Volumes of revolution - the disc method

When the region under $f(x)\ge 0$ is revolved about the $x$-axis, each “slice” of the solid is a disc of radius $f(x)$ and thickness $dx$: $dV = \pi\bigl[f(x)\bigr]^2\,dx \implies \boxed{V = \pi\int_a^b\bigl[f(x)\bigr]^2\,dx}.$

Revolution about the $x$-axis: each slice is a disc of radius $f(x)$ and thickness $dx$.

Example — Paraboloid about the $x$-axis

$V = \pi\int_0^4(\sqrt{x})^2\,dx = \pi\int_0^4 x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = 8\pi.$

Example — Solid of revolution of $y=xe^{-x}$

$V = \pi\int_1^5(xe^{-x})^2\,dx = \pi\int_1^5 x^2 e^{-2x}\,dx.$ One integrates by parts twice. First application: $f'=e^{-2x}\to f=-\tfrac{1}{2}e^{-2x}$; $g=x^2\to g'=2x$: $= \pi\left[-\frac{x^2}{2}e^{-2x}\right]_1^5 + \pi\int_1^5 x\,e^{-2x}\,dx.$ Second application on the remaining integral ($f'=e^{-2x}\to f=-\tfrac12 e^{-2x}$; $g=x\to g'=1$): $\int_1^5 x\,e^{-2x}\,dx = \left[-\frac{x}{2}e^{-2x}\right]_1^5 + \left[-\frac{1}{4}e^{-2x}\right]_1^5.$ Assembling and evaluating one obtains a small number (because $e^{-10}\approx 5\times 10^{-5}$). Note: choosing $f'=x$ would have led to something more complicated.

Example — Revolution of $\sin x$ about the $x$-axis

$V = \pi\int_0^\pi\sin^2 x\,dx = \pi\int_0^\pi\frac{1-\cos(2x)}{2}\,dx = \frac{\pi}{2}\left[x-\frac{\sin(2x)}{2}\right]_0^\pi = \frac{\pi^2}{2}.$ We use the linearisation formula $\sin^2 x = \dfrac{1-\cos(2x)}{2}$ seen in the chapter on trigonometry.

Links

Topics: Integrale
Concepts: Integrale definito · Metodo dei dischi · Volume di rotazione
Methods: Integrale parti
Skills: Calcolare · Integrare