Introduction

Welcome to the FiveHive article for Unit 8.10 of AP Calculus! 

In this article, we will be exploring yet another major section of the applications of integral calculus. Namely, we will continue exploring the disc method for finding the volume of solids of revolution, but when revolved around axes other than the $x$ or $y$ axes.

As usual, we will only cover the topics included in the CED for unit 8.10.

Define the Region

Similar to when revolving around the normal coordinate axes, the first step is to define the region that is going to be revolved. 

This is going to be a little bit more complicated than before, as now you have to take into account a second function as a boundary. Once again, drawing out the functions on a coordinate plane really helps with visualization. 

For example, let us consider the functions depicted in the following image. If we are considering the volume of the solid made from revolving $x^2$ around $2x$, we define the region that is revolved as the area that is enclosed by the two functions (shaded in green). Once again, we would consider the shape of the cross-section to decide what type of integral to evaluate. 

Define the Radius

The radius of the solid of revolution is simply defined as the function on top (or right) minus the function on the bottom (or left). It is important to pay attention to this detail, as for integration with respect to $x$, the radius is defined as top - bottom, and for integration with respect to $y$, the radius is defined as right - left. 

Writing and Solving the Volume Integral

By this point, we have everything that we need to set up an integral to solve for the volume. As discussed before, the volume of a solid of revolution is simply going to be

${\int }_a^{b}A(x)dx$ 

Or

${\int }_a^{b}A(y)dy$ 

The bounds of the integral are dependent on the variable we are integrating with respect to, but we will get deeper into the details later. 

Let us consider another example. The graph shows the functions $-x^2+4$ and $y=1$. We want to find the volume of a solid that is created when revolving the region around $y=1$. 

Since the shape of the cross-section is a circle, we will integrate the area of a circle. To find the radius of the circle, we will subtract the function on the bottom from the function on top. That is:

$r(x)=(-x^2+4)-1$.

Therefore, the area of the circle created is $A(x)=\pi \mathring{(}x{)}^2$.

To find the bounds, we set the two equations equal to each other and solve. Through this process, we are finding the points of intersection to establish the bounds for integration:

$-x^2+4=1$ (1)

$-x^2+3=0$ (2)

Using the quadratic equation, we find that the two points at which they are equal are $(-1.732,1)$ and $(1.732,1)$.

Now we set up the integral. Since the cross-sections are perpendicular to the $x-$axis, we integrate with respect to $x$. 

$V(x)={\int }_{-1.732}^{1.732}\pi r(x{)}^{2}dx$

$\therefore V(x)={\int }_{-1.732}^{1.732}\pi [-x^2+3{]}^{2}dx$

Once we integrate, we find that the $V(x)\approx 52.237$. 

Now let us consider a situation where we have to differentiate with respect to $y$. In this case, we want to find the volume of the solid that is created when the shaded region between $\sqrt[3]{x}$, $y=0$, and $x=8$ is rotated around $x=8$. 

NOTE: If integrating with respect to $y$, it is important to remember to change the functions into $f(y)$ form and to change the limits to $y$ values rather than $x$ values. 

The radius of the solid created is the function on the right minus the function on the left. That is:

$r(y)=8-y^3$.

$\therefore A(y)=\pi r(y{)}^2$

In this case, the bounds are easy to find, as we are integrating from $y=0$ and $y=\sqrt[3]{8}=2$.

That said, we can now integrate. Since the cross-sections are perpendicular to the $y-$axis, we integrate with respect to $y$. 

$V(y)={\int }_0^2\pi [8-y^3{]}^{2}dy$

Once we integrate, we find that $V(y)\approx 258.508$.

Strategy

If this seems confusing, the strategy to approach these problems is extremely similar to the strategy for approaching problems that revolve around the $x$- or $y$- axes. The only difference is that you have to think about the order of subtraction for the radius expression. 

Practice

That was a great unit! Now, it is time to practice all the skills gathered from this article.