FALSE: finite volume implies finite lateral surface area

Given: Gabriel's horn, obtained by rotating $r(x)=1\mathrm{/}x$ for $x\ge 1$, and its compact truncations $1\le x\le T$.

If $a\le b$ and $f:[a,b]\to [0,\mathrm{\infty })$ is continuous, then its solid of revolution about the $x$-axis is compact and Jordan measurable and has volume $\pi {\int }_a^{b}f(x{)}^{2}dx$ (The disc formula for the volume of a solid of revolution).

The improper integral ${\int }_1^{\mathrm{\infty }}{x}^{-p}dx$ converges for rational $p>1$ and diverges for $p\le 1$ (Improper integrals over unbounded intervals, The improper $p$-test for rational exponents).

If $a<b$ and $r:[a,b]\to [0,\mathrm{\infty })$ is $C^1$ on a neighbourhood of $[a,b]$, positive on $(a,b)$, and vanishes at most at the endpoints, then the surface obtained by rotating $r$ about the axis has area $2\pi {\int }_a^{b}r(s)\sqrt{1+r^{\mathrm{\prime }}(s{)}^2}ds$ (The surface of revolution has area $2\pi {\int }_a^{b}r(s)\sqrt{1+r^{\mathrm{\prime }}(s{)}^2}ds$).

Products, sums, scalar multiples, and quotients with nonzero denominator obey their usual derivative rules (Sums, scalar multiples, products and quotients: $(f+g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)+g^{\mathrm{\prime }}(c)$, $(\alpha f{)}^{\mathrm{\prime }}(c)=\alpha f^{\mathrm{\prime }}(c)$, $(fg{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)g(c)+f(c)g^{\mathrm{\prime }}(c)$, and $(f\mathrm{/}g{)}^{\mathrm{\prime }}(c)=(f^{\mathrm{\prime }}(c)g(c)-f(c)g^{\mathrm{\prime }}(c))\mathrm{/}g(c{)}^2$ when $g(c)\mathrm{\ne }0$).

If $a<b$ and integrable $f,g:[a,b]\to \mathbb{R}$ satisfy $f(x)\le g(x)$ for every $x$, then ${\int }_a^{b}f\le {\int }_a^{b}g$ (If $f\le g$ on $[a,b]$ and both are integrable then ${\int }_a^{b}f\le {\int }_a^{b}g$; and $m(b-a)\le {\int }_a^{b}f\le M(b-a)$).