I would like to create my first paper on math, this time in geometry, and the intuitive definition of the theorem would be that:
A 2D disk with radius $r$ and a continuous function (for example, a mountain) that forms a disk with radius $r$ seen from the top, fixed at the exact centre, will always have a greater surface than the initial disk it is placed above.
Here there are two images: a red disk and a red mountain. The point would be that the mountain always has greater surface/area than the disk for a fixed area (marked as red):
The Disk:
The Mountain:
I am thinking of demonstrating it with integrals and polar coordinates. I'll also highen it to $n$ dimensions.
Thanks!
Let $\Omega\subset\Bbb{R}^n$ be any non-empty open set, and $f:\Omega\to\Bbb{R}$ a $C^1$ function. Then, for any (Lebesgue-measurable) set $E\subset \Omega$, the $n$-dimensional surface area of the portion of the graph of $f$ which lies over $E$ is given by \begin{align} S(\Gamma_{f,E})=\int_E\sqrt{1+\|\nabla f\|^2}\,dx. \end{align} THis is a standard formula from advanced calculus (the 1D analogue is $\int_a^b\sqrt{1+[f’(x)]^2}\,dx$ for the length of the graph of a function $f:[a,b]\to\Bbb{R}$). Compare with the $n$-dimensional area of the region $E$ itself, which is simply \begin{align} \lambda(E)&=\int_E1\,dx. \end{align} Which is bigger? Obviously it is the first because of monotonicity of integrals (i.e $\|\nabla f\|^2\geq 0$ and therefore $\sqrt{1+\|\nabla f\|^2}\geq 1$, so integrating over $E$ still keeps the same inequality). So, we have \begin{align} \text{$n$-dimensional surface area of graph of $f$ lying over $E$}&\geq \text{$n$-dimensional area of $E$}, \end{align} and the proof is simply a one-line inequality. In simpler words, the mountain has larger surface area than the disk. But as you can see above, the statement holds quite generally, and is not specific to disks.
In fact the proof tells us that equality holds for all sets $E$ if and only if $\nabla f=0$ identically on $\Omega$, which happens if and only if the function $f$ is constant (on each connected component of $\Omega$).
Final Remarks.
I’m not sure what level you’re at mathematically, but seeing as this is such a simple inequality, it certainly doesn’t warrant a paper. If this is some sort of high-school project, maybe talk about where the formula for surface area comes from? And why such an inequality is useful/interesting. Now, I actually can’t think of anything useful to say about this particular inequality since it is too obvious, but the isoperimetric inequality is a much more non-trivial and interesting inequality.
Ok, to make things slightly non-trivial, one should weaken hypotheses on the function $f$. For example, continuous $f$ or something else. In this case, a-priori it’s not even clear how to define the surface area, and this could be an interesting thing to write a short paper about (but beware, you can quickly run into very abstract measure-theoretic technicalities).