Calculating Volume of an Arbitrary Pyramid

In summary: If a shape is not flat, it can be approximated by a set of shapes that are flat except for a small part of their sides (where they join up with each other). If you make the sides of those shapes straighter and straighter, the approximation gets better and better. If you make them straight enough, you can get the exact answer. And that's the idea behind integration.
  • #1
Guest432
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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) to 0 at the apex (assumed to lie at a height [PLAIN]http://mathworld.wolfram.com/images/equations/Pyramid/Inline6.gif).
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?
 
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  • #2
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.
 
  • #3
Jonathan Scott said:
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.

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
 
  • #4
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:

$$ A(z) = A_b \frac{(h-z)^2}{h^2} $$
 
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  • #5
Jonathan Scott said:
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:

$$ A(z) = A_b \frac{(h-z)^2}{h^2} $$
Jonathan Scott said:
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:

$$ A(z) = A_b \frac{(h-z)^2}{h^2} $$

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).
 
  • #6
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).
 
  • #7
Jonathan Scott said:
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).
I see, is there some mathematical law that defines this, or is this just intuition?
 
  • #8
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.
 

What is the formula for calculating the volume of an arbitrary pyramid?

The formula for calculating the volume of an arbitrary pyramid is V = (1/3) * B * h, where B is the area of the base and h is the height of the pyramid.

How do I find the area of the base of an arbitrary pyramid?

To find the area of the base of an arbitrary pyramid, you need to know the shape of the base. For example, if the base is a square, you can find the area by multiplying the length of one side by itself.

Do I need the height of the pyramid to calculate its volume?

Yes, the height of the pyramid is necessary to calculate its volume. Without the height, you will not have all the required information to use the volume formula.

Can I use any unit of measurement for calculating the volume of an arbitrary pyramid?

Yes, you can use any unit of measurement for calculating the volume of an arbitrary pyramid as long as you are consistent with the units used for the base and height. For example, if the base is measured in square meters, the height should also be in meters.

Can I use the formula for calculating the volume of a regular pyramid for an arbitrary pyramid?

No, the formula for calculating the volume of a regular pyramid (V = (1/3) * B * h) can only be used for pyramids with a regular base, such as a square or triangle. For an arbitrary pyramid, you will need to use a different formula that takes into account the shape of the base.

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