Mass from a density
Problem 5.237 · easy
A thin disk of radius 4 cm has density \( \displaystyle \rho(r) = r + 5 \) 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
- \[ \int\limits_{0}^{4} 2 \pi r \left(r + 5\right)\, dr = \frac{368 \pi}{3} \]Add up the rings.✓ Proved
Answer \( \frac{368 \pi}{3} \approx 385.37 \)
✓ 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 solution correctly sets up the mass integral using polar coordinates, where the area element is 2πr dr. The integration limits and density function are applied correctly.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly sets up the mass integral using polar coordinates, where the area element is 2πr dr. The integration limits and density function are applied correctly.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: pass 2026-10-04 — The setup correctly identifies the differential mass element for a disk with radial symmetry, and the integration bounds and result are correct.gpt-oss:20b: pass 2026-10-04
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 2026-10-04 with SymPy 1.14.0.