Area and surface area with parametric curves
Problem 8.250 · medium
Find the area under the curve \( \displaystyle x = t^{3},\ y = t^{2} \) for \( \displaystyle 0 \le t \le 1 \) (above the \( \displaystyle x \)-axis).
- x increases with t here, so A = ∫ y dx = ∫ y(t) x′(t) dt.Reviewed
- \[ t^{2} \frac{d}{d t} t^{3} = 3 t^{4} \]y·x′(t).✓ Proved
- \[ \int\limits_{0}^{1} 3 t^{4}\, dt = \frac{3}{5} \]Integrate.✓ Proved
Answer \( \frac{3}{5} \)
✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify
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 |
| 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 | the curve as a 20,000-segment polyline: trapezoids under it, or the frusta it sweeps |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-17gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: pass 2026-10-15gpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: pass 2026-10-14gpt-oss:20b: pass 2026-10-14
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/parametric_area_surface, checked 2026-10-17 with SymPy 1.14.0.