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Gabriel's Horn

 
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Sep17-09, 05:00 PM   #1
 

Gabriel's Horn


I'm relatively new to calculus, and this question was bugging me, so I have decided to ask it.
We have the function [tex]y=1/x[/tex] with domain [tex]x\geq1[/tex] and we rotate the curve around the x-axis in order to form a solid of revolution. (Gabriel's Horn)
The integral is [tex]V=\pi\int^{z}_{1}1/x^2 dx[/tex] and we evaluate to [tex]V=\pi(1-1/z)[/tex]
Take the limit as z approaches infinity: [tex]\lim_{z \to \infty}\pi(1-1/z)=\pi[/tex]

Apparently my teacher says that the volume is a finite amount, which is to say [tex]\pi[/tex]. However isn't pi an irrational number? Meaning that its digits keep going without end? If pi is not finite, then how can the volume be finite? Wouldn't the volume keep getting minutely closer to pi as x tends to infinity?

Thanks!
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Sep17-09, 05:36 PM   #2
 
What he means when he says it is "finite" is that it converges to a real value (such as [itex]\pi[/itex]), instead of diverging to [itex]\infty[/itex].
Sep17-09, 05:42 PM   #3
 
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Hi tpingt! Welcome to PF!
Quote by tpingt View Post
… Take the limit as z approaches infinity: [tex]\lim_{z \to \infty}\pi(1-1/z)=\pi[/tex]

Apparently my teacher says that the volume is a finite amount, which is to say [tex]\pi[/tex]. However isn't pi an irrational number? Meaning that its digits keep going without end? If pi is not finite, then how can the volume be finite? Wouldn't the volume keep getting minutely closer to pi as x tends to infinity?
Yes, π is irrational, and so its digits keep going without end and without repetition …

but that doesn't make it infinite …

it doesn't even make it more than 4 …

or more than 3.2 …

or more than 3.15 …

or … well, you get the idea.
Sep17-09, 05:54 PM   #4
 

Gabriel's Horn


Thanks for the great explanations guys, I understand it now! :)
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