The question is as follows
$f(x)=x^3$, $y=1$, $x=2$, About the x-axis
So I know that I am going to be using the disk method as their is only one function and there is no need to find $x$ in terms of $y$ in order to use the shell method.
Setting up the integral is a bit tricky because we have two boundaries that are on different axes, but I believe it should look something like this. $$\pi \int_0^1\left(2-x^3\right)^2-\left(1-x^3 \right)^2dx$$
I have solved the integral and gotten $\frac{5\pi}{2}$ but this incorrect. I know that the first squared term is correct but I am having difficulty finding how to arrive at the correct second squared term.
Here is the region in question (it's always good with a drawing when things become confusing):
So, in order to find this volume, we use what some call the disc method (where you use one disc for the entire solid of revolution, and then use the same method to calculate the volume that us cut away in the center), and some call the washer method (as the cross sections of the solid of revolution for any given $x$ looks like a washer): $$ \int_1^2 \left(\pi(x^3)^2 - \pi\cdot 1^2\right)\,dx = \frac{120\pi}7 $$
We can, of course, instead express $y$ in terms of $x$ and use what some call the shell method (as each value of $y$ here gives a cylindrical shell centered around the $x$-axis): $$ \int_1^{8} 2\pi y(2 - \sqrt[3]y)\,dy $$ which gives the same result.