calculussoft-question
I was just reading about Gabriel Horn and just wanted to know if there were others. Meaning with infinity surface area but finite volume, or infinit volume but finite surface area.
I have tried to find a cross section of the Klien bottle, but the equations I have found for it are above my level of math. Studying for the AP BC Exam.
Try doing what you did to find the circumference of a circle. It's like you approximated the circumference with a bunch of vertical line segments. But no matter how many you make and how small you make them, they still just have a total lenght of $4R$. They're not doing a good job of approximating the curve. To do that, you need to make the ones closer to the top and bottom slanted and therefore longer.
The total length of these segments is just $4R$, not $2\pi R$, and even using more of them at a smaller size would not change this. Try to see the next higher dimension analogy.
Now, if you used the rectangles that these segments define (by adding in horizontal segments) you do get a good approximation of the area, which does get better as the height of these segments gets smaller. The missing area omitted by such rectangles will approach zero as $n\to\infty$. But again, this is not the case with the lengths, which are a constant $4R$ regardless of $n$.
I'm not sure about your setup, but in any case you should be integrating in a direction perpendicular to the $r$ in your cross section.
I'll use a setup that's hopefully what you had in mind: the tip of the cone is at the origin, and the cone is lying on its side. Suppose the radius of the base is $r$ (this is the $y$ value at the end of the triangle you're rotating) and the height is $h$ (this is the $x$ value at the end of the triangle).
Then, consider a circular cross section at some point $x$. The radius is the corresponding $y$ value, which by similar triangles is $\frac{rx}{h}$. So, the area is $$ A(x) = \frac{\pi r^2}{h^2} x^2, $$ and the volume is $$ V = \int_0^h \frac{\pi r^2}{h^2} x^2\ dx = \frac{\pi r^2 h^3}{3h^2} = \frac{1}{3} \pi r^2h. $$ Note the cone lies on its side, so the $x$ values we integrate over range from $0$ to the "height" of the cone, $h$.
Best Answer
Well, first one has the following fact:
If one makes the corresponding "Gabriel's Horn" by rotating the region under $\frac{1}{x^p}$ from $x=1$ to $x=\infty$ around the $x$-axis, then:
The volume is finite iff $p>\frac{1}{2}$
The surface area is finite iff $p > 1$ so in particular
The Horn has finite volume and infinite surface area for any $p$ between $\frac{1}{2}$ and $1$, where $p$ cannot be $\frac{1}{2}$ but it can be $1$.
The proof is simply several uses of the comparison theorem.
One cool thing that pops out is that the standard Gabriel's horn $(p=1)$ actually gives the minimal volume Gabriel's horn which still has infinite surface area. If you shrink the volume at all (i.e., $p$ goes up), then the surface area becomes finite.
There are, of course, variations on this idea. For example, one can add some small bumps to the graph of $\frac{1}{x}$. Note that bumps will tend to increase the surface area (so it will stay infinite), but if the bumps are not too frequent or too high, then the volume will still converge. (Concretely, imagine rotation $\frac{\sin x}{x}$ with $1\leq x < \infty$ around the $x$ axis).
Finally, there are some other shapes which are in no way related to Gabriel's horn. For example, the Menger Sponge has finite volume but infinite surface area. You start with a cube of volume $1$ (so, very finite), and start cutting pieces out. Of course, this only lowers the volume, so the volume is finite at the end of the day. On the other hand, cutting pieces out increases the surface, and at the end up the day it's infinite.