Gabriel's Horn

The cross-section at $x$ is a disk of radius $1/x$, so the volume of the horn truncated at $R$ is the improper integral $V(R) = \int_1^R \pi\left(\frac{1}{x}\right)^2 dx = \pi\int_1^R x^{-2}\,dx$.

The antiderivative is $-1/x$, so $V(R) = \pi\left[-\frac{1}{x}\right]_1^R = \pi\left(1 - \frac{1}{R}\right)$. As $R \to \infty$, $1/R \to 0$, so $V(R) \to \pi$ — the whole horn has finite volume, exactly $\pi$.

The convergence of $\int_1^{\infty} x^{-2}\,dx$ is the textbook case of the $p$-integral test ($p = 2 > 1$). The blue curve in the lower panel clinging to the dashed $\pi$ line is exactly this convergence.

Rotating the curve $y = f(x) = 1/x$ about the $x$-axis gives surface area $A(R) = 2\pi\int_1^R f(x)\sqrt{1 + f'(x)^2}\,dx$. With $f'(x) = -1/x^2$ this is $A(R) = 2\pi\int_1^R \frac{1}{x}\sqrt{1 + \frac{1}{x^4}}\,dx$.

Since $\sqrt{1 + 1/x^4} > 1$ for every $x \ge 1$, the integrand exceeds $\frac{2\pi}{x}$. Hence $A(R) > 2\pi\int_1^R \frac{dx}{x} = 2\pi\ln R$.

As $R \to \infty$, $2\pi\ln R \to \infty$, so $A(R) \to \infty$ — the surface area is infinite. The volume decays like $x^{-2}$ ($p = 2 > 1$, convergent) while the area decays like $\frac{1}{x}\sqrt{1 + 1/x^4} \sim \frac{1}{x}$ ($p = 1$, divergent) — this one order of decay is the entire paradox.

Filling: by problem 1 the whole horn has volume $\pi$, so pouring in $\pi$ of paint fills it — a finite amount suffices.

Coating: a uniform coat of thickness $\delta > 0$ over the surface occupies volume at least $\min(\delta, \tfrac12) \cdot A(R)$ (justified in the next step). By problem 2, $A(R) \to \infty$, so the coating volume diverges to $\infty$ for every $\delta > 0$ — no finite paint coats the whole surface.

Justifying the bound — use tubular coordinates $(x, \varphi, t)$ ($0 \le t \le \delta$) along the outward normal.

The horn's principal curvatures are $\kappa_m = \frac{r''}{(1+(r')^2)^{3/2}} = \frac{2/x^3}{(1+1/x^4)^{3/2}} \ge 0$ in the meridian direction and $\kappa_p = \frac{1}{r\sqrt{1+(r')^2}} \ge 0$ in the parallel direction (with $r = 1/x$), and the volume element is $dV = (1 + t\kappa_m)(1 + t\kappa_p)\,dA\,dt$. The normal segments are disjoint for $t < 1/\max\kappa_m$: normals of the convex meridian curve do not meet before the focal distance $1/\kappa_m \ge 1$, and normals with different $\varphi$ keep $\rho > 0$ so they cannot meet.

Since $(1+t\kappa_m)(1+t\kappa_p) \ge 1$ for $t \ge 0$, with $\delta_0 = \min(\delta, \tfrac12)$ we get $V_{\text{coat}}(\delta) \ge V_{\text{coat}}(\delta_0) = \int_0^{\delta_0}\!\!\int_S (1+t\kappa_m)(1+t\kappa_p)\,dA\,dt \ge \delta_0 \cdot A(R)$.

Why it is not a paradox: 'filling' spends the volume $\pi$, while 'coating' spends thickness times area — two different quantities.

One being finite and the other infinite is no contradiction.

This is exactly where the physical intuition that paint has thickness breaks down.