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Volume of an octagonal dome by using calculus

  1. Dec 16, 2015 #1
    On this picture we see a octagonal dome. I am trying to calculate the volume of this object by integral calculus but I can't find a way. How would you calculate this?
    https://dl.dropboxusercontent.com/u/17974596/Sk%C3%A6rmbillede%202015-12-17%20kl.%2002.14.48.png [Broken]
    I am majoring in math-econ but i will try to understand geometrical challenges a bit. Hence my lack of overview in calculus.
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Dec 16, 2015 #2
    Hi dane:

    I suggest integrating vertically, say the h dimension. That is, consider the dome to consist of a parallel hexagons of thickness dh. Calculate the area A(h) for a hexagon at height h, and integrate ∫0H A(h) dh, from zero, the base, to the peak height H.

    Hope this is helpful.

  4. Dec 16, 2015 #3
    thank you very much answer. I will try to do so. How sure are you of this approach?
  5. Dec 16, 2015 #4
    Hi dane:

    I am sure this will give the correct answer if you know enough about the geometry to calculate A(h). That is, you need to know the shape of the curve of the dome, that is, how the length of a side of a hexagon varies with h. You will also need to know how to calculate the area of a hexagon given the length of a side.

  6. Dec 16, 2015 #5
    I think this is a great start and I get google the things you mentioned above. Thank you.
  7. Dec 17, 2015 #6
    Let's divide the botum into 8 pieces.
    v=∫0r a(t=Solve(equaton for the cirkle)dy

    Here's what I did. I define the area of the octagon by it's apothem and call it r: a1( r) is the function. I divide the octagon in 8 pieces and each piece have the area (1/8)a1( r) which I call a( r)=(1/8)a1( r). Then I integrate from v=∫0r a(y)dy, where y is isolated from the y^2-x^2=r^2 is the circle. And I assume that a octagon with apothem r is unique?
  8. Dec 17, 2015 #7
    Hi dane:

    I am not sure I understand your description of what you are doing.

    I think you may have a typo and wrote "botum" rather than "bottom".

    You divide the bottom hexagon into to 8 parts. I assume you intend each piece to be a triangle with its height equaling the apothem. The area of the bottom will be 8 times the area of this triangle. The area of the triangle is 1/2 the apothem times the side of the octagon.

    You said, "where y is isolated from the y^2-x^2=r^2 is the circle." I do not know what x and y are or what the "circle" is.

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