Mass and center of mass of a lamina
Problem 11.216 · medium
A lamina occupies the triangle with vertices \( \displaystyle (0, 0) \), \( \displaystyle (1, 0) \), \( \displaystyle (0, 1) \) with density \( \displaystyle \rho(x, y) = x^{2} + y^{2} \). Find its mass and center of mass.
- \[ \int\limits_{0}^{1}\int\limits_{0}^{1 - x} \left(x^{2} + y^{2}\right)\, dy\, dx = \frac{1}{6} \]Mass m = ∬ ρ dA.✓ Proved
- \[ \left[\begin{matrix}\int\limits_{0}^{1}\int\limits_{0}^{1 - x} x \left(x^{2} + y^{2}\right)\, dy\, dx\\\int\limits_{0}^{1}\int\limits_{0}^{1 - x} y \left(x^{2} + y^{2}\right)\, dy\, dx\end{matrix}\right] = \left[\begin{matrix}\frac{1}{15}\\\frac{1}{15}\end{matrix}\right] \]Moments M_y = ∬ xρ dA and M_x = ∬ yρ dA.✓ Proved
- \[ \left[\begin{matrix}\frac{2}{5}\\\frac{2}{5}\end{matrix}\right] \]x̄ = M_y/m, ȳ = M_x/m.✓ Proved
Answer \( m = \frac{1}{6},\ (\bar x, \bar y) = \left(\frac{2}{5}, \frac{2}{5}\right) \)
✓ 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 over the specified triangular region, and the final calculations for the center of mass are correct.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly sets up the integrals for mass and moments over the specified triangular region, and the final calculations for the center of mass are correct.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly sets up the integrals for mass and moments over the specified triangular region, and the algebraic results 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/lamina_mass_center, checked 2026-10-04 with SymPy 1.14.0.