Last night I had the idea to validate the formula for volume of a cone using line integrals. The idea is to consider traversing the x-axis from the origin to some point $(h,0)$, where $h$ is the height of the cone. Then, we would apply some density function $\delta$ such that the density of some point $(\alpha,0)$ is equal to the area of the cross-sectional circle at that point.
Perhaps this would make more sense written out. Consider the general form of a line integral:
$$\int_C\delta(x,y)\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\;dt$$
Now consider the following curve to integrate along. (Note it is simply the axis of revolution)
$$C:\;\vec r(t)=\langle t,0\rangle$$
Note the following range of $t$ corresponding to the height $h$ of the cone:
$$t\in[0,h]$$
Now note the following density function that gives the area of the circular cross section at some point $(x,y)$:
$$\delta(x,y)=\frac{\pi r^2}{h^2}x^2$$
Substituting our values:
$$V(r,h)=\int_0^h\frac{\pi r^2}{h^2}t^2\sqrt{1^2+0^2}\;dt$$
$$V(r,h)=\frac{\pi r^2}{h^2}\cdot\left[\frac{t^3}{3}\right]_0^h$$
$$V(r,h)=\frac{1}{3}\pi r^2h$$
Is the disc method just a simplification of line integrals in the same way that rectification integrals are?
Notice that your calculation is independent of $y$. All single-variable integrals can be considered as line integrals over the $x$-axis, but in the same way that any interval in $\mathbb{R}$ can be considered as a 'curve' in $\mathbb{R}^2$ or indeed any $\mathbb{R}^n$. So your 'line integral' is exactly a single-variable integral, and it would be unfair to say that the disk method is derived from line integrals. Sure, if you had a curve that didn't lie on the $x$-axis, a line integral would prove more useful, but in this case it wasn't required. Line integrals are useful for integrating over curves that aren't intervals, but in the case of an interval, it is equivalent to a single-variable integral.