Line integrals of scalar functions
Problem 12.320 · medium
Evaluate \( \displaystyle \int_C x + y + z \, ds \), where \( \displaystyle C \) is the segment from the origin to (1, 2, 2).
- Parametrize: r(t) = <t, 2*t, 2*t>, 0 ≤ t ≤ 1; ds = ‖r′(t)‖ dt.Reviewed
- \[ \sqrt{\left(\frac{d}{d t} t\right)^{2} + 2 \left(\frac{d}{d t} 2 t\right)^{2}} = 3 \]‖r′(t)‖.✓ Proved
- \[ \int\limits_{0}^{1} \left. x + y + z \right|_{\substack{ x=t\\ y=2 t\\ z=2 t }} 3\, dt = \frac{15}{2} \]∫ f(r(t)) ‖r′(t)‖ dt.✓ Proved
Answer \( \frac{15}{2} \)
✓ 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
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 2026-10-14 — The parametrization correctly models the line segment from the origin to (1, 2, 2). The calculation of the magnitude of the derivative and the subsequent integration are 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.