Given a curve $y\=f\left(x\right)$ that is continuous on $x \= \left[a\, b\right]$, we can divide $\left[a\, b\right]$ into $n$ subintervals $\left[x_i \, x_{i\+1}\right]$, each of width $\mathrm{\Δ} x$.

Then, we can form a rectangle from the $x$-axis to the curve, of width $\mathrm{\Δ} x$ and height $f\left(x\right)$.

Rotating this rectangle about the $x$-axis, a vertical disk is formed with radius $f\left(x_i^{\ast }\right)$ and width $\mathrm{\Δ} x$, where $x_i^{\ast }\in \left[x_i \, x_{i\+1}\right]$.

The volume of this disk is given by $V_{\mathrm{disk}} \= \mathrm{\π}{\left(f\left(x_i^{\ast }\right)\right)}^2\mathrm{\Delta }x$.  The volume of the entire solid can be found by summing the volumes of all $n$ disks, while letting them become infinitely thin by having $n \to \infty$.

Therefore, the volume of the solid of revolution is:

$V_{\mathrm{solid}} \= \underset{n\to \infty }{lim}\sum _{i\=1}^n\mathrm{\pi }{\left(f\left(x_i^{\ast }\right)\right)}^2\mathrm{\Delta }x \= {\int }_a^b\mathrm{\pi }{\left(f\left(x\right)\right)}^2 \ⅆx$