Introduction

Welcome to the FiveHive article for Unit 8.12 of AP Calculus! 

In this article we will cover yet another major section of the applications of  integral calculus. Namely we will be covering calculating volume using the washer method revolving around other axes

As usual, we will only cover the topics included in the CED for unit 8.12.

Define the radii

Consider two functions $f(x)=x^2$ and $g(x)=\sqrt{x}$ and the intersection area is rotated around the horizontal line $y=2$. Find the area of the solid created through the rotation.

Step 1: Determine which axis is being integrated

If the region is being revolved around a straight horizontal line (i.e $y=2$), you will integrate with respect to the $x$ axis. If it is a straight vertical line (i.e $x=2$), you must first change the function to be in respect to $x$ (i.e $x=y^2$) and then integrate with respect to $y$. In this case, we are integrating with respect to $x$ axis.

Step 2: Determine the big and small “radius”

The big radius is the radius going from the axes to the bottom of the region. The small radius is the radius going from the axes to the top of the region. For example, let region T be defined as the region between $f(x)=x^2$ and $g(x)=\sqrt{x}$ as shown. The ‘big radius’ would be equal to $2-x^2$ because $x^2$ in red is at the bottom of the region. The small radius would be equal to $2-\sqrt{x}$ because $\sqrt{x}$ in blue is at the top of the region.

Step 3: Apply the washer method 

The final step is to apply the washer method formula which is  $\pi {\int }_a^b[(R(x){)}^2-(r(x){)}^2]dx$. For our example we get $\pi {\int }_a^b(2-x^2{)}^2-(2-\sqrt{x}{)}^{2}dx$ which evaluates to $\frac{31\pi }{30}$.

Practice Problems