The Washer Method

Introduction
We have learned the process for calculating the area of functions using integration, but what about calculating volume? Incorporating the third dimension (z-axis) seems rather challenging but calculating volume of three-dimensional solids using integration is a rather interesting concept. This concept uses cross-sections and certain methods such as the one we will be looking at today.
Washer Method?
The washer method can be used with two different scenarios. The first scenario just being a single function of \(y=f(x)\). The second scenario being the two functions \(y=f(x)\) and \(y=g(x)\). In my example, I am using two functions to show the washer method. My first function is \(y=x^3\) and the second function is \(y=\sqrt[3]{x}\).

First, appreciate how pretty the functions look. It looks like a nice flower petal. Also, pay attention to the region that the functions are bounded by. The graph shows that the functions are bounded from 0 to 1. This interval, \(0≤x≤1\) is going to be bounds of our integration. Next, picture the area between the functions being rotated around the x-axis. This would give us the solid that we are trying to create and compute. This method, the washer method, gets its name from when the area is rotated around its axis of rotation, then the resulting solid of revolution has a disk with a hole in the middle, like a washer! The washer method is used to calculate the volume of a given solid by dividing the solid into washers, and then adding up the volumes of each washer.

Approximating and Computing with the Washer Method
As stated previously, the washer method is used by dividing the solid into however many washers and then adding each of the volumes up. In this case, the solid will be divided into 10 washers, and each washer will have a height of 0.1 inches, for our bounds of integration are from 0 to 1. By taking the formula for the volume of a cylinder \[Volume = (\pi(r_{1})^2h)-(\pi(r_{2})^2h)\] We can calculate the volume of our washers. However, this will only approximate our volume, thus integrating to find the volume is necessary for a more accurate answer. Below is the table for our approximation.

This approximation gave us a \(Volume = 1.411in^2\)

Next, to calculate the accurate volume of the solid, we need to integrate. Using the formula from earlier, \(Volume=(π(r_1 )^2 h)-(π(r_2 )^2 h)\), we can rearrange this to become: \[\pi\int_a^b (r_{1}-r_{2})^2dr\] In this equation, \(r_{1}\) is the function \(y=\sqrt[3]{x}\). and \(r_{2}\) is the function \(y=x^3\) which are integrated from the range \(0≤x≤1\) . Substituting the functions and the bounds in, we get the integral: \[\pi\int_0^1(\sqrt[3]{x})^2-(x^3)^2dx\] Evaluating the integral we then get: \[\pi\int_0^1(\sqrt[3]{x})^2-(x^3)^2dx \rightarrow\pi\int_0^1((x\sqrt[3]{x})^2-\frac{2}{5}(x)^\frac{5}{3})-(x^6)dx \rightarrow \pi(\frac{3}{5}x^\frac{5}{3})-\frac{1}{7}x^7\] \[(\pi(\frac{3}{5}(1)^\frac{5}{3}-\frac{1}{7}(1)^7)-(\pi(\frac{3}{5}(0)^\frac{5}{3}-\frac{1}{7}(0)^7) = \frac{16\pi}{35}\] \[=1.436in^2\] Why these Functions?
There are a few different reasons for picking the function that I did. First, picking a function was rather difficult for me. I am not sure if it was because I was overthinking the topic, or wanted something unique, but after going through my previous calculus exams I found these functions that I believe were friendly to someone just starting the subject. I also believe that the integration for the functions were easier to compute. This would then allow for someone who is just starting to learn this concept focus more on the concept than the integration. Finally, the functions make a really pretty design that looks like a flower petal. And if you look in quadrant three, the design is the same! I believe this example would allow someone to grasp the concept, while not having to worry about the calculus involved.