the graph F(x)=1/(x^2) simple enough graphing no problem calculator does that just fine Please Register or Log in to view the hidden image! rotate it about the x axis you get gabriels horn. http://mathworld.wolfram.com/GabrielsHorn.html infinite surface area yet finite volume. You can fill it up with paint but you cant paint it. But when this object is full of paint is there not paint touching every place on the inside wall there by the inside is painted?
It's the rotated graph of 1/x that gives you Gabriel's horn, not 1/x^2. Rotating 1/x^2 gives you something that has finite surface area and finite volume. Don't take the paint argument too far. In order to conclude that you can't paint it given a finite bucket of paint, you have to assume your paint has some kind of thickness when applied. In order to assume you can actually fill the horn as is, you have to assume your paint can fit down arbitrarily small cracks. These assumptions don't mix well together. Infinite surface area on a finite volume is hoopy enough without mentioning paint.
Hmm I think I'm beginning to see a little small beam of light... Please Register or Log in to view the hidden image!
There's a problem when you are trying to compare a 2-dimensional quantity (the surface area) with a 3 dimensional one (the volume). The 2-d thing can be romoved from the 3-d thing without affecting the 3-d things volume, since the 2-d thing has no volume itself.. Suppose you have a mathematically perfect, but finite, can of paint. By mathematically perfect paint, I'm assuming that when we actually paint a surface, it has no thickness. So to paint a disc on the wall, you just remove a 2-dimensional disc of your paint and slap it on. Removing this disc has absolutely no effect on the the volume of our remaining paint can since this disc has no volume itself. If you remove a (countably) infinite number of discs like this, you can cover an infinite surface area, and your can of paint is still full. Thinking the other way, Imagine that you cut the surface of the horn at every integer value of x. So the section from x=1..2 forms a piece, the section from x=2..3 forms a piece and so on. Now you can collapse this dissected horn by moving all pieces so they are in line with x=1 (think of a folding telescope). Our horn will now fit nicely into our finite backpacks (or bucket of paint). You can make a lower dimensional version of this easily enough and get a finite area with an infiniteyly long boundary. Draw the graph of 1/x^2 from x=1 to infinity. Assuming you didn't run out of ink or die of old age, you have an infinitely long line. Now find the area between this line and the x-axis...
hmmm strange my last post did not post anyways. My class was working on the 1/x^2 problem and we finished the vol but did not get finished with the SA. I'll have to finish the math on that one to see if it does converge.
