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B Volume of a Pyramid

  1. Jul 12, 2016 #1
    I am trying to gain some understanding from this article regarding deriving the volume of an arbitrary pyramid.

    "An arbitrary pyramid has a single cross-sectional shape whose lengths scale linearly with height. Therefore, the area of a cross section scales quadratically with height, decreasing from Inline4.gif at the base ([PLAIN]http://mathworld.wolfram.com/images/equations/Pyramid/Inline5.gif) [Broken] to 0 at the apex (assumed to lie at a height [PLAIN]http://mathworld.wolfram.com/images/equations/Pyramid/Inline6.gif). [Broken] The area at a height http://mathworld.wolfram.com/images/equations/Pyramid/Inline7.gif above the base is therefore given by"

    NumberedEquation1.gif

    Can you describe this to me mathematically but easily?
     
    Last edited by a moderator: May 8, 2017
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  3. Jul 12, 2016 #2

    Jonathan Scott

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    The article seems quite clear, so unless you explain exactly which terms are giving a problem, it's difficult to answer.

    It might be easier if you think downwards from the apex. The width of the pyramid in any direction at a given level is then proportional to how far down you've gone, so the area at a given level is proportion to the square of how far down you've gone. The area at the bottom sets the constant of proportionality.
     
  4. Jul 12, 2016 #3
    Why does the area at any given level proportionate to the square of the height? Why does the (h-z)^2 get divided by the h^2
     
  5. Jul 12, 2016 #4

    Jonathan Scott

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    If something is twice as wide in both directions, it has four times the area. In general, for different sizes of the same shape, the area is proportional to the square of the size.

    If ##k## is the constant of proportionality we have the following at any given level ##z## which is a distance ##(h-z)## down from the apex:

    $$ A(z) = k \, (h-z)^2 $$

    At the base, the area is equal to the base area ##A_b## and ##z = 0##, so we have

    $$ A_b = k \, h^2 $$

    By simple algebra, that means ## k = A_b / h^2 ##, giving the general rule:

    $$ A(z) = A_b \frac{(h-z)^2}{h^2} $$
     
  6. Jul 13, 2016 #5
    By saying $$ A(z) = k \, (h-z)^2 $$, why do you have $$ (h-z)^2 $$, I would understand if it was a square based pyramid (area = s^2), but the article refers to any arbitrary pyramid (say, pentagonal base).
     
  7. Jul 13, 2016 #6

    Jonathan Scott

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    For any flat shape, if the size is increased by some factor in all directions, the total area increases by the square of that size. More generally, if the size is increased proportionally in any direction, lengths in that direction and the total area are both multiplied by the same amount, so if the size is increased in two different directions by the same amount (as occurs in this case) lengths are multiplied by that amount and then by the same amount again, giving the square of the original amount.

    (If you find it helps your understanding, imagine that the shape is full of tiny squares, each of which will scale up in the same way).
     
  8. Jul 13, 2016 #7
    I see, is there some mathematical law that defines this, or is this just intuition?
     
  9. Jul 13, 2016 #8

    Jonathan Scott

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    It's very basic standard mathematics, just involving the concept of multiplication and the definition of area. The area of a rectangle is given by its length times its width, so if the length is multiplied by some constant and the width by another constant then the area is multiplied by the product of the two constants. If the two constants are the same, the area is multiplied by the square of the constant. The area of any flat shape can be approximated as closely as you like by a set of rectangles, so the same rule also applies to the shape as a whole.

    Edit: The bit about "flat shape" is true in any case you'll encounter normally anyway, but just in case any mathematical nerds are watching I'll mention that mathematicians have some weird stuff such as "space-filling curves" that don't follow the conventional rules.
     
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