Home›Calculus 1›Volumes by cylindrical shells›Problem 5.103

Volumes by cylindrical shells

Problem 5.103 · easy

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

Answer \( 4 \pi \)

Lines: 1 proved, 1 reviewed. 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.

The full receipt

Line	Status	Checked by	Detail
1	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	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: pass — The setup correctly identifies the radius, height, and thickness for the shell method. The integration bounds and result are correct.

Every verdict on record (4)

qwen3.6:27b-mlx: pass 2026-09-26 — The setup correctly identifies the radius, height, and thickness for the shell method. The integration bounds and result are correct.
gpt-oss:20b: pass 2026-09-26
qwen3.6:27b-mlx: fail (error) 2026-09-26 — The solution sets up the integral for the volume but fails to evaluate it. The integral of 4πx from 0 to 1 is 2π, not 4π. The stated answer is incorrect.
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 2026-09-26 with SymPy 1.14.0.