Mass from a density
Problem 5.307 · easy
A thin disk of radius 1 cm has density \( \displaystyle \rho(r) = 3 r + 3 \) g/cm² at distance \( \displaystyle r \) from its center. Find its mass.
- A thin ring at radius r has area about 2πr·dr, so its mass is 2πr·ρ(r)·dr.Reviewed
- \[ \int\limits_{0}^{1} 2 \pi r \left(3 r + 3\right)\, dr = 5 \pi \]Add up the rings.✓ Proved
Answer \( 5 \pi \approx 15.708 \)
✓ Nihil obstat Lines: 1 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.
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 | numerical quadrature |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The setup correctly identifies the area element for a thin ring as 2πr dr and applies the given density function. The integration bounds and result are correct.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: pass 2026-10-06 — The setup correctly identifies the area element for a thin ring as 2πr dr and applies the given density function. The integration bounds and result are correct.gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: pass 2026-10-06 — The solution correctly sets up the mass integral using the area element for a thin ring, 2πr dr, and integrates over the correct radius bounds.
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/mass_density, checked 2026-10-06 with SymPy 1.14.0.