**Can you paint a surface with infinite area with a finite quantity of paint?** For sure… let’s do it!

Consider the 3D surface given in cylindrical coordinates as \[

S(\rho,\varphi):\begin{cases}

x &= \rho \cos \varphi\\

y &= \rho \sin \varphi\\

z &= \frac{1}{\rho}\end{cases}\] for \((\rho,\varphi) \in [1,\infty) \times [0, 2 \pi)\). The surface is named Gabriel’s horn.

### Volume of Garbiel’s horn

The volume of Gabriel’s horn is \[

V = \pi \int_1^\infty \left( \frac{1}{\rho^2} \right) \ d\rho = \pi\] which is finite.

### Area of Garbiel’s horn

The area of Gabriel’s horn for \((\rho,\varphi) \in [1,a) \times [0, 2 \pi)\) with \(a > 1\) is: \[

A = 2 \pi \int_1^a \frac{1}{\rho} \sqrt{1+\left( -\frac{1}{\rho^2} \right)^2} \ d\rho \ge 2 \pi \int_1^a \frac{d \rho}{\rho} = 2 \pi \log a.\] As the right hand side of inequality above diverges to \(\infty\) as \(a \to \infty\), we can conclude that the area of Gabriel’s horn is infinite.

### Conclusion

Gabriel’s horn could be filled with a finite quantity of paint… therefore painting a surface with infinite area. Unfortunately the thickness of the paint coat is converging to \(0\) as \(z\) goes to \(\infty\), leading to a paint which won’t be too visible!