Conceptual Question of Area and Volume

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SUMMARY

The area under the curve of the function y=1/x from 1 to infinity diverges, resulting in an infinite area (A = ∫_1^{∞} (1/x) dx = ∞). In contrast, when this area is revolved around the y-axis, the volume converges to π (V = ∫_1^{∞} π(1/x)² dx = π). This illustrates that while volume is one dimension higher than area, it does not imply that volume is always larger; the convergence rates of the respective integrals differ significantly.

PREREQUISITES
  • Understanding of integral calculus, specifically improper integrals.
  • Familiarity with the concept of volume of revolution in calculus.
  • Knowledge of logarithmic functions and their properties.
  • Basic understanding of limits in mathematical analysis.
NEXT STEPS
  • Study the properties of improper integrals and their convergence.
  • Learn about the method of disks and washers for calculating volumes of revolution.
  • Explore the relationship between dimensions in geometry, particularly in calculus.
  • Investigate logarithmic functions and their applications in calculus.
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Students of mathematics, educators teaching calculus, and anyone interested in the concepts of area and volume in mathematical analysis.

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Why the area under the curve of the function y=1/x form 1 to inf is infinite. But if we take this area and revolve it around the y-axis we obtain a volume of pi ?
 
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Because the volume of this rotated solid converges while its surface area diverges. Just because Volume is "one-dimension more" than Area doesn't mean it's "larger." For example, a 0.1m by 0.1m square has an area of 0.01 m^2, but a 0.1 x 0.1 x 0.1 cube has a volume of 0.001 m^3. In your problem, the V_n simple converges quicker than A_n.

Your topic heading said "conceptual question," but here's the math explanation as well if you're curious:

[tex]A = \int_1^{\infty} \frac{dx}{x} \ = \ {lim}_{b \rightarrow \ \infty} ( \ ln|x|_1^{b} \ ) = b - 0 = b = \infty.[/tex]

On the other hand,

[tex]V = \int_1^{\infty} \pi(\frac{1}{x})^2 * dx = \pi \int_1^{\infty}\frac{dx}{x^2} = \pi * {lim}_{b \rightarrow \ \infty} [ \ -x^{-1}|_1^{b} \ ] = 0 - (-\pi) = \pi.[/tex]
 
Well that makes perfect sense. I knew about the mathematical explanation but was looking for the other explanation you provided. Thank you very much sir.
 

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