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Volumes by cylindrical shells

Problem 5.83 · easy

The region under \( \displaystyle y = \frac{1}{x} \) for \( \displaystyle 0 \le x \le 2 \) 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.
\[ \int\limits_{0}^{2} 2 \pi\, dx = 4 \pi \]
V = ∫ 2πx f(x) dx.✓ Proved

Answer \( 4 \pi \)

Lines: 1 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in 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: pass
qwen3.6:27b-mlx: fail (error) — The solution sets up the integral for revolving y=x around the y-axis, ignoring the given function y=1/x. The integrand should be 2πx(1/x) = 2π, not 2πx.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-09-28 — The solution sets up the integral for revolving y=x around the y-axis, ignoring the given function y=1/x. The integrand should be 2πx(1/x) = 2π, not 2πx.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: fail (error) 2026-09-27 — The solution uses the shell method formula but fails to include the function f(x) = 1/x in the integrand, effectively integrating 2πx instead of 2π. This results in an incorrect volume calculation.
gpt-oss:20b: fail (style) 2026-09-27 — [domain objection, downgraded to style] The problem is ill‑posed: the curve y=1/x is undefined at x=0, so the region under the curve from 0 to 2 does not exist (the area diverges). The solution ignores this and uses f(x) without specifying it, leading to an incorrect volume of 4π. The setup and calculation are therefore incorrect.
qwen3.6:27b-mlx: fail (error) 2026-09-26 — The solution sets up the integral for revolving y=x around the y-axis, not y=1/x. The height of the shell should be 1/x, making the integrand 2π, not 2πx.
gpt-oss:20b: pass 2026-09-26

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 2026-09-28 with SymPy 1.14.0.