What am I doing wrong in finding the volume of a square based pyramid

[Psinter]

1. I need to find the volume of a square based pyramid which one side of its base is 756ft. Its height is 481ft and the material density is 150lb/ft^3It has to be using calculus.



2. Homework Equations : I know that the formula would be 1/3*(b^2*h), but I need to use calculus to reach that formula. Basically a proof.



3. The Attempt at a Solution : The book gives me an example which basically answers me the whole exercise and the internet too, but I don't really get it. Look at this page and its explanation on example 2: http://tutorial.math.lamar.edu/Classes/CalcI/MoreVolume.aspx What I tried to do, because it seems logical, is sum the area of the squares as they approach infinite from 0 to its height (481ft).

I don't really get it, shouldn't I just divide the pyramid in infinite squares and sum their areas to obtain the pyramid volume? Also I don't get the 2D drawing; is the length of s 2 times one side of the square?

 

[eumyang]

The book gives me an example which basically answers me the whole exercise and the internet too, but I don't really get it. Look at this page and its explanation on example 2: http://tutorial.math.lamar.edu/Classes/CalcI/MoreVolume.aspx What I tried to do, because it seems logical, is sum the area of the squares as they approach infinite from 0 to its height (481ft).

I don't really get it, shouldn't I just divide the pyramid in infinite squares and sum their areas to obtain the pyramid volume? Also I don't get the 2D drawing; is the length of s 2 times one side of the square?

No, s is just the length of the side of the square, not two times it. If you can show is how you set up the integral, we can tell you what you're doing wrong.

 

[Psinter]

Oh, never mind, I found it. The integral for the volume would be: integral[(756/481)(481-x)]^2 dx which would be the same as integral[(length/height)(height - x)]^2 dx

I'm checking right now how I concluded that because I have about 5 papers full of numbers and rubbish calculations. Hehe.

EDIT: Found the whole process. What happens is that as we go up to the top, the size of the square changes and so does its area. To calculate the size of the square at a specific point we do:

height/length = (height - xcoordinate)/newsquarelength

and we solve for newsquarelength:

newsquarelength = (length/height)(height - xcoordinate)

where:
length/height = ratio of the triangle; the length is the same as saying the base of the triangle.

height - x = location of the new square

Since the area of a square is its side elevated the power of 2, we just have to elevate newsquarelength to the power of 2 in our integral:

integral[(height/length)(height - xcoordinate)]^2

In my case, this would become:

integral[(756/481)(481 - x)]^2 dx From 0 to 481

And this would be the volume of the pyramid.

No, s is just the length of the side of the square, not two times it. If you can show is how you set up the integral, we can tell you what you're doing wrong.

Thanks for making that clear, I was trying to visualize how they drew that triangle and thought that they expanded into a single plane 2 triangles of the pyramid and that's why I thought that it was 2 times the side of the square, but now I know its not like that.