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Homework Help: Volume by revolution

  1. Jan 28, 2012 #1
    I have to find the volume of the solid whose area is bound by the equations [tex]y=-x+3 \ and \ y=x^2-3x[/tex] as it revolves around the x and y axis. I approached it by finding the center of mass and then using Pappus's theorem:

    [tex]M_{x}=\int^{3}_{-1} \frac{1}{2}((-x+3)^2-(x^2-3x)^2)\,dx = \frac{64}{15}[/tex]
    [tex]M_{y}=\int^{3}_{-1} x(-x+3-x^2+3x)\,dx = \frac{32}{3}[/tex]
    [tex]A=\int^{3}_{-1} -x+3-x^2+3x \, \ dx = \frac{32}{3}[/tex]
    So the center of mass coordinates are:
    [tex]\overline{x} = \frac{M_{y}}{A} = \frac{\frac{32}{3}}{\frac{32}{3}} = 1[/tex]
    [tex]\overline{y} = \frac{M_{x}}{A} = \frac{\frac{64}{15}}{\frac{32}{3}} = \frac{2}{5}[/tex]
    Revolving around x:
    [tex]d=(2\pi)(\frac{2}{5}) = \frac{4\pi}{5}[/tex]
    [tex]V=dA=(\frac{4\pi}{5})(\frac{32}{3}) = \frac{128\pi}{15} \approx 26.8083[/tex]

    Revolving around y:
    [tex]d=(2\pi)(1) = 2\pi[/tex]
    [tex]V=dA=(2\pi)(\frac{32}{3}) = \frac{64\pi}{3} \approx 67.0206[/tex]

    However the answers in the book are 63.9848 and 72.9478 respectively. Can anyone tell me where I went wrong or if I am even approaching the problem correctly? Thank you.
    Last edited: Jan 28, 2012
  2. jcsd
  3. Jan 28, 2012 #2


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    First of all: Are you instructed to use Pappus's Theorem ?

    What precisely does Pappus's Theorem state?

    No matter what method you use, this problem includes some tricky details.
  4. Jan 28, 2012 #3
    I have not been instructed to use any particular method, I just need to find a solution to the problem. For these types of problems I generally use disks or washers to find the volume, but I guess the "tricky" part of this problem is that the bound area crosses the the axis on which it's supposed to rotate, as u can see:
    http://www4c.wolframalpha.com/Calculate/MSP/MSP3271a035bi2ii1400fa000012ie2ia8673b6ch5?MSPStoreType=image/gif&s=50&w=381&h=306&cdf=Coordinates&cdf=Tooltips [Broken]

    Pappus's second theorem states that the volume of a solid of revolution generated by the revolution of a region about an external axis is equal to the product of the area of the region and the distance traveled by the region's geometric centroid.

    Last edited by a moderator: May 5, 2017
  5. Jan 28, 2012 #4


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    Yes, this is the tricky part … but it's easily handled. The trickier part is to revolve this around the y-axis. Only part of the area on the left is overlapped by the area on the right.
    The part of Pappus's Theorem you seem to be ignoring is: That the axis must be an external axis to the region being rotated. So, when you calculate the centroid, only include the portion of the region above the x-axis . For the other part only the portion to the right of the y-axis , then do the little portion (that's not overlapped) left of the y-axis.
  6. Jan 28, 2012 #5
    Yeah you are right, it is only external. I figured it out by cutting it into smaller parts like you said and then using disks and washers. Thank you for your help.
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