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Gabriel's horn has finite improper volume $π$

Example

Gabriel's horn is obtained by revolving $y = 1 / x$ for $x \geq 1$ about the $x$ -axis. Its improper volume exists and equals $π$ .

Facts & Assumptions

Given: For $R > 1$ , the truncation obtained by revolving $f (x) = 1 / x$ on $[1, R]$ .

[F1]

A solid of revolution with profile $f$ has volume $π \int_{a}^{b} f (x)^{2} d x$ (The disc formula for the volume of a solid of revolution).

[F2]

For rational $p > 1$ , $\int_{1}^{\infty} x^{- p} d x = 1 / (p - 1)$ (The improper $p$ -test for rational exponents).

Verification

technique · direct

1.1

By [F1], the truncation has volume $π \int_{1}^{R} x^{- 2} d x = π (1 - 1 / R)$ .

2.1

By [F2] with $p = 2$ , the improper integral tends to $1$ , so the truncated volumes tend to $π$ .

step 1.1F2

3.1

Thus the horn has finite improper volume $π$ . No assertion about its lateral surface area is used here.

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

21 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

APEX Calculus II, §7.4, Example 216 (standard reference, not scraped)

Gabriel's horn has finite improper volume ππ