<center> <img src = "https://i.imgsafe.org/94/94d6f641ea.png"/></center>
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# Paradox and Math 

## Be Careful!

Starting with this post, next 3 to 4 posts will be about paradoxes in Math. In fact, most of the topics that  are considered as mathematical paradox are not ___real___ paradoxes anymore; there exist numerous solutions and logical reasonings for those problems. However, it is worth to discuss about them. 

In this post, topic is about Gabriel's Horn, a mathematical object having finite volume but __infinite surface area__ . 

## 1. Torricelli's Discovery

In 1643, [Evangelista Torricelli](https://en.wikipedia.org/wiki/Evangelista_Torricelli) made his discovery of the strange nature of the acute hyperbolic solid, which we would now call the rectangular hyperboloid. It is generated by rotating the rectangular hyperbola 

<center> <img src="http://latex.codecogs.com/gif.latex?y&space;=&space;\frac{1}{x}&space;\&space;(x&space;>&space;1)" title="y = \frac{1}{x} \ (x > 0)" /> </center>

by <img src="http://latex.codecogs.com/gif.latex?360^{\circ}&space;(=2\pi&space;)" title="360^{\circ} (=2\pi )" align = "center"/> about the <img src="http://latex.codecogs.com/gif.latex?x" title="x" align = "center"/>-axis. 

<center> <img src = "https://i.imgsafe.org/92/9249c48d6c.png"/></center>

This is called [Gabriel's Horn](https://en.wikipedia.org/wiki/Gabriel%27s_Horn#cite_ref-1) or sometimes Torricelli's trumpet. He showed that this infinite solid has a finite volume. To today’s post-calculus eyes this single fact is not shocking but it does become rather more surprising when we realize that not only is its length infinite, but so is its surface area. We will use calculus and modern-day notation to prove both results

- the volume is finite and

- the surface area is infinite.

## 2. Finite Volume 

The volume of the solid formed by rotating the area between the curves of <img src="http://latex.codecogs.com/gif.latex?y=f(x)" title="y=f(x)" align = "center"/>,  the lines <img src="http://latex.codecogs.com/gif.latex?x = a" title="x" align = "center"/>, <img src="http://latex.codecogs.com/gif.latex?x = b" title="x" align = "center"/> and the <img src="http://latex.codecogs.com/gif.latex?x" title="x" />-axis is given by 

<center> <img src="http://latex.codecogs.com/gif.latex?V&space;=\pi&space;\int_{a}^{b}&space;f^2(x)dx" title="V =\pi \int_{a}^{b} f^2(x)dx" /> </center>

Since Gabriel's Horn is infinite solid, we should take <img src="http://latex.codecogs.com/gif.latex?b" title="b" /> to infinity, <img src="http://latex.codecogs.com/gif.latex?\infty" title="\infty" /> . The calculation gives 

<center><img src="http://latex.codecogs.com/gif.latex?\begin{align*}&space;V&space;&=&space;\lim_{b&space;\rightarrow&space;\infty}&space;\pi&space;\int_{1}^{b}&space;\frac{1}{x^2}dx&space;\\&space;&=&space;\lim_{b&space;\rightarrow&space;\infty}&space;\pi&space;\left[&space;-\frac{1}{x}&space;\right]_{1}^{b}&space;\\&space;&=&space;\lim_{b&space;\rightarrow&space;\infty}&space;\pi&space;\left(1&space;-&space;\frac{1}{b}&space;\right&space;)&space;=&space;\pi&space;\end{align*}" title="\begin{align*} V &= \lim_{b \rightarrow \infty} \pi \int_{1}^{b} \frac{1}{x^2}dx \\ &= \lim_{b \rightarrow \infty} \pi \left[ -\frac{1}{x} \right]_{1}^{b} \\ &= \lim_{b \rightarrow \infty} \pi \left(1 - \frac{1}{b} \right ) = \pi \end{align*}" /> </center>

which is clearly finite. 

## 3. Infinite Surface Area?!

The area of the solid formed by rotating the curve <img src="http://latex.codecogs.com/gif.latex?y&space;=&space;f(x)" title="y = f(x)" align = "center"/> along the <img src="http://latex.codecogs.com/gif.latex?x" title="x" />-axis from <img src="http://latex.codecogs.com/gif.latex?x=a" title="x=a" /> to <img src="http://latex.codecogs.com/gif.latex?b" title="b" /> is given by 

<center> <img src="http://latex.codecogs.com/gif.latex?A&space;=&space;2\pi&space;\int_{a}^{b}&space;f(x)&space;\sqrt{1&space;&plus;&space;(f'(x))^2}&space;dx" title="A = 2\pi \int_{a}^{b} f(x) \sqrt{1 + (f'(x))^2} dx" /> </center>

given that the function is __continuously differentiable__ on interval <img src="http://latex.codecogs.com/gif.latex?(a,b)" title="(a,b)" align = "center"/> . Luckily, out function <img src="http://latex.codecogs.com/gif.latex?y&space;=&space;1/x" title="y = 1/x" align = "center"/> is continuously differentiable in <img src="http://latex.codecogs.com/gif.latex?f'(x)&space;=&space;-1/x^2" title="f'(x) = -1/x^2" align = "center"/> . Since 

<center> <img src="http://latex.codecogs.com/gif.latex?\sqrt{1&space;&plus;&space;(f'(x))^2}&space;=&space;\sqrt{1&space;&plus;&space;\frac{1}{x^4}}&space;>&space;1" title="\sqrt{1 + (f'(x))^2} = \sqrt{1 + \frac{1}{x^4}} > 1" /> </center> 

for all <img src="http://latex.codecogs.com/gif.latex?x&space;\in&space;(1,&space;\infty)" title="x \in (1, \infty)" align = "center"/> , we have the inequality 

<center> <img src="http://latex.codecogs.com/gif.latex?\begin{align*}&space;2\pi&space;\int_{1}^{b}&space;f(x)\sqrt{1&plus;(f'(x))^2}&space;&>&space;2\pi&space;\int_{1}^{b}&space;f(x)&space;dx&space;\\&space;&=&space;2\pi&space;\int_{1}^{b}&space;\frac{1}{x}&space;dx&space;\\&space;&=&space;2&space;\pi&space;\ln&space;b&space;\end{align*}" title="\begin{align*} 2\pi \int_{1}^{b} f(x)\sqrt{1+(f'(x))^2} &> 2\pi \int_{1}^{b} f(x) dx \\ &= 2\pi \int_{1}^{b} \frac{1}{x} dx \\ &= 2 \pi \ln b \end{align*}" /> </center> 

where <img src="http://latex.codecogs.com/gif.latex?\ln" title="\ln" /> denotes natural logarithm. Now taking the limit both sides give 

<center> <img src="http://latex.codecogs.com/gif.latex?A&space;\geq&space;\lim_{b&space;\rightarrow&space;\infty}&space;2\pi&space;\ln&space;b&space;=&space;\infty" title="A \geq \lim_{b \rightarrow \infty} 2\pi \ln b = \infty" /> </center>

so that the surface area of Gabriel's horn is infinite. 

## 4. Painter's Paradox and Solution

So where is the paradox? Imagine the Gabriel's horn as a infinite cup. Then since volume is finite, we can fill the cup with finite amount, <img src="http://latex.codecogs.com/gif.latex?\pi" title="\pi" align = "center"/> liter of paint. But at the same time, the paint would not be sufficient to coat its inner surface; as the surface area of a cup is infinite! This ironic situation is called the painter's paradox. 

<center> <img src = "https://i.imgsafe.org/9f/9f8e5cfce1.png"/></center>


Is an infinite amount of paint really needed to paint the inner surface of the Gabriel's horn? Well, the answer is NO. The painter's paradox is __resolved by__ realizing that a finite amount of paint can in fact coat an infinite surface area, 

<center> <b> <i> It simply needs to get thinner at a fast enough rate. </i></b> </center>

Painting a surface means applying a coat of paint on the surface. The thickness of coat, is NOT uniform all over the surface, it should become thinner as the horn goes to infinity. So what is exactly a fast enough rate? Remember the area formula

<center> <img src="http://latex.codecogs.com/gif.latex?A(t)&space;=&space;2\pi&space;\int_{1}^{t}&space;f(x)\sqrt{1&space;&plus;&space;(f'(x))^2}&space;dx" title="A(t) = 2\pi \int_{1}^{t} f(x)\sqrt{1 + (f'(x))^2} dx" /> </center>

for the surface of revolution defined on interval <img src="http://latex.codecogs.com/gif.latex?x&space;\in&space;[1,&space;t]" title="x \in [1, t]" align = "center" /> ? By [Fundamental Theorem of Calculus](https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus), the rate of change is equal to 

<center> <img src="http://latex.codecogs.com/gif.latex?\begin{align*}&space;\frac{dA}{dt}&space;&=&space;2\pi&space;f(t)\sqrt{1&plus;(f'(t))^2}\\&space;&&space;=&space;\frac{2\pi}{t}&space;\sqrt{1&space;&plus;&space;\frac{1}{t^4}}&space;\\&space;&=&space;2\pi&space;\frac{\sqrt{t^4&space;&plus;&space;1}}{t^3}&space;\\&space;&\sim&space;\frac{2\pi}{t}&space;\end{align*}" title="\begin{align*} \frac{dA}{dt} &= 2\pi f(t)\sqrt{1+(f'(t))^2}\\ & = \frac{2\pi}{t} \sqrt{1 + \frac{1}{t^4}} \\ &= 2\pi \frac{\sqrt{t^4 + 1}}{t^3} \\ &\sim \frac{2\pi}{t} \end{align*}" /></center> 

Thus if we set the thickness of paint to be __inversely proportional__ to the length of the throat of the horn, we can actually coat the surface with finite amount of paint! However, in real world, this is impossible; since the thickness of coating can not go infinitesimally small due to physical limitations. 


## 5. What about the Converse?
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Now a natural question arises. 

<center> <b> <i>  Does there exist a surface of revolution that has a finite surface area but an infinite volume? </i></b> </center>

The answer is unfortunately NO...&#128533;

Let <img src="http://latex.codecogs.com/gif.latex?f(x)" title="f(x)" align = "center"/>  be continuously differentiable function on <img src="http://latex.codecogs.com/gif.latex?[1,&space;\infty)" title="[1, \infty)" align = "center"/> . Let <img src="http://latex.codecogs.com/gif.latex?A,&space;V" title="A, V" align = "center"/> be the surface area and volume of solid of revolution over <img src="http://latex.codecogs.com/gif.latex?x" title="x" align = "center" />-axis respectively. If the surface area  <img src="http://latex.codecogs.com/gif.latex?A" title="A" /> is finite,

<center> <img src="http://latex.codecogs.com/gif.latex?\begin{align*}&space;\lim_{t&space;\rightarrow&space;\infty}&space;\left(&space;\sup_{x&space;\geq&space;t}&space;f^2(x)&space;\right&space;)&space;&=&space;\limsup_{t&space;\rightarrow&space;\infty}&space;\left(&space;f^2(1)&space;&plus;&space;\int_{1}^{t}&space;(f^2(x))'dx&space;\right)\&space;(\because&space;\text{F.T.C})&space;\\&space;&\leq&space;f^2(1)&space;&plus;&space;\int_{1}^{\infty}|(f^2(x))'|dx&space;\&space;(\because&space;(f^2)'&space;\leq&space;|(f^2)'|)&space;\\&space;&=&space;f^2(1)&space;&plus;&space;\int_1^{\infty}&space;2|f(x)||f'(x)|&space;dx\&space;(\because&space;(f^2)'&space;=&space;2ff'))\\&space;&\leq&space;f^2(1)&space;&plus;&space;\int_1^{\infty}&space;2|f(x)|&space;\sqrt{1&space;&plus;&space;(f'(x))^2}&space;dx\&space;(\because&space;|f'|&space;=&space;\sqrt{(f')^2}&space;\leq&space;\sqrt{1&space;&plus;&space;(f')^2})&space;\\&space;&=&space;f^2(1)&space;&plus;&space;\frac{A}{\pi}&space;\end{align*}" title="\begin{align*} \lim_{t \rightarrow \infty} \left( \sup_{x \geq t} f^2(x) \right ) &= \limsup_{t \rightarrow \infty} \left( f^2(1) + \int_{1}^{t} (f^2(x))'dx \right)\ (\because \text{F.T.C}) \\ &\leq f^2(1) + \int_{1}^{\infty}|(f^2(x))'|dx \ (\because (f^2)' \leq |(f^2)'|) \\ &= f^2(1) + \int_1^{\infty} 2|f(x)||f'(x)| dx\ (\because (f^2)' = 2ff'))\\ &\leq f^2(1) + \int_1^{\infty} 2|f(x)| \sqrt{1 + (f'(x))^2} dx\ (\because |f'| = \sqrt{(f')^2} \leq \sqrt{1 + (f')^2}) \\ &= f^2(1) + \frac{A}{\pi} \end{align*}" /></center>

Therefore, we can deduce that 
<center> <img src="http://latex.codecogs.com/gif.latex?M&space;=&space;\sup_{x&space;\ge&space;1}&space;|f(x)|" title="M = \sup_{x \ge 1} |f(x)|" /> </center> 
is finite, using the fact that <img src="http://latex.codecogs.com/gif.latex?f" title="f" align  = "center"/> is continuous over <img src="http://latex.codecogs.com/gif.latex?[1,&space;\infty)" title="[1, \infty)" align = "center"/> . Now, the volume:

<center> <img src="http://latex.codecogs.com/gif.latex?\begin{align*}&space;V&space;&=&space;\pi&space;\int_{1}^{\infty}&space;f^2(x)dx&space;\\&space;&=&space;\pi&space;\int_{1}^{\infty}&space;f(x)\cdot&space;f(x)dx&space;\\&space;&\leq&space;M\pi&space;\int_1^{\infty}&space;f(x)dx&space;\\&space;&=&space;M\pi&space;\int_1^{\infty}&space;f(x)&space;\cdot&space;1&space;dx&space;\\&space;&\leq&space;M\pi&space;\int_1^{\infty}&space;f(x)&space;\sqrt{1&space;&plus;&space;(f'(x))^2}&space;dx&space;=&space;\frac{MA}{2}&space;\end{align*}" title="\begin{align*} V &= \pi \int_{1}^{\infty} f^2(x)dx \\ &= \pi \int_{1}^{\infty} f(x)\cdot f(x)dx \\ &\leq M\pi \int_1^{\infty} f(x)dx \\ &= M\pi \int_1^{\infty} f(x) \cdot 1 dx \\ &\leq M\pi \int_1^{\infty} f(x) \sqrt{1 + (f'(x))^2} dx = \frac{MA}{2} \end{align*}" /> </center>

should necessarily be finite! 

Actually, the converse also does not holds for arbitrary rectifiable solids, because of [isoperimetric inequality](https://en.wikipedia.org/wiki/Isoperimetric_inequality) in 3D space but proving the generalized fact needs more mathematical concepts. 

## 6. Conclusion
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The painter's paradox, arises when our intuition of painting deviates from the mathematical concept of area. Also the concept of infinity should be treated rigorously, not just using simple simulations using real world applications (like painting) . 

## 7. Citations

[1] [Image Source Link](https://brilliant.org/wiki/gabriels-horn/)

[3] [Gabriel's Horn Explained Wiki Article](https://en.wikipedia.org/wiki/Gabriel%27s_Horn#cite_ref-1)

[4] [On Painter’s Paradox: Contextual and Mathematical Approaches to Infinity](https://link.springer.com/article/10.1007/s40753-015-0004-z) (Article)

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All other images are self-made using Mathematica.