Q.Find, by integration, the volume of the solid generated by revolving about the $x$-axis, the region enclosed by $y=2x^2$, $y=0$ and $x=1$.

When a plane region is rotated one full turn ($360^\circ=2\pi$ radians) about a fixed axis lying in its own plane, it sweeps out a solid of revolution. The formulas below restrict to revolution about the $x$-axis or the $y$-axis, with the revolved region lying, respectively, above the $x$-axis ($y\ge0$) or to the right of the $y$-axis ($x\ge0$).

Derivation (disc method). Partition $[a,b]$ as for the Riemann integral. At each sample point $x_i$, the thin vertical strip of height $y_i=f(x_i)$ and width $\Delta x$ sweeps out, on revolution about the $x$-axis, an (approximately) cylindrical disc of radius $y_i$ and height $\Delta x$, hence volume $\pi y_i^2\,\Delta x$ (using "volume of a cylinder $=\pi r^2h$"). Summing over all strips and passing to the limit $n\to\infty,\ \Delta x\to0$ gives the volume of the whole solid.

Volume formulas.

• About the $x$-axis, for the region bounded by $y=f(x)$, the $x$-axis, and $x=a,\ x=b$: $\displaystyle V=\pi\int_a^b y^2\,dx$.
• About the $y$-axis, for the region bounded by $x=f(y)$, the $y$-axis, and $y=c,\ y=d$: $\displaystyle V=\pi\int_c^d x^2\,dy$.

Standard solids recovered from these formulas (all derivable by integration, not just quoted): a sphere of radius $a$ from revolving the semicircular region under $y=\sqrt{a^2-x^2}$ about the $x$-axis, $V=\tfrac43\pi a^3$; a right circular cone of base radius $r$, height $h$ from revolving the triangular region under $y=\tfrac{r}{h}x$, $V=\tfrac13\pi r^2h$; a spherical cap of height $h$ cut from a sphere of radius $r$, $V=\pi h^2\!\left(r-\tfrac{h}{3}\right)$; an ellipsoid from revolving the ellipse $\tfrac{x^2}{a^2}+\tfrac{y^2}{b^2}=1$ about its major axis, $V=\tfrac43\pi ab^2$ (about the $x$-axis) or $\tfrac43\pi a^2b$ (about the $y$-axis, i.e. the minor axis case if $a>b$).