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Very hard modelling and problem solving Calculus Question!

  1. Mar 7, 2013 #1
    A light-weight "pop-up" tent consists of six flexible plastic struts that are inserted into pockets sewn into the joins of the fabric panels. The resulting shape has hexagonal horizontal cross-sections, while vertical cross-sections through the centre are semi-circular.

    Derive a formula for the volume of the tent as a function of its height.

    How do i do it? I tried doing a symmetrical trapezium on the graph and revolved it around half way to get the tent look, not a sphere, but the top part (the roof of the tent) was in the shape of a trapezium (shaped like a diamond? like theres edges), not half a sphere like it should be, and the base being a hexagon. If you can imagine it? Anyways, need help!

    Attached is the question and the picture of the tent.

    Attached Files:

  2. jcsd
  3. Mar 7, 2013 #2


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    Can you use a volume integral?

    It is sufficient to consider 1/6 or 1/12 of the tent, of course.

    The result looks very nice.
    Last edited: Mar 7, 2013
  4. Mar 10, 2013 #3
    You can avoid explicit volume integration if you're clever (and get it right, which perhaps I haven't):

    Note that the base of a 1/6 of the tent is an equilateral triangle, sides of length [itex]l [/itex] (which is h at its maximum), and area[itex]\frac{\sqrt{3}}{4}l [/itex]. http://gyazo.com/6ce4bd547e7a5a47fc89a45ad586f79b Note that
    [tex]l^2+x^2=h^2 [/tex]
    [tex]l=\sqrt{h^2-x^2} [/tex]

    [tex] \int_{x=0}^{x=h}\frac{\sqrt{3}}{4}ldx=\frac{\sqrt{3}}{4}\int_{x=0}^{x=h}\sqrt{h^2-x^2}dx[/tex]
    [tex]x=h \cos(u)[/tex]
    [tex]dx=-h \sin(u)du[/tex]
    [tex]\frac{\sqrt{3}}{4}\int_{x=0}^{x=h}\sqrt{h^2-h^2 \cos^2(u)}(-h \sin(u))du=\frac{-h^2 \sqrt{3}}{4}\int_{x=0}^{x=h}(\sin^2(u) )du[/tex]
    [tex]x=h, u=0[/tex]
    [tex]x=0, u=\frac{\pi}{2}[/tex]
    [tex]\frac{h^2 \sqrt{3}}{4}\int_{0}^{\frac{\pi}{2}}(\sin^2(u) )du=\frac{h^2 \sqrt{3}}{4}\frac{\pi}{4}[/tex]

    Multiply by 6 for the whole thing.
    Last edited: Mar 10, 2013
  5. Mar 11, 2013 #4


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    Well, two parts of the volume integral are trivial, so you just have to calculate an easy 1-dimensional integral there as well.

    I think there is an ##l## missing in your formula of the triangle area.
    I used a different coordinate for the non-trivial integral.
  6. Mar 11, 2013 #5
    You're right- the integral is much simpler, just a polynomial.
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