Gabriel's Horn with a twist

delmotteo

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Gabriel's Horn is a famous problem in calculus where a finite volume is bound by an infinite surface area.
The surface area is obtained by rotating the graph of the function y = 1/x around the X axis and the volume is what's inside it.
Both the volume and the surface area can be obtained by integration using the appropriate limits of integration. The paradox arises when we integrate between a strictly positive value (1, for example) and +infinity. (see details here: https://en.wikipedia.org/wiki/Gabriel's_horn#:~:text=Gabriel's horn (also called Torricelli's,surface area but finite volume).

I understand the math and I enjoy the paradox.
What I don't understand is the following: The integral of y = 1/x between a positive number and +infinity is also infinite. This integral represents the area below the curve until the X axis. This area can be seen as a slice of the volume obtained by rotation. My understanding is that the volume is obtained by integrating (which is kind of adding) an infinite number of slices. So how come adding an infinite number of slices, each of them having an infinite area, gives a finite volume? Is it because what is being integrated/added is not an area, but a volume? Assuming integration in cylindrical coordinates, the volume is obtained by multiplying the (infinite) area by rdθ, which is infinitely small and therefore the dV may not be infinite?
 
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This area can be seen as a slice of the volume obtained by rotation...So how come adding an infinite number of slices, each of them having an infinite area, gives a finite volume?
I would carefully think about the lines above. Recall oo*0 is not oo.
This question had me thinking for a good 30 seconds!
 
Good question indeed!
If you take a thin slice parallel to the XY plane and (temporarily) fix the thickness [imath]\Delta z[/imath] of the slice the resulting volume will not be infinity because for large [imath]x > \frac{1}{\Delta z}[/imath] the radius of the rotated figure will be less than [imath]\Delta z[/imath].
 
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