Mass and center of mass of a lamina
Problem 11.264 · medium
A lamina occupies the rectangle \( \displaystyle 0 \le x \le 1 \), \( \displaystyle 0 \le y \le 1 \) with density \( \displaystyle \rho(x, y) = 2 x \). Find its mass and center of mass.
- \[ \int\limits_{0}^{1}\int\limits_{0}^{1} 2 x\, dy\, dx = 1 \]Mass m = ∬ ρ dA.✓ Proved
- \[ \left[\begin{matrix}\int\limits_{0}^{1}\int\limits_{0}^{1} 2 x^{2}\, dy\, dx\\\int\limits_{0}^{1}\int\limits_{0}^{1} 2 x y\, dy\, dx\end{matrix}\right] = \left[\begin{matrix}\frac{2}{3}\\\frac{1}{2}\end{matrix}\right] \]Moments M_y = ∬ xρ dA and M_x = ∬ yρ dA.✓ Proved
- \[ \left[\begin{matrix}\frac{2}{3}\\1 \cdot \frac{1}{2}\end{matrix}\right] = \left[\begin{matrix}\frac{2}{3}\\\frac{1}{2}\end{matrix}\right] \]x̄ = M_y/m, ȳ = M_x/m.✓ Proved
Answer \( m = 1,\ (\bar x, \bar y) = \left(\frac{2}{3}, \frac{1}{2}\right) \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. 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 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ 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 double integrals |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly sets up the integrals for mass and moments, computes them accurately, and applies the center of mass formulas correctly.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: pass 2026-10-06 — The solution correctly sets up the integrals for mass and moments, computes them accurately, and applies the center of mass formulas correctly.gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: pass 2026-10-06 — The solution correctly sets up the integrals for mass and moments over the specified rectangular region with the given density function. The calculations and final results are correct.
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/lamina_mass_center, checked 2026-10-06 with SymPy 1.14.0.