Q.Find, by integration, the volume of the container which is in the shape of a right circular conical frustum of height $2\text{ m}$, whose two circular ends have radii $1\text{ m}$ and $2\text{ m}$ (as shown in Fig 9.46).

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 $y=c,\ y=d$: $\displaystyle V=\pi\int_c^d x^2\,dy$. …