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Fillable but not paintable?
I thought it would be fun to post this question. I know the answer but just to keep you guys busy on a question I post instead of me being busy on answering your questions... :-)
Imagine a funnel that has a surface of revolution obtained from revolving y = 1/x , between x=1 and x=infinity - about the x-axes.
The volume of that funnel of course is
V = Integral_1^infinity pi y^2 dx
= pi * Integral_1^infinity 1/x^2 dx
= pi .
The surface area is
A = integral_1^infinity 2 pi y sqrt(1+ y'^2) dx
= 2 pi * integral_1^infinity 1/x * sqrt(1+ 1/x^4) dx
But this integral diverges.
Hence the volume is finite but the surface area is infinite.
This begs the question: does this mean that you can fill this funnel with a finite amount of paint but that finite amount of paint does not suffice to paint the inside wall...?
@Jim indeed decreasing the thickness of the layer of (mathematical) paint sufficiently rapidly will ensure a finite amount is needed. But you cannot paint the outside say with a constant non-zero thickness layer. The funny thing with this shape (aka Gabriel's Horn) is this finite volume vs infinite surface area. Remarkable.
1 Answer
- JimLv 47 years agoFavorite Answer
No it doesn't. The thickness of the layer of paint would be zero - so any non-zero volume of paint would keep you going forever! The mathematical model doesn't quite fit the reality....nice try though!
Source(s): Jim. Dept. Of Physics - imperial College, London.
