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Asymptotic Functions vs. Finite Functions

The equations I chose for my example are \(e^{-x}\) and \(\ln x\). I used the shell method from 0 to 1 with ten divisions. I used the right hand rule for the approximation because if I used the left hand rule, the innermost shell would be infinitely long. Each shell is .1 inches thick and its height above the outermost shell is its radius plugged into \(e^{-x}\). Its depth below the outermost shell is its radius plugged into \(\ln x\).
 The reason I chose these equations is to show how big of a difference an asymptote makes. The upper bounds of the graph cross the y-axis at y=1. The lower bounds never cross the y-axis. Looking at a graph of a function with an asymptote, you see the function get very close to the asymptote and is nearly touching it. The difference between the asymptote and the function gets smaller and smaller, but never reaches zero. My model shows how big of a difference crossing the y-axis is to almost crossing it. The height of each shell gets slightly bigger going inward, but if you were to see the actual height of the function at 0, it would not be too big of a difference from the height at .1. The depth of my model, however, gets much bigger with each shell going inward. In fact, the actual depth of the function at 0 is negative infinity. My functions, \(e^{-x}\) and \(\ln x\), make the same lines, just mirrored and rotated a bit. Even though they make the same lines, their behavior near the y-axis is very different.

 The volume of my model can be shown by \[π * (\sum_{n=2}^{10} ((.1n)^2 * (e^{-.1n} - ln(.1n)) – (.1n-.1)^2 * (e^{-.1n} - ln(.1n))) + .1^2 * (e^{-.1} – ln(.1)))\] Pi is multiplied to the entire thing at the end because of the distribution rule. Inside the summation, the volume of each shell is computed by taking the volume of the cylinder the shell makes and subtracting from it the volume of the cylinder with radius .1 smaller than it. The summation adds the volume of each shell together and then the volume of the innermost cylinder with radius .1 is added to that. After multiplying the whole thing by pi, you get 2.876.
 The actual volume of the equations rotated around the y axis is calculated very differently. I personally always mirror equations so that they are rotated around the x-axis instead of the y-axis because I find that easier to solve. The new equations are \(e^x\) and \(-\ln x\). They are so similar to the originals because, like I said, they are the same line just rotated and mirrored. Three integrals are needed. \[\int_{-∞}^0 (e^x)^2dx\] \[\int_0^.36788 1dx\] \[\int_.36788^1 \ln(x)^2dx\] The original model was only from 0 to 1, so there is a maximum height of 1 for the integrated functions. The value of 1 at each original equation is 0 and .36788. The first integral happens to equal exactly .5, the second integral equals .36788, and the third integral equals .1606. Adding each of these and multiplying by π gives you the volume of the functions rotated about the x-axis, 3.231. Even though the first integral starts at negative infinity, it does have a definite area. Considering the actual function goes on until infinity, a difference in volume of .355 is not that big for a model made using the right hand rule.

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