MA391 Shells Trey Jacks

Throughout one's mathematical journey, we come across this constant use of approximations. Approximations help us apply concepts in real-world activities since it becomes very difficult to show off some expressions (infinite decimal approximations for example). While you are in Calculus One, we are introduced to this concept of integrating a function. Integrating a function, let's say f(x)), is a method of finding the area underneath that curve on a graph. This application of area is very useful, especially when looking at graphs like: Distance vs. Time, Velocity vs. Time, and Acceleration vs. Time.

The next natural thought after coming up with the area is “How can we extend this 2-D method into a 3-D world?”. Instead of looking at objects with a 2-D lens, once we get into Calculus 2 we can view things as 3-D. This leads up to going from finding the area of an object to finding the volume instead. How do we do this? We want to look at a function and rotate it along an axis, capturing the image of the function as it rotates. Below is a graph from Geogebra that shows the rotation done with a triangle, where the grey, flat, circle shows the pathing of the rotation.

Next we want to get a formula for this volume. We derive this from the formula of a circle, where we then have different cases. One case is when we want to integrate the function by slicing it into different washers built upon itself. Another case is using a bunch of cylindrical shells and adding them all up next to each other. Below we will write out the exact formulas we use in each case:

If you have the height of the function and the radius, it may we best to just by the Shell Method:

Shell Method:
$\int_{a}^{b}2\pi*r*y(x)dx$, where y(x) is the height of the shell and r is the radius

Next is the Washer method, which is very similar to above but is more useful when we have a circle around another, having two different radii’s.

Washer Method:
$\int_{c}^{d}\pi*x^2dx$, where $x^2 = R_1^2 - R_2^2$ where $R$'s represent the outter and inner radius respectfully.

Let's look at an example. For the object we chose, we are using the function:

$y(x) = \frac{x}{\sqrt{x^3+1}}$ on the interval [2, 12], rotated about the y-axis. For this, we will showcase the shell method! First let us identify what the radius is. As we discussed above, we want to make the radius x, so that we can manipulate what the radius is at a given x. Next, we just have to notice that the y(x) is the one given. Now, let us see the calculation!

Math Time:
$V = \int_{2}^{12}2\pi*x*\frac{x}{\sqrt{x^3+1}}dx$

Now that we have the setup, we need to just integrate! We will use a u-substitution here! Since we have this higher power x in our square root, it might be nice to try it here. So for our problem, let: $u=x^3+1, du = 3x^2 dx$. After making this substitution, we need to change our bounds to show this! We need to solve: $u=2^3+1, u=12^3+1$ => u=9, 1729. Thus, we now have an integral like:

$V =\int_{9}^{1729}2\pi*x*\frac{1}{3\sqrt{u}}du$
This then evaluates out to:
$V = \frac{4\pi}{3}({\sqrt{1729}-\sqrt{9}})$ which is approximatly 161.61 units cubed.

Now that we have seen the exact evaluation, let us look at what the approximation might look like! Below is an image generated on OnShape where we look at the adding up of shells for the curve!

The top view is to show out the hollow center since we are not starting when x=0, we start when x=2. To find the volume of our approximation, we just add up the value of the cyclindrals shells we have created! Below is the layout of the summation!

\[V = \sum_{n=2}^{12} 2\pi*\frac{n}{\sqrt{n^3+1}} \approx 168.15\] units cubed

As shown, this approximation is slightly off from the actual volume, but this is primarily due to the fact we only used a small number of shells. As you use more shells and do the summation, the approximation becomes more accurate, leading you to get to the actual volume from before!

The reason for this function is because of the shape of the graph on the 2-D plane. It reminds me of a carnival ride of going around a carousel, with each layer getting faster. Besides this, the actual approximation shows on the outside very clearly what we are trying to accomplish. One can see that we have these steps along the side, making our approximation shown. These reasons helped me come to a conclusion that this function would be very useful to showcase an approximation with the Shell Method since you can explicitly see the steps we are taking with the differences in heights of each shell.