Volume of a solid of revolution (disk method)

Partition $[a,b]$ into $n$ subintervals of width $\Delta x=(b-a)/n$ with sample points $x_i^*$. The solid restricted to $[x_{i-1},x_i]$ is approximately a cylindrical disk of radius $f(x_i^*)$ and thickness $\Delta x$, so its volume is approximately $\pi[f(x_i^*)]^2\,\Delta x$.

Summing these disk volumes over all $n$ subintervals gives the Riemann sum $\sum_{i=1}^n \pi[f(x_i^*)]^2\,\Delta x$, which approximates the true volume $V$.

As $n\to\infty$, continuity of $f$ on $[a,b]$ makes the approximation error vanish, and since $x\mapsto \pi[f(x)]^2$ is continuous, the Riemann sum converges by definition to $\pi\int_a^b[f(x)]^2\,dx$. Hence $V=\pi\int_a^b[f(x)]^2\,dx$.

As a consistency check, take $f(x)=\dfrac{r}{h}x$ on $[0,h]$, whose rotation about the $x$-axis is exactly a cone of base radius $r$ and height $h$: $V=\pi\int_0^h \left(\dfrac{r}{h}x\right)^2 dx = \pi\dfrac{r^2}{h^2}\cdot\dfrac{h^3}{3} = \dfrac{1}{3}\pi r^2 h$, exactly the classical cone volume formula.