This week, we are going over solids of known Cross-Sections! This is another method of integration, going along with the theme of last week. The solids of known Cross-Sections is very similar to the process of integration by Shells and Washers, since these are merely cases of solids of known Cross-Sections.
In order to find the volume of a known Cross-Sections, we merely take the area of that shape on the 2-D plane, and then extend it to the 3-D plane by multiplying the area by the height, where the height represents how much you extend into the 3-D plane. Below is a simulation of what the cross-section actually looks like, whose creator is Michael Andrejkovics (a link to their Geogebra will be below!).
Now that we have the foundation down, let us build! The function I chose to showcase solids of known Cross-Sections is not necessarily difficult, but a twist on a plain example of using. The function is: $f(x) = \sin(x)*\pi*e$. I originally wanted to use a more complex function, but the math behind the volume and the integration seemed too difficult for a post wanting to show off this technique for a first time viewer. Next was to decide which shape to use. As seen in the post above, they used triangles in order to find the solid of known Cross-Sections. This did not seem interesting enough for me to dive into. I then decided on my favorite shape, a pentagon.
A pentagon has always been my favorite shape ever since elementary school. Along with this, I remember in my MA341, Topics In Geometry course having to prove the construction of a regular pentagon inscribed in a circle. This task was difficult for me at the time and I kept getting wrapped up into smaller details that did not matter. Coming back to this helps me beat this fear I once had.
$\it{Math \ Time!}$
The integration of solids of known Cross-Sections is done by the following formula, which should look very familiar to last weeks posts.
$\int_{a}^{b} A dx$, where A is the area of the object being integrated.
Here is where the fun of pentagons comes in! The area for a regular pentagon is given by the following: $A = .25 * \sqrt{(5(5+2\sqrt(5))}) * s^2$, where s is the side length of the pentagon.
For our function, I decided to make the height of the function the y-value, so the evaluation of $\pi * e * \sin(x)$.
Using some good ole geometry, we can find the side length of a regular pentagon with this height by solving: $\tan(36^{\circ}) = \frac{h}{.5s}$
After solving for the side, we have that the area of the Cross-Section is: $A = .25 * \sqrt{(5(5+2\sqrt(5))}) * (\frac{(\pi * e * \sin(x))}{2*\sqrt{5-2\sqrt{5}}})^2$. Fun, right? Yes! Finally, going back to our integral, our only task is to evaluate the integral of $\sin(x)^2$.
$V = \frac{.25 * \sqrt{(5(5+2\sqrt(5))})* e^2 * \pi^2}{4(5-2\sqrt{5})} \int_{0}^{\pi} \sin(x)^2 dx$
This allows us to arrive at the volume is: $V \approx 93.3417 \ units^3$.
To approximate this volume, I used the x-axis to be the width of the pentagonal prism, letting the width be $\frac{\pi}{12}$. Below is what the actual 3-D model is!
We want to finish this off by seeing how close our approximation is! The approximation can be found with the given sum, which I also build in the google sheet below!
$V_{approx}$ = \[ \sum_{n = 1}^{11} \frac{\sqrt{5(5+2\sqrt{5}})\pi^3 * e^2 * \sin(\frac{n}{12})^2}{12*16*(5-2\sqrt{5})} \] $\Rightarrow \ V_{approx} \approx 83.22804633 \ units^3$
As shown, there is some difference between our approximation and the actual integration, but this is due to use not taking smaller intervals. If we had the $x$ values be less than the $\frac{\pi}{12}$ we chose and used a summation of all of those prisms, we would have a value closer to the actual integration. For our intents and purposes, this is sufficient to show that we can approximate this value with the method used.
Here is the link to Michael Andrejkovics GeoGebra, which has many different examples of integration! https://www.geogebra.org/u/mr.+andrejkovics
Now that we have the foundation down, let us build! The function I chose to showcase solids of known Cross-Sections is not necessarily difficult, but a twist on a plain example of using. The function is: $f(x) = \sin(x)*\pi*e$. I originally wanted to use a more complex function, but the math behind the volume and the integration seemed too difficult for a post wanting to show off this technique for a first time viewer. Next was to decide which shape to use. As seen in the post above, they used triangles in order to find the solid of known Cross-Sections. This did not seem interesting enough for me to dive into. I then decided on my favorite shape, a pentagon.
A pentagon has always been my favorite shape ever since elementary school. Along with this, I remember in my MA341, Topics In Geometry course having to prove the construction of a regular pentagon inscribed in a circle. This task was difficult for me at the time and I kept getting wrapped up into smaller details that did not matter. Coming back to this helps me beat this fear I once had.
$\it{Math \ Time!}$
The integration of solids of known Cross-Sections is done by the following formula, which should look very familiar to last weeks posts.
$\int_{a}^{b} A dx$, where A is the area of the object being integrated.
Here is where the fun of pentagons comes in! The area for a regular pentagon is given by the following: $A = .25 * \sqrt{(5(5+2\sqrt(5))}) * s^2$, where s is the side length of the pentagon.
For our function, I decided to make the height of the function the y-value, so the evaluation of $\pi * e * \sin(x)$.
Using some good ole geometry, we can find the side length of a regular pentagon with this height by solving: $\tan(36^{\circ}) = \frac{h}{.5s}$
After solving for the side, we have that the area of the Cross-Section is: $A = .25 * \sqrt{(5(5+2\sqrt(5))}) * (\frac{(\pi * e * \sin(x))}{2*\sqrt{5-2\sqrt{5}}})^2$. Fun, right? Yes! Finally, going back to our integral, our only task is to evaluate the integral of $\sin(x)^2$.
$V = \frac{.25 * \sqrt{(5(5+2\sqrt(5))})* e^2 * \pi^2}{4(5-2\sqrt{5})} \int_{0}^{\pi} \sin(x)^2 dx$
This allows us to arrive at the volume is: $V \approx 93.3417 \ units^3$.
To approximate this volume, I used the x-axis to be the width of the pentagonal prism, letting the width be $\frac{\pi}{12}$. Below is what the actual 3-D model is!
We want to finish this off by seeing how close our approximation is! The approximation can be found with the given sum, which I also build in the google sheet below!
$V_{approx}$ = \[ \sum_{n = 1}^{11} \frac{\sqrt{5(5+2\sqrt{5}})\pi^3 * e^2 * \sin(\frac{n}{12})^2}{12*16*(5-2\sqrt{5})} \] $\Rightarrow \ V_{approx} \approx 83.22804633 \ units^3$
As shown, there is some difference between our approximation and the actual integration, but this is due to use not taking smaller intervals. If we had the $x$ values be less than the $\frac{\pi}{12}$ we chose and used a summation of all of those prisms, we would have a value closer to the actual integration. For our intents and purposes, this is sufficient to show that we can approximate this value with the method used.
Here is the link to Michael Andrejkovics GeoGebra, which has many different examples of integration! https://www.geogebra.org/u/mr.+andrejkovics
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