lengthier. Why doesn't integrating with just dx work???...it seems as though it would be the same concept of volume, just with the surface area instead.
    Last edited: Mar 24, 2008
  2. jcsd
  3. Mar 24, 2008 #2
    it is somewhat subtle to see why [itex]2\pi f(x)dx[/itex] doesn't capture the surface area of revolution, but to give an explanation in words, you're trying to interpolate the length of a 2d curve (f(x) plotted vs. x) but assuming each dx chunk has a contribution proportional to f(x), which is not true. In order to truly approximate its length, you need to capture both the change in dy and the change in dx by integrating with respect to dl = [itex]\sqrt{dx^2+dy^2}[/itex], which is the length of the distance between two interpolating points with x values dx apart.

    As a simple but extreme example, assume there is a vertical asymptote to the graph of f(x). If you try to find the surface area using [itex]\int 2\pi f(x)dx[/itex], you don't capture the vertical parts of the surface area, which become quite significant near the vertical asymptote.
    Last edited: Mar 24, 2008
  4. Mar 24, 2008 #3
    I'm beginning to understand; you've made a good logical argument there. Why isn't this taken into account for finding the volume revolved around an axis?
  5. Mar 24, 2008 #4
    Well that's a little more difficult to explain elegantly, but i'll give it a shot: Basically since you're talking about the cylinder approximation to volumes of rotation ([itex]V=\int \pi f(x)^2 dx[/itex]), each cylinder's volume is independent of the cylinder to the left and right, i.e. it only depends on the radius of the cylinder f(x) and its height dx. Let's say [itex]f(x)=\frac{1}{x^2}[/itex] and let's look from x=1 to 100 and rotated around the x axis. If I divide the volume up into chunks dx wide, each of these chunks has its own volume expressed completely in terms of the properties of that chunk without regard for chunks around it: [itex]V_{dx}=\pi f(x)^2 dx[/itex]. This is obvious because I can "mix up" those chunks and the volume of every chunk as well as the entire volume will stay the same. This is essentially saying the volume of the chunks does not depend on dy.

    What if I rearranged those chunks so that if there were n chunks, I exchange the 1st chunk with the nth chunk, the 3rd with the (n-2)th, the 5th with the (n-4)th and so on? The volume obviously remained the same, but the surface area becomes dramatically different because f(x) is much larger at x=1 than x=100 so the overall effect is to make an object with a series of large grooves but with the same volume, with the surface area of these grooves being related to how large adjacent cylinders are. Does this make sense?
    Last edited: Mar 24, 2008
  6. Mar 25, 2008 #5
    Yeah...you've laid it out clearly; if each piece has a volume, rearranging the pieces wouldn't change the total volume, but it could change the surface area. Therefore, the surface area is also dependent on the dy factor as well.
    Last edited: Mar 25, 2008
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