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Homework Help: Volume of rotated graph

  1. Mar 24, 2006 #1
    Hi. There are a few problems on my homework that involved the volume of a rotated solid. I do not know how to do these, but I'm trying to devise a method. This is what I figure:

    [tex]\int_{a}^{b} f(x) dx[/tex] is the area under the graph.

    [tex]\frac {\int_{a}^{b} f(x) dx} {b-a}[/tex] is the average height.
    This is where logic comes in:
    I figure you can use the mean value (average height) as the radius of an "average cylinder" of the function. Therefore, the volume would be [tex]\pi r^{2} (b-a)[/tex], where [tex]r[/tex] is [tex]\frac {\int_{a}^{b} f(x) dx} {b-a}[/tex].

    Is this correct? I will simplify once i confirm my Latex is correct.

    Alright, with this, is it okay to conclude that the over volume is [tex]V = \frac {\pi (\int_{a}^{b} f(x) dx)^2} {b-a}[/tex]?

    On a sidenote, I make mistakes quite often with Latex, is there an offline generator to check my syntax with?
    Last edited: Mar 24, 2006
  2. jcsd
  3. Mar 24, 2006 #2

    Tom Mattson

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    Do you mean a volume of revolution? It looks like it.

    Not that I know of, but you do have 24 hours to edit your posts.
  4. Mar 24, 2006 #3


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    No. Consider a sphere of radius 1, centered at the origin, so r as you've defined it would be [itex]\pi /4[/itex], since the area under f will just be the area of half of a circle of radius 1, which is [itex]\pi /2[/itex], and then you divide by b-a = 1-(-1) = 2 to get r. So

    [tex]\pi r^2(b-a) = \pi ^3/8[/tex]

    On the other hand, you know the volume is [itex]4\pi /3[/itex].
  5. Mar 25, 2006 #4
    Mind helping me figure out how you actually do this then?
  6. Mar 25, 2006 #5


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    If you used the distance from the axis of rotation to the centroid of the figure instead of the "average radius" then you would have Pappus' theorem.
    Here's a reference on volumes of revolution:
    Last edited by a moderator: Apr 22, 2017
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