Paradox? An infinite set having finite volume?

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Homework Help Overview

The discussion revolves around finding the volume of a solid formed by rotating a region R, defined by the curve e^{-x} and the coordinate axes, about the y-axis. The original poster expresses confusion regarding the volume of an infinite set having finite volume.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the evaluation of an integral related to the volume, questioning the validity of the integral's limits and the behavior of the integrand as it approaches zero. There are suggestions to use integration by parts and to consider limits for terms that appear undefined.

Discussion Status

There is an ongoing exploration of the integral's evaluation, with some participants suggesting methods to handle the divergence and others questioning the assumptions about limits and undefined terms. No consensus has been reached, but multiple approaches are being considered.

Contextual Notes

Participants note that the integral may diverge, yet the region itself is finite, raising questions about the implications of this paradox. The discussion includes considerations of limits and the behavior of logarithmic functions near zero.

mrdoe
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Paradox? An infinite set having finite volume??

Homework Statement


Find the volume, in terms of k, of the solid made from R rotating about the y-axis, if R is defined as the region bounded by [tex]e^{-x}[/tex] and the coordinate axes.
2. The attempt at a solution
Obviously, [tex]\displaystyle\int^{1}_{0} (\ln (y))^2\pi dx =[/tex] undefined... but the region R itself has a finite volume. So if you rotate it shouldn't it have a finite volume?
 
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I assume that you mean
[tex] \displaystyle\int^{1}_{0} (\ln (y))^2\pi dy =[/tex]
because you are integrating over y.

The integral can be done, but it is not trivial to find the primitive.

My suggestion would be to "try"
[tex]y \ln^2(y)[/tex]
which upon differentiation gives you in any case the integrand back... then try to cancel out the unwanted term by adding something.

As for the evaluation of the integral, it looks like it is diverging. However, if you take an appropriate limit, you will still get the correct (finite) answer. The trick here is that your logs will only occur in terms like
[tex]y \ln y, y \ln^2 y, \cdots[/tex]
and that the linear term goes to zero faster than the logarithm goes to infinity, as y -> 0.
 


You don't really have to guess the primitive. The systematic way to do the integral is to do integration by parts twice. Start with u=ln(y)^2 dv=dy.
 


Yeah I'm getting the following for indefinite integral...

[tex]\pi x\ln^2 x - 2\pi x\ln x + 2\pi x[/tex]

But then if you try that with 0 it's undefined so... would the answer be [tex]2\pi[/tex]?
 


CompuChip already said this, but I'll repeat. Things like x*ln(x) may be undefined at x=0, but they do have a limit as x->0. You do have to check that limit is 0 before you can ignore those terms. You can't ignore something just because it's 'undefined'.
 

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