Volumes by cylindrical shells
Problem 5.181 · easy
The region under \( \displaystyle y = \frac{1}{x} \) for \( \displaystyle 0 \le x \le 1 \) is revolved about the y-axis. Use shells to find the volume.
- A shell at x has radius x, height f(x) and thickness dx, so its volume is 2πx f(x) dx.
- \[ \int\limits_{0}^{1} 2 \pi\, dx = 2 \pi \]V = ∫ 2πx f(x) dx.✓ Proved
Answer \( 2 \pi \)
Lines: 1 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. A reviewer's objection to the explanation stands; see the receipt. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | 20,000 thin shells (annular slabs) summed numerically give the same volume |
Reviewers
gpt-oss:20b: fail (error) — The setup ignores that y=1/x blows up at x=0, so the shell integral ∫₀¹ 2πx·(1/x)dx = ∫₀¹ 2πdx diverges to infinity. The solution also omits the function f(x) in the integrand and incorrectly writes the integral as ∫₂π dx. Thus the conclusion that the volume is 2π is mathematically incorrect.qwen3.6:27b-mlx: fail (error) — The solution sets up the shell method formula but fails to substitute the function f(x) = 1/x into the integrand. The integral evaluated is of 2πx, not 2πx(1/x) = 2π, which leads to an incorrect volume of 2π instead of the correct value of π.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution sets up the shell method formula but fails to substitute the function f(x) = 1/x into the integrand. The integral evaluated is of 2πx, not 2πx(1/x) = 2π, which leads to an incorrect volume of 2π instead of the correct value of π.gpt-oss:20b: fail (error) 2026-10-04 — The setup ignores that y=1/x blows up at x=0, so the shell integral ∫₀¹ 2πx·(1/x)dx = ∫₀¹ 2πdx diverges to infinity. The solution also omits the function f(x) in the integrand and incorrectly writes the integral as ∫₂π dx. Thus the conclusion that the volume is 2π is mathematically incorrect.qwen3.6:27b-mlx: inconclusive 2026-09-29 — reviewer returned a non-objectgpt-oss:20b: pass 2026-09-29qwen3.6:27b-mlx: inconclusive 2026-09-29 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution sets up the integral for revolving y=1 around the y-axis, not y=1/x. The height of the shell should be f(x) = 1/x, making the integrandgpt-oss:20b: pass 2026-09-29
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/volume_shells, checked 2026-10-04 with SymPy 1.14.0.