The shell and washer methods both give $8\pi \mathrm{/}3$ for a rotated parabolic cap

Put $f_1(x)=1$ and $f_2(x)=(x-1{)}^2$ on $[0,2]$; both are continuous and nonnegative and $f_2\le f_1$ there. By [F1] the solids $S_1$ and $S_2$ obtained by revolving $0\le y\le f_i(x)$ about the $y$-axis are compact and Jordan measurable with contents $2\pi {\int }_0^{2}xdx=4\pi$ and $2\pi {\int }_0^{2}x(x-1{)}^{2}dx=4\pi \mathrm{/}3$.

At height $0\le y\le 1$, the washer radii are $1+\sqrt{y}$ and $1-\sqrt{y}$. By [F2], its area is $4\pi \sqrt{y}$, and ${\int }_0^{1}4\pi \sqrt{y}dy=8\pi \mathrm{/}3$.

In cylindrical terms, writing $\rho$ for the distance to the $y$-axis, the solid $S$ of the Example is $\{f_2(\rho )\le y\le f_1(\rho ),\rho \le 2\}$, while $S_i=\{0\le y\le f_i(\rho ),\rho \le 2\}$. Hence $S\cup S_2=S_1$, and $S\cap S_2$ is the set $\{y=f_2(\rho )\}$, the graph of a continuous function on the closed disc of radius $2$, which has content zero by [F4]. The boundary of the bounded set $S$ lies in the union of the boundaries of $S_1$ and $S_2$, which have content zero by [F5] and step 1.1, so [F5] makes $S$ Jordan measurable.

By [F3] applied to $S$ and $S_2$, $\mathrm{cont}(S_1)=\mathrm{cont}(S)+\mathrm{cont}(S_2)$, so step 1.1 gives $\mathrm{cont}(S)=4\pi -4\pi \mathrm{/}3=8\pi \mathrm{/}3$; this is the shell value $2\pi {\int }_0^{2}x(1-(x-1{)}^2)dx$, the shell height being the difference of the two profiles.

The two independent descriptions therefore give the same volume $8\pi \mathrm{/}3$.