Line integrals of scalar functions
Problem 12.314 · medium
Evaluate \( \displaystyle \int_C x^{2} + y^{2} + z \, ds \), where \( \displaystyle C \) is the segment from (0, 1, 0) to (1, 0, 0).
- Parametrize: r(t) = <t, 1 - t, 0>, 0 ≤ t ≤ 1; ds = ‖r′(t)‖ dt.Reviewed
- \[ \sqrt{\left(\frac{d}{d t} 0\right)^{2} + \left(\frac{d}{d t} t\right)^{2} + \left(\frac{d}{d t} \left(1 - t\right)\right)^{2}} = \sqrt{2} \]‖r′(t)‖.✓ Proved
- \[ \int\limits_{0}^{1} \sqrt{2} \left. x^{2} + y^{2} + z \right|_{\substack{ x=t\\ y=1 - t\\ z=0 }}\, dt = \frac{2 \sqrt{2}}{3} \]∫ f(r(t)) ‖r′(t)‖ dt.✓ Proved
Answer \( \frac{2 \sqrt{2}}{3} \)
✓ 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 | f at each segment's midpoint times its length, over a 20,000-segment polyline |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The parametrization correctly models the line segment, the arc length element is calculated correctly, and the integration yields the stated answer.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-17 — The parametrization correctly models the line segment, the arc length element is calculated correctly, and the integration yields the stated answer.gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: pass 2026-10-15 — The parametrization correctly describes the line segment from (0, 1, 0) to (1, 0, 0). The calculation of the arc length element ds and the subsequent integration are correct.gpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: pass 2026-10-14 — The parametrization correctly describes the line segment from (0, 1, 0) to (1, 0, 0). The magnitude of the derivative is correctly calculated as sqrt(2), and the resulting integral evaluation is correct.gpt-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/line_integral_scalar, checked 2026-10-17 with SymPy 1.14.0.