Volumes of revolution - the cylindrical shell method

When the region under $f(x)\ge 0$ is revolved about the $y$-axis, the disc method would require inverting $f$. The cylindrical shell method avoids the inversion: one “unrolls” each little cylinder into a rectangle of width $2\pi x$, height $f(x)$ and thickness $dx$: $dV = 2\pi x\cdot f(x)\,dx \implies \boxed{V = 2\pi\int_a^b x\cdot f(x)\,dx}.$

Unrolling: the cylindrical shell (on the left) becomes a rectangle (on the right) with base $2\pi x$ and height $f(x)$.

Example — Volume of the cone with shells

Line $y = h(1-x/r)$ revolved about the $y$-axis: $V = 2\pi\int_0^r x\cdot h(1-x/r)\,dx = 2\pi h\left[\frac{x^2}{2}-\frac{x^3}{3r}\right]_0^r = \frac{1}{3}\pi r^2 h.$

Example — Paraboloid about the $y$-axis (shells)

$V = 2\pi\int_0^4 x\sqrt{x}\,dx = 2\pi\int_0^4 x^{3/2}\,dx = 2\pi\left[\frac{2x^{5/2}}{5}\right]_0^4 = \frac{128\pi}{5}.$

Example — Discs vs shells comparison

About the $x$-axis (discs): $V = \pi\int_0^2 x^4\,dx = \pi\cdot\dfrac{32}{5} = \dfrac{32\pi}{5}$.

About the $y$-axis (shells): $V = 2\pi\int_0^2 x\cdot x^2\,dx = 2\pi\int_0^2 x^3\,dx = 8\pi$.

The volumes are different because the solids are different (different axes of revolution). In both cases the “natural” method is the most efficient: discs for the revolution about the axis on which the function is expressed, shells for the other axis.

Links

Topics: Integrale
Concepts: Integrale definito · Metodo dei gusci · Volume di rotazione
Skills: Calcolare · Integrare