Line integrals of scalar functions
Problem 12.418 · medium
Evaluate \( \displaystyle \int_C z \, ds \), 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
- \[ \int\limits_{0}^{1} \left. z \right|_{\substack{ x=t\\ y=2 t\\ z=2 t }} 3\, dt = 3 \]∫ f(r(t)) ‖r′(t)‖ dt.✓ Proved
Answer \( 3 \)
✓ 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 is correct for the segment from the origin to (1, 2, 2). The magnitude of the derivative vector is correctly calculated as 3. The integral setup and evaluation are correct.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-18 — The parametrization is correct for the segment from the origin to (1, 2, 2). The magnitude of the derivative vector is correctly calculated as 3. The integral setup and evaluation are correct.gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: pass 2026-10-16 — The parametrization is correct for the segment from the origin to (1, 2, 2). The calculation of the magnitude of the derivative vector is correct, and the resulting integral evaluation is accurate.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: pass 2026-10-15 — The parametrization is correct for the segment from the origin to (1, 2, 2). The magnitude of the derivative vector is correctly calculated as 3. The integral setup and evaluation are correct.gpt-oss:20b: pass 2026-10-15
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 2026-10-18 with SymPy 1.14.0.