*This specific case involves solids of revolution about the $x$-axis
I've just started reading on solids of revolution, so forgive me if this is trivial. I'm wondering if this is a valid derivation of the formula used in the disk method.
Supposing we have some function $f(x)$ which we desire to rotate about the $x$-axis between bounds $x_1$ and $x_2$, then per the mean value theorem of integrals, there exists some value for $f(x)$ which we can use to construct a single cylinder of identical volume to the solid of revolution.
Now let's look at the formula for a cylinder:
$$V_C=\pi r^2h$$
Clearly height $h$ is the distance between our lower and upper bounds, which we'll denote $x_1$ and $x_2$, respectively. Thus:
$$V_C=\pi r^2(x_2-x_1)$$
Let's rewrite $\pi r^2$ as base $b$.
$$V_C=b(x_2-x_1)$$
Finally, let's denote the average base area per the mean value theorem as $\bar{b}$. Note that $r^2$ at any point $x$ is given by $f^2(x)$, thus:
$$\bar{b}=\frac{\pi}{x_2-x_1}\int_{x_1}^{x_2}f^2(x)dx$$
Substituting $\bar{b}$ for $b$:
$$V_C=\frac{\pi}{x_2-x_1}\int_{x_1}^{x_2}f^2(x)dx\; *\;(x_2-x_1)$$
Cancelling $(x_2-x_1)$:
$$V_C=\pi\int_{x_1}^{x_2}f^2(x)dx$$
Which is the formula used for the disk method.
Is this a valid approach, and if so, can I improve my explanation in any way?