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Volume of Revolution

  1. Jan 16, 2004 #1
    Included is my attempt at the following question. I get an answer of 10pi, whereas the right answer is (10pi)/3 from my text book. Here is the question:

    Rotate the triangle described by (-1,0),(0,1),(1,0) around the axis x=2 and calculate the volume of the solid.

    I basically changed the problem to the following and continued as it is the same

    Rotate the triangle described by (1,0),(2,1),(3,0) around the y-axis and calculate the volume of the solid

    Thanks for the help

    Attached Files:

  2. jcsd
  3. Jan 17, 2004 #2
    It is so small to understand the pic u quoted so i'm giving u my solution
  4. Jan 17, 2004 #3
    I hope u will take it from here

    http://in.geocities.com/mathsforjee/index.htm [Broken]
    Last edited by a moderator: May 1, 2017
  5. Jan 17, 2004 #4
    I don't quite understand it and I dont understand why mines wrong :(

    I'll keep trying to figure it out.
  6. Jan 17, 2004 #5


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    You may not have had it yet but this looks like an exercise in using Pappus' theorem: the volume of a solid of revolution is the area of a cross section times the circumference of the circle generated by the centroid of that cross section.

    In this case, the cross section is a triangle with base of length 2 and height 1: area= (1/2)(2)(1)= 1.
    The centroid (for a triangle only) is the "average" of the vertices:
    ((-1+0+1)//3,(0+1+0)/3)= (0, 1/3). The distance from (0, 1/3) to the line x= 2 is 2- 1/3= 5/3. The centroid "travels in" (generates) a circle of radius 5/3 and so circumference (10/3)pi.

    The volume of the figure is (1)(10/3)pi= (10/3)pi.
  7. Jan 17, 2004 #6
    Is this true for all kind of figure I never came across that theorem is there any link where i can go for reference
  8. Jan 18, 2004 #7


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    Pappus' theorem is true of any "solid of revolution". You should be able to find it in any calculus textbook (that includes multiple integrals.)
  9. Jan 18, 2004 #8
    I haven't learned that theorem so I don't really think I should use it.

    I'm trying to use horizontal rectangles for my area so I formulated the following integral which is how I was taught how to do questions like this.

    \int \pi r_o^2 - \pi r_i^2 dr



    and the limits of integration are from y=0 to y=1

    so we have this

    \int_0^1 \pi (-y+3)^2 - \pi (y+1)^2 dy

    it makes perfect sense to me but then once you work it out you get 10 pi. which is wrong

    So can someone point out what I did wrong and how to fix it so I dont do it again.

    Thanks very much.
    Last edited: Jan 18, 2004
  10. Jan 18, 2004 #9
    Here is the general formula

    http://in.geocities.com/mathsforjee/GM.html [Broken]
    Last edited by a moderator: May 1, 2017
  11. Jan 18, 2004 #10
    does mine not make sense for some reason

    i am calculating the area of the circle when the farthest line is rotated and then subtracting the area of the circle when the closest line is rotated
  12. Jan 18, 2004 #11
    Yes it do makes sense dont you have gone to the previous post.

    Thats what u have to do and its general too

    Your way do make sense
  13. Jan 18, 2004 #12

    When you rotate a point about a line u get a circle


    When you rotate a line

    It would be somewhat like a truncated cone
  14. Jan 18, 2004 #13
    but i get the wrong answer so can you point out whats wrong with my formulation?

    keep in mind i moved the triangle to (1,0),(2,1),(3,0) because its the same problem correct?
  15. Jan 18, 2004 #14

    I'm rotating horizontal rectangles around the y-axis therefore each rectangle will make a circle
  16. Jan 18, 2004 #15
    Ok It would form rings as of saturn
  17. Jan 18, 2004 #16
    U can also do it analytically With no integration
  18. Jan 18, 2004 #17
    I still want to know whats wrong with this answer because it does not yield 10pi/3

    [tex]\int_0^1 \pi (-y+3)^2 - \pi (y+1)^2 dy[/tex]
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