#### TL;DR Gabriel's wedding cake describes a geometric figure with a finite volume but infinite surface are. In the author's words imagine ***"a cake you can eat, but cannot frost"***. A few interesting properties of this cake (see image below): - Gabriel's wedding cake describes a particular geometric figure which has **infinite surface area but finite volume** - the rotation of an infinitely large section of the $xy$ plane about the x axis generates an object of finite volume was considered a **paradox because the section lying in the $xy$ plane has an infinite area, any other section parallel to it has a finite area. ** This paradox was first discovered in the 17th century and it originated a **dispute on the nature of infinity involving many of the key thinkers of the time, including Thomas Hobbes, John Wallis and Galileo Galilei.**  How to transform the the "continuous horn" into a "discrete cake". Another similar object is the infinite Gift 🎁: The infinite gift is an interesting object where the side of the nth box is $\frac{1}{√n}$. As $n\rightarrow +∞$, the gift has infinite surface area and length but finite volume! - [Learn more here: Infinit Gift](https://twitter.com/fermatslibrary/status/1166324490760065024)  You can learn more about Gabriel's Horn here: - [Fermat's Library - Gabriel's Horn](https://fermatslibrary.com/s/gabriels-horn) - [Wolfram - Gabriel's Horn](https://mathworld.wolfram.com/GabrielsHorn.html) Three-dimensional illustration of Gabriel’s Horn. This horn shape has an interesting, counterintuitive property: **a finite volume but an infinite surface area.**  Create your own Gabriel Horn with Mathematica: ```mathematica x[u_, v_] := u y[u_, v_] := Cos[v]/u z[u_, v_] := Sin[v]/u Manipulate[ParametricPlot3D[{{x[u, v], y[u, v], z[u, v]}}, {u, 1, umax}, {v, 0, 2*Pi}, PlotRange -> {{0, 20}, {-1, 1}, {-1, 1}}, Mesh -> {Floor[umax], 20}, Axes -> False, Boxed -> False], {{umax, 20}, 1.1, 20}] ``` Source: [Fouriest Seriess.](http://fouriestseries.tumblr.com/post/90296337418/gabriels-horn-and-the-painters-paradox) Determining the surface of the horn is a slightly more complex than the volume. The area for a volume of revolution is given by: \begin{eqnarray} S &=& 2\pi \int_1^N y \sqrt{1+ \left(\frac{dy}{dx} \right)^2} dx\\ \end{eqnarray} In this scenario we know that $y = \frac{1}{x}$ and we can then write: \begin{eqnarray*} \frac{dy}{dx} &=& -\frac{1}{x^2} \end{eqnarray*}We can then rewrite (1): \begin{eqnarray} S &=& 2\pi \int_1^N \frac{1}{x} \sqrt{1+ \frac{1}{x^4} } dx \nonumber \\ &=& 2\pi \int_1^N \frac{\sqrt{x^4+ 1}}{x^3} dx\\ \end{eqnarray}using integration by parts we can rewrite the integra in equation (2) as: \begin{eqnarray} \int \frac{\sqrt{x^4+ 1}}{x^3} dx &=&- \frac{1}{2x^2}\sqrt{x^4+1} + \int \frac{x}{\sqrt{x^4+1}}dx \end{eqnarray}to compute the integral in the right side of equation (3) we perform the following substition: \begin{eqnarray*} u = x^2 ; \frac{du}{dx} = 2x \end{eqnarray*}we have then that: \begin{eqnarray*} \int \frac{x}{\sqrt{x^4+1}}dx &=& \frac{1}{2} \int \frac{1}{\sqrt{u^2+1}}du\\ &=& \frac{1}{2} \ln{\left(u + \sqrt{u^2+1}\right)}+k\\ \end{eqnarray*}and re-doing the substition we finally have that: \begin{eqnarray} \int \frac{x}{\sqrt{x^4+1}}dx &=& \frac{1}{2} \ln{\left(x^2 + \sqrt{x^4+1}\right)}+k \end{eqnarray}Taking (3) and (4) into account we can rewritte the surface area (2): \begin{eqnarray} S &=& 2\pi \left[ - \frac{1}{2x^2}\sqrt{x^4+1} + \frac{1}{2} \ln{\left(x^2 + \sqrt{x^4+1}\right)}\right]_1^N \nonumber\\ &=& \pi \left(- \frac{1}{N^2}\sqrt{N^4+1} + \ln{\left(N^2 + \sqrt{N^4+1}\right)} \nonumber \\ +\sqrt{2}- \ln{ \left( 1+\sqrt{2}\right)} \right) \end{eqnarray} As $N \rightarrow \infty$ the dominent term is $\ln{\left(N^2 + \sqrt{N^4+1}\right)}$ and taking the properties of the logarithm into accoun we can then write that: \begin{eqnarray*} \lim_{x\to\infty} S &\simeq& \lim_{x\to\infty} \left(\pi \ln{N} \right) &=& \infty \end{eqnarray*} Thus we determined with modern calculous that **Gabriel's horn has a finite Volume and an intfinite area!** Paradox! **Cavalieri's Principle:** states that if two objects are between two parallel planes and if every plane cross-section parallel to the given planes have equal areas then the two objects must have equal volume. Figure 1. below is a great illustration of different objects with matching cross-sectional areas. They all have the same volume.  **Figure 1:** Simple illustration of different objects with the same volume. #### We will now use Cavalieri's principle and apply it to Gabriel's horn. Figure 2 shows shows the intersection of a cylinder with an hyperbola in the $xy$-plane at point $(x_0,y_0)$. We can easily compute the curved surface area of the cylinder: $$V_C = (2\pi y_0)x_0$$ and we know that $y= \frac{1}{x}$ thus: $$V_C = 2\pi$$  **Figure 2:** Illustration of a horn with open lip (1,1) and it's intersection with a cylinder at point $(x_0,y_0)$. The area of the cylinder is constant as the point of contact varies. Now imagine that you can cut the cylinder along the $x$-axis and unravel it so that it becomes a horizontal rectangle at point $y_0$, see the left panel of Figure 3. We can construct a new object that has surface are $2\pi$ at point $y_0$, see right panel of Figure 3.  **Figure 3:** Applying Cavalieri's principle to Gabriel's horn. Using Cavalieri's principle we know that two objects between two parallel planes with the same matching cross-sectional areas have the same volume. We know that the cylinder that we just constructed is a succession of $2\pi$ planes from $y=0, y=1$. Thus the volume of the cylinder is $V_C = 2\pi*1 = 2\pi$. Now we can use Cavalieri's principle and infer the volume of Gabriel's horn $V_{GH}$. $$V_{GH} = V_C = 2\pi.$$ Thanks to Cavalieri's principle have shown that although the horn seems to have an infinite length it has a finite volume!