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Theorem of Pappus

  1. Apr 11, 2008 #1
    The theorem of pappus seems too simple that i do not get it.
    ok
    suppose I have the 1. (x cm., y c.m) meaning i have the x and y center of mass (the centroids). the center of mass is of some function
    2. y(x) how do i find the volume when i revolve this region about the line 3. y(x) = mx+b

    thank you
     
  2. jcsd
  3. Apr 12, 2008 #2

    HallsofIvy

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    No, the center of mass is not "some function". The center of mass is a single point: of some 2 dimensional region you have not mentioned.

    And since you haven't mentioned the region, you haven't mentioned the area of the region which is the the cross-sectional area of the volume swept out. Pappus' theorem says that if you have a region of area A, rotated about a line, the volume of the figure created is 2[itex]\pi[/itex]RA where R is the distance from the centroid of the region.

    Calculate the distance, R, from the centroid of the given region to the line y= mx+ b, calculate the area of the region, and then use [itex]V= 2\pi RA[/itex].
     
    Last edited: Apr 12, 2008
  4. Apr 12, 2008 #3
    Ok here is a sample of what I mean.
    Given the centroid of the region enclosed by the x-axis and
    y = Sqrt[a^2-x^2] (a semicircle) is located at (0,4a/3Pi).
    find the volume of the solid generated by revolving this region about the y= x-a.

    So I know Volume = 2Pi*area*distance from centroid to line y=x-a

    Area = Pi * a^2 right?
    distance to from centroid to line y = x-a is what??

    thank you
     
  5. Apr 13, 2008 #4

    HallsofIvy

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    The distance from a point, [itex](x_0, y_0)[/itex], to a line Ax+ By+C= 0 is given by
    [tex]\frac{|Ax_0+ By_0+ C|}{\sqrt{A^2+ B^2}}[/itex]

    The area of a circle is [itex]\pi r^2[/itex]. The area of a semi- circle is half that.
     
  6. Apr 13, 2008 #5
    Hi Thank you very much. quick question what does
    |Ax +By+c| mean?
    I do not understand what the tabs means?

    can you please give me a quick example.
    In my case

    y= x-a
    so |1x-1y-a|
    does this equal x^2-y^2-a^2"
    thank you
     
  7. Apr 13, 2008 #6

    HallsofIvy

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    | | is absolute value. But let's back up a little. WHY are you doing this problem? You don't know Pappus' theorem, you got the wrong area for a semi- circle and you have the wrong value for the centroid! If you are taking a multi-variable Calculus course, you should know all those things before you attempt a problem like this. I recommend you go back, re-read the problem, take a deep breath, and start all over again.
     
  8. Apr 13, 2008 #7
    We skipped this stuff when I was in calc II and I really want to get it so I am doing a couple of problems for fun. I know the area of a semicircle.
    The problem says
    "the centroid of the region enclosed by the x-axis and the semicircle
    y= Sqrt[a^2-x^2] lies at the point (0,4a/3Pi). Find the volume of the solid generated by revolving this region about the line y = x-a"

    Volume = 2Pi A *b(distance)

    A= Pi*a^2/2
    b = (1x0-1y0-a)/Sqrt[(1+1)] = |0-(4a/3Pi)-a|/Sqrt[2] =

    (2*Pi*Pi*a^2/2) * ((4a/3Pi)+a) /Sqrt[2]

    is that right?
    thank you!!
     
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