The equation I chose was \(\log_{10}x\) whose cross section is squares with side length x. My model is eleven horizontal square layers from y=0 to y=1.1. Each layer's side length is its smaller x value on the function. 
The reason I chose \(\log_{10}x\) is to demonstrate the square cube law. The bottom layer's side length is \(\frac{1}{10}\) the side length of the top layer, but the top layer has 100 times the area of the bottom layer. This is because of the square-cube law. If a shape's nth dimension grows by a factor, k, to make a similar object, its mth dimension grows by a factor of \(\sqrt[n]{k^{m}}\). My model's first layer increased by a factor of 10 in the 1st dimension. Area is used in the second dimension, so the increase in area is \(\sqrt[1]{10^{2}}\) or 100. The model's layers have constant height, so its volume increases by the same factor as its area. If its height was proportional to its side length, though, the volume would increase by a factor of \(\sqrt[1]{10^{3}}\) or 1000. This works for any two dimensions with either dimension being bigger or smaller. It is much easier to demonstrate it with the first three dimensions, though, because those are the dimensions we can easily picture in our heads.
If you know about integrals, you know that integrals essentially add up the infinetesimally small areas of lines under a function to find the total area under the function. You know the height of the line is the y value on the function. The volume of a solid with known cross-sectional area can be calculated similarly. My solid has a cross-sectional area of a square whose side length is the x value of the function. My solid can be divided horizontally an infinite number of times until each slice has an infinitesimally small height. If the length and depth of each slice can be figured out, the volume of each slice can be added up to find the total volume. And in my solid, the cross-sectional area of each slice is its x value squared, so its length and depth is each slice's x value. So, the volume of the solid can be figured out by squaring each x value and multipling it by the height, dy, across the integral.
To calculate the volume of the real solid, I used the logarithm's inverse, \(10^x\) so that the integral can be taken using dx instead of dy. The integral is taken from 0 to 1.1. \[\int_0^{1.1}(10^x)^2dx\] This integral turns out to equal 34.198.
The equation to calculate the volume of the model is \(\sum_{n=0}^{10} ((10^{.1*n})^2*.1)\) which turns out to be 26.926. The equation takes each layer's x value, squares it, and multiplies it by its height, .1, to get the volume of each layer and then adds them all together. It makes sense that the approximation is less than the real volume because the approximation uses the lesser of the two values of each layer.
The reason I chose \(\log_{10}x\) is to demonstrate the square cube law. The bottom layer's side length is \(\frac{1}{10}\) the side length of the top layer, but the top layer has 100 times the area of the bottom layer. This is because of the square-cube law. If a shape's nth dimension grows by a factor, k, to make a similar object, its mth dimension grows by a factor of \(\sqrt[n]{k^{m}}\). My model's first layer increased by a factor of 10 in the 1st dimension. Area is used in the second dimension, so the increase in area is \(\sqrt[1]{10^{2}}\) or 100. The model's layers have constant height, so its volume increases by the same factor as its area. If its height was proportional to its side length, though, the volume would increase by a factor of \(\sqrt[1]{10^{3}}\) or 1000. This works for any two dimensions with either dimension being bigger or smaller. It is much easier to demonstrate it with the first three dimensions, though, because those are the dimensions we can easily picture in our heads.
If you know about integrals, you know that integrals essentially add up the infinetesimally small areas of lines under a function to find the total area under the function. You know the height of the line is the y value on the function. The volume of a solid with known cross-sectional area can be calculated similarly. My solid has a cross-sectional area of a square whose side length is the x value of the function. My solid can be divided horizontally an infinite number of times until each slice has an infinitesimally small height. If the length and depth of each slice can be figured out, the volume of each slice can be added up to find the total volume. And in my solid, the cross-sectional area of each slice is its x value squared, so its length and depth is each slice's x value. So, the volume of the solid can be figured out by squaring each x value and multipling it by the height, dy, across the integral.
To calculate the volume of the real solid, I used the logarithm's inverse, \(10^x\) so that the integral can be taken using dx instead of dy. The integral is taken from 0 to 1.1. \[\int_0^{1.1}(10^x)^2dx\] This integral turns out to equal 34.198.
The equation to calculate the volume of the model is \(\sum_{n=0}^{10} ((10^{.1*n})^2*.1)\) which turns out to be 26.926. The equation takes each layer's x value, squares it, and multiplies it by its height, .1, to get the volume of each layer and then adds them all together. It makes sense that the approximation is less than the real volume because the approximation uses the lesser of the two values of each layer.
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