Mass and center of mass of a lamina
Problem 11.214 · medium
A lamina occupies the triangle with vertices \( \displaystyle (0, 0) \), \( \displaystyle (3, 0) \), \( \displaystyle (0, 3) \) with density \( \displaystyle \rho(x, y) = x^{2} + y^{2} \). Find its mass and center of mass.
- \[ \int\limits_{0}^{3}\int\limits_{0}^{3 - x} \left(x^{2} + y^{2}\right)\, dy\, dx = \frac{27}{2} \]Mass m = ∬ ρ dA.✓ Proved
- \[ \left[\begin{matrix}\int\limits_{0}^{3}\int\limits_{0}^{3 - x} x \left(x^{2} + y^{2}\right)\, dy\, dx\\\int\limits_{0}^{3}\int\limits_{0}^{3 - x} y \left(x^{2} + y^{2}\right)\, dy\, dx\end{matrix}\right] = \left[\begin{matrix}\frac{81}{5}\\\frac{81}{5}\end{matrix}\right] \]Moments M_y = ∬ xρ dA and M_x = ∬ yρ dA.✓ Proved
- \[ \left[\begin{matrix}\frac{6}{5}\\\frac{6}{5}\end{matrix}\right] \]x̄ = M_y/m, ȳ = M_x/m.✓ Proved
Answer \( m = \frac{27}{2},\ (\bar x, \bar y) = \left(\frac{6}{5}, \frac{6}{5}\right) \)
✓ 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 algebraic results are verified as 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 algebraic results are verified as 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 computed values match the stated answer.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.