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Homework Help: Integrating for volume of a hexagonal pyramid

  1. Jan 28, 2010 #1
    1. The problem statement, all variables and given/known data
    A) Find the volume of a truncated pyramid with a regular hexagonal base.
    perimeter of bottom = 1.2 km
    perimeter of top = 480 m
    height = 200m
    B) If density = 1500 kg/m^3 how much work was done in construction?

    2. Relevant equations

    3. The attempt at a solution
    I first split the hexagon into 6 equilateral triangles found the area to be about 102.9 m^2, and...well thats pretty much all i got. i cant figure out how to integrate it correctly to find a volume.
  2. jcsd
  3. Jan 28, 2010 #2


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    Welcome to PF!

    Hi Jack4761! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    Slice the pyramid horizontally into very thin slices of height dh, calculate the volume of each slice, then integrate over h from h = 0 to 200. :wink:
  4. Jan 28, 2010 #3


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    Remember that if you can figure out the area of a cross-section at height h, call it A(h), you can calculate the volume by

    [tex]V = \int_0^H A(h)\ dh[/tex]
  5. Feb 9, 2011 #4
    Please, leading questions, only.

    I have a problem similar to this, except that the overall shape is not a pyramid, it's a dome. Each cross-section is a hexagon, with the corners being 'r' from the center of the hexagon.

    I began by finding the area of the hexagon by splitting the base into 6 triangles, each of which has height, h = r (sqrt3)/2 and base, b = r/2

    Area then would be: [r2(sqrt3)]/8

    Now the above function [tex]V = \int_0^H A(h)\ dh[/tex] will not work for me, precisely, because y varies as a function of x in a circular fashion.

    x2 + y2 = r2

    I'd like to integrate with respect to the Y-axis, right, so I solve for x.

    x = sqrt(r2-y2)

    So, (and this is where I may be wrong?!?), I use the integral, from 0 to h (which is r), of the function sqrt(r2 - [(r2sqrt3)/8]2 dr
    Last edited: Feb 9, 2011
  6. Feb 9, 2011 #5


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    Check that. The base of an equilateral triangle is the same length as the other sides.

    I think you are confusing the radius of the hexagon's circumscribing circle with the radius of the dome. The radius of the hexagon's circumscribing circle depends on its height.
  7. Feb 9, 2011 #6
    Thanks for the help.

    I've refigured the area of a hexagon to be [3(sqrt3)r2]/2

    As for the confusion of the radius and height, they are equal, as this dome is composed of semi-circular supports. Think of it as a tent with three half circle poles. The height would be the radius as well.

    My problem is this, I take the integral from zero to h (height) of the above formula. which yields [r3(sqrt3)]/2... I'm looking to prove that r3(sqrt3) is the answer, so somehow I'm only taking into consideration the volume of half of the 'dome'...
  8. Feb 9, 2011 #7


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    Good, that's better.

    No, your problem is you haven't understood what I was trying to tell you. Suppose you pass a horizontal plane through the dome. It will intersect your tent in a hexagon. If the plane is near the top of the dome it will be a small hexagon and near the bottom it will be a large one. Think of the circle that circumscribes this cross section hexagon. Near the bottom that circle radius will be almost r, but near the top it will be near 0. This radius for the hexagon cross sections varies with the height. You need to get it in terms of that variable.

    I think part of your problem is with your poor choice of notation. I would suggest you might use a for the radius of the dome, r for the variable radius of the circumscribing circle of the hexagons, and h for the integration variable with h going from 0 to a. You need to get r in terms of h to proceed. Then the volume formula should work nicely.
  9. Feb 10, 2011 #8
    For a circle:
    the integral from A to B of pi(f(x)2) dx
    where f(x) is the rate of change of r
    (from area = pi(r)2)

    For a hexagon: (vertical integration)
    the integral from 0 to r of [3(sqrt3)/2][g(y)2]
    (from area = [3(sqrt3)/2](r2)
    Where g(y) is the rate of change of r, in this case, the equation of a circle, solved for x (vertical integration), g(y) = x = sqrt(r2-y2)

    This works out just fine, to prove the area is equal to r3sqrt3

    My problem was mainly this, instead of substituting the equation of a circle into the formula for the area, I was trying to substitute the area into the equation of a circle. Writing out the circle methodology (even though I could see it in my head) above, then the hexagon method below helped out because I could visualize better that in the circle's case, they inserted the equation for the rate of change as the radius. So when I did the same, inserting the equation for the rate of change of the radius (the equation of a circle) into the formula for the area of a hexagon, things went smoothly.

    Thanks for your time and help, LCKurtz, I'll try to work on my variable labeling.
    (I'll try to scan and post my working later for those who might care.)
    Last edited: Feb 10, 2011
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