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Integrating to find volume of an unusual shape

  1. Apr 28, 2009 #1
    • Member warned about not using the template
    is there an exact way to calculate the volume of a shape where one shape is revolved around a parabolic curve (obviously with limits).

    i dont know much about this stuff but im pretty sure u need to integrate it.
    the shape revolved is a rectangle (vertical) with a triangle connected at the vertices

    for a birds eye view there are 3 (given) parabolic curves. the area between the top and middle curves represent the area on the top of the shape. the area between the middle and bottom represent the area on the bottom of the shape.

    i have to find a way to calculate it.
    i have no idea.
    i tried integrating the curves to find the areas for the top and base and averaging them and multiplying by the height. (basically trapeoid rule sort of)
    but this is only an approximation!!!!!
    CAN ANYONE GIVE ME AN ACCURATE WAY OR A BETTER APPROXIMATION!!!!

    help!!!!
    need help urgently
     
  2. jcsd
  3. Dec 9, 2014 #2

    OldEngr63

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    I think that the answer is "yes, there is a way," but I cannot understand your exact situation. Please post a figure so that we can work the correct problem.
     
  4. Dec 9, 2014 #3

    haruspex

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    Not sure what you mean by revolving an area around a curve. One normally revolves entities around axes. Do you mean it is translated along a curve?
    What is the relationship between the parabolas? E.g. do they all intersect, in symmetric fashion, at the same two points, or are they identical and 'parallel'?
    My guess is that the parabolas are something like y = a x2+ci, i = 1, 2 , 3, and intersections of the solid with planes orthogonal to the x axis produce a constant shape.
    Oh, and I don't understand
    Do you mean it's connected to two adjacent vertices, forming a pentagon?
     
  5. Dec 9, 2014 #4

    OldEngr63

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    @ haruspex: That is why I asked for a picture.
     
  6. Dec 9, 2014 #5
    Hey guys, this thread was created in 2009 so brandy is most likely not going to reply to anything here.
     
  7. Dec 9, 2014 #6

    haruspex

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    How true, brandy will have surely evaporated.
     
  8. Dec 9, 2014 #7

    Ray Vickson

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    Just as a matter of interest: any guesses as to why did thread suddenly re-appeared?
     
  9. Dec 9, 2014 #8
    My best guess is that maybe OldEngr was browsing "Unanswered Threads" and this one somehow snuck its way in.
     
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