So, limA→∞∫A1lnxxdx=∞ All good.
But, if you take the volume of revolution of the curve, rotated about the x axis 2π radians, from 1 to A, the volume remains finite as A→∞ (it approaches 2π):
limA→∞π∫A1(lnxx)2dx=2πPretty cool result, can anyone help me get my head round this intuitively? You’d assume an infinite area rotated around the x axis would produce an infinite volume… right?
Answer
Yes, this was considered a pretty strange phenomenon when Torricelli first constructed such an example in 1643. (Torricelli himself found it so incredible that he offered two different proofs that the volume of his shape was finite, the better to convince himself and the reader).
I think there are two main mental hurdles that one must conquer in order to shake the impression that this is paradoxical:

An infinite sequence of positive numbers can have finite sum.

Volumes of revolution with the same finite crosssectional area can have as large or small volume as you want.
The first of these has occupied minds since Zeno’s paradoxes. Most of us are eventually able to come to sort of peace with it, if only by bludgeoning ourselves with the many examples of it.
For the second, just compare the volumes of cylinders you get by rotating a rectangle of unit area around one of its sides. The sides of the rectangle is a by 1a for some a, and calculating its volume we get πa2⋅1a=πa which we can make anything we want just by choosing a appropriately. What’s going on is that a small piece of area close to the rotation axis contributes less volume than the same area longer away, so by choosing a long thin rectangle that hugs the axis tightly, we can make the volume as small as we want to.
Putting these two together: Choose a convergent series of positive numbers
b1+b2+b3+⋯=c<∞
and for each bi pick the cylinder with volume bi and crosssectional area 1. Stack all of these cylinders after one another, and you get an infinitely long volume of revolution with infinite crosssection but finite volume, namely c.
This is perhaps not as beautiful as your nice smooth logxx, but is arguably easier to wrap one's head around because one needs only deal with discrete series rather than continuous integrals.
Attribution
Source : Link , Question Author : Nev , Answer Author : hmakholm left over Monica