∫Calc Practice

Centers of mass and centroids

Problem 5.206 · medium

Find the centroid of the region bounded by \( \displaystyle y = x \) and \( \displaystyle y = x^{2} \).
  1. \[ \int\limits_{0}^{1} \left(- x^{2} + x\right)\, dx = \frac{1}{6} \]
    The area.✓ Proved
  2. \[ \int\limits_{0}^{1} x \left(- x^{2} + x\right)\, dx = \frac{1}{12} \]
    The moment about the y-axis.✓ Proved
  3. \[ \int\limits_{0}^{1} \left(- \frac{x^{4}}{2} + \frac{x^{2}}{2}\right)\, dx = \frac{1}{15} \]
    The moment about the x-axis: ∫ ½(top² − bottom²) dx.✓ Proved
  4. \[ \left[\begin{matrix}1 \cdot \frac{1}{2}\\\frac{2}{5}\end{matrix}\right] = \left[\begin{matrix}\frac{1}{2}\\\frac{2}{5}\end{matrix}\right] \]
    Divide each moment by the area.✓ Proved
Answer \( (\bar{x}, \bar{y}) = \left(\frac{1}{2}, \frac{2}{5}\right) \)

✓ 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
LineStatusChecked byDetail
1✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0numerical quadrature of each moment

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: pass
Every verdict on record (4)
  • qwen3.6:27b-mlx: pass 2026-10-04
  • gpt-oss:20b: pass 2026-10-04
  • qwen3.6:27b-mlx: pass 2026-10-04
  • 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/center_of_mass, checked 2026-10-04 with SymPy 1.14.0.