The cylindrical-shell formula for a solid of revolution about the $y$-axis

Let $A:=\{(x,z):a\le \parallel (x,z){\parallel }_2\le b\}$ and $h(x,z):=f(\parallel (x,z){\parallel }_2)$. The annulus is closed and bounded, hence compact by [F12], and its boundary lies in the two circle boundaries of closed discs; [F2] and [F6] make those circles content zero and then make $A$ Jordan measurable. The triangle inequality in [F7], applied in both orders, gives $\mathrm{\mid }\mathrm{\parallel }u{\mathrm{\parallel }}_2-\mathrm{\parallel }v{\mathrm{\parallel }}_2\mathrm{\mid }\le \mathrm{\parallel }u-v{\mathrm{\parallel }}_2$, so the norm is continuous; [F8] then makes $h$ continuous. Fact [F1] applied between the graphs $0$ and $h$ identifies the resulting solid with $S_y(f)$ and its volume with ${\int }_{A}h$.

If $a=b$, the annulus $A$ is the boundary circle of the closed disc of radius $a$, so [F2] and [F6] give it content zero. Fact [F10] bounds $h$ on $A$, and the single-set upper bound in [F9], which transfers the rectangle monotonicity of [F5] to Jordan-set integrals, gives $0\le {\int }_{A}h\le ({\max }_{A}h)\mathrm{cont}(A)=0$. The integral $2\pi {\int }_a^{b}xf(x)dx$ is also zero, so the theorem holds in this case. Henceforth assume $a<b$.

For a partition $a=r_0<\cdots <r_N=b$, let $A_i$ be the closed subannulus with radii $r_i,r_{i+1}$, and let $m_i,M_i$ be the minimum and maximum of $f$ on $[r_i,r_{i+1}]$, which exist by [F10]. Fact [F2], the boundary criterion [F6], and additivity [F11] give $\mathrm{cont}(A_i)=\pi (r_{i+1}^2-r_i^2)$. The $A_i$ cover $A$ and have pairwise disjoint interiors, so [F9] bounds ${\int }_{A}h$ between ${\sum }_i{m}_i\pi (r_{i+1}^2-r_i^2)$ and ${\sum }_i{M}_i\pi (r_{i+1}^2-r_i^2)$.

Uniform continuity from [F3] makes ${\max }_i(M_i-m_i)$ tend to zero with the mesh. Hence the difference between the upper and lower annular sums in step 3.1 is at most $\pi (b^2-a^2){\max }_i(M_i-m_i)$ and tends to zero.

Since $r_{i+1}^2-r_i^2=(r_i+r_{i+1})(r_{i+1}-r_i)$, each annular sum differs by a vanishing mesh error from a tagged Riemann sum for $2\pi xf(x)$. By [F4], steps 3.1 and 4.1 therefore squeeze ${\int }_{A}h$ to $2\pi {\int }_a^{b}xf(x)dx$. Together with step 2.1, the argument permits $a=0$, zeros of $f$, and $a=b$.