Gabriel's Horn

Rotating y=1/xy = 1/x (x≥1x \ge 1) makes a horn with finite volume π\pi but infinite surface area — fillable with π\pi of paint, yet unpaintable. The painter's paradox.

V=π∫1∞dxx2=πV = \pi\int_1^{\infty} \frac{dx}{x^2} = \pi
A≥2π∫1Rdxx=2πln⁡R→∞A \ge 2\pi\int_1^R \frac{dx}{x} = 2\pi\ln R \to \infty

Examples & Exercises

Prove completely, via the disk method for volumes of revolution, that the whole of Gabriel's horn has volume π\pi: V=π∫1∞1x2 dx=πV = \pi\int_1^{\infty} \frac{1}{x^2}\,dx = \pi.

V(R)=π∫1Rx−2 dx=π(1−1R)→R→∞πV(R) = \pi\int_1^R x^{-2}\,dx = \pi\left(1 - \frac{1}{R}\right) \xrightarrow[R\to\infty]{} \pi
R=50R = 50: V(50)=π(1−1/50)≈3.0788V(50) = \pi(1 - 1/50) \approx 3.0788 — already hugging π≈3.1416\pi \approx 3.1416.
Show proof
  1. The cross-section at xx is a disk of radius 1/x1/x, so the volume of the horn truncated at RR is the improper integral V(R)=∫1Rπ(1x)2dx=π∫1Rx−2 dxV(R) = \int_1^R \pi\left(\frac{1}{x}\right)^2 dx = \pi\int_1^R x^{-2}\,dx.
  2. The antiderivative is −1/x-1/x, so V(R)=π[−1x]1R=π(1−1R)V(R) = \pi\left[-\frac{1}{x}\right]_1^R = \pi\left(1 - \frac{1}{R}\right). As R→∞R \to \infty, 1/R→01/R \to 0, so V(R)→πV(R) \to \pi — the whole horn has finite volume, exactly π\pi.
  3. The convergence of ∫1∞x−2 dx\int_1^{\infty} x^{-2}\,dx is the textbook case of the pp-integral test (p=2>1p = 2 > 1). The blue curve in the lower panel clinging to the dashed π\pi line is exactly this convergence.

Using the surface-of-revolution formula, show A(R)=2π∫1R1x1+1x4 dx≥2πln⁡RA(R) = 2\pi\int_1^R \frac{1}{x}\sqrt{1 + \frac{1}{x^4}}\,dx \ge 2\pi\ln R and prove completely that Gabriel's horn has infinite surface area.

A(R)=2π∫1R1x1+1x4 dx>2π∫1Rdxx=2πln⁡R→∞A(R) = 2\pi\int_1^R \frac{1}{x}\sqrt{1 + \frac{1}{x^4}}\,dx > 2\pi\int_1^R \frac{dx}{x} = 2\pi\ln R \to \infty
R=10R = 10: lower bound 2πln⁡10≈14.46762\pi\ln 10 \approx 14.4676, quadrature ≈15.1774\approx 15.1774 — above the bound, and both grow without end as RR increases.
Show proof
  1. Rotating the curve y=f(x)=1/xy = f(x) = 1/x about the xx-axis gives surface area A(R)=2π∫1Rf(x)1+f′(x)2 dxA(R) = 2\pi\int_1^R f(x)\sqrt{1 + f'(x)^2}\,dx. With f′(x)=−1/x2f'(x) = -1/x^2 this is A(R)=2π∫1R1x1+1x4 dxA(R) = 2\pi\int_1^R \frac{1}{x}\sqrt{1 + \frac{1}{x^4}}\,dx.
  2. Since 1+1/x4>1\sqrt{1 + 1/x^4} > 1 for every x≥1x \ge 1, the integrand exceeds 2πx\frac{2\pi}{x}. Hence A(R)>2π∫1Rdxx=2πln⁡RA(R) > 2\pi\int_1^R \frac{dx}{x} = 2\pi\ln R.
  3. As R→∞R \to \infty, 2πln⁡R→∞2\pi\ln R \to \infty, so A(R)→∞A(R) \to \infty — the surface area is infinite. The volume decays like x−2x^{-2} (p=2>1p = 2 > 1, convergent) while the area decays like 1x1+1/x4∼1x\frac{1}{x}\sqrt{1 + 1/x^4} \sim \frac{1}{x} (p=1p = 1, divergent) — this one order of decay is the entire paradox.

Resolve the 'painter's paradox': the horn can be filled with π\pi of paint since its volume is π\pi, but prove that coating its surface with a uniform coat of thickness δ>0\delta > 0 needs infinitely much paint.

fill: V=π<∞,coat: Vcoat≥δ⋅A(R)→∞\text{fill: } V = \pi < \infty, \qquad \text{coat: } V_{\text{coat}} \ge \delta \cdot A(R) \to \infty
R=5R = 5: the volume π(1−1/5)≈2.5133\pi(1 - 1/5) \approx 2.5133 is settling toward π\pi, but the area already exceeds 2πln⁡5≈10.11242\pi\ln 5 \approx 10.1124.
Show proof
  1. Filling: by problem 1 the whole horn has volume π\pi, so pouring in π\pi of paint fills it — a finite amount suffices.
  2. Coating: a uniform coat of thickness δ>0\delta > 0 over the surface occupies volume at least δ⋅A(R)\delta \cdot A(R) (thickness δ\delta times each surface patch, summed). By problem 2, A(R)→∞A(R) \to \infty, so δ⋅A(R)→∞\delta \cdot A(R) \to \infty for every δ>0\delta > 0 — no finite paint coats the whole surface.
  3. Why it is not a paradox: 'filling' spends the volume π\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.

10. Integration