MHB Solid of revolution vs. area below 1/x

Click For Summary
The area under the curve of 1/x from 1 to infinity diverges, resulting in an infinite area, as shown by the integral A = ∫(1 to ∞) (1/x) dx = ∞. In contrast, the volume of the solid of revolution formed by rotating this curve around the x-axis converges to a finite value, calculated as V = π ∫(1 to ∞) (1/x²) dx = π. This phenomenon, often perceived as paradoxical, was noted by the ancient mathematician Pappo di Alessandria. The discussion highlights the distinction between area and volume in calculus, emphasizing that infinite area can correspond to finite volume. Understanding this concept is crucial in advanced calculus and mathematical analysis.
SweatingBear
Messages
119
Reaction score
0
How come the area below the graph of $$\frac 1x$$ between $$[1, \infty)$$ does not exist, but the solid of revolution below the same graph in that same interval does exist? I do not see the logic.
 
Physics news on Phys.org
sweatingbear said:
How come the area below the graph of $$\frac 1x$$ between $$[1, \infty)$$ does not exist, but the solid of revolution below the same graph in that same interval does exist? I do not see the logic.

The calculus demonstrates that the area... $\displaystyle A = \int_{1}^{\infty} \frac{dx}{x} = \infty$ (1)

...and the volume of its the solid of revolution is finite and is... $\displaystyle V= \pi\ \int_{1}^{\infty} \frac{dx}{x^{2}} = \pi$ (2)

It may seem incredible but this 'paradox' was discovered by Pappo di Alessandria in the fourth century after Christ [!]...Kind regards $\chi$ $\sigma$
 
Thank you both for your replies.
 
Thread 'Problem with calculating projections of curl using rotation of contour'
Hello! I tried to calculate projections of curl using rotation of coordinate system but I encountered with following problem. Given: ##rot_xA=\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}=0## ##rot_yA=\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}=1## ##rot_zA=\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}=0## I rotated ##yz##-plane of this coordinate system by an angle ##45## degrees about ##x##-axis and used rotation matrix to...

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 24 ·
Replies
24
Views
3K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
6
Views
3K
Replies
11
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K