Line integrals and work
Problem 12.110 · medium
Find the work done by \( \displaystyle \mathbf{F} = \langle - y, x \rangle \) along the line segment from (0, 0) to (2, 1).
- Parametrize: r(t) = ⟨2*t, t⟩ for 0 ≤ t ≤ 1.Reviewed
- \[ \left[\begin{matrix}\frac{d}{d t} 2 t\\\frac{d}{d t} t\end{matrix}\right] = \left[\begin{matrix}2\\1\end{matrix}\right] \]r'(t).✓ Proved
- \[ 0 \]F(r(t)) · r'(t).✓ Proved
- \[ \int\limits_{0}^{1} 0\, dt = 0 \]W = ∫ F · dr.✓ Proved
Answer \( 0 \)
Lines: 3 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt.
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 |
| 4 | ✓ 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 | a 20,000-chord polygon along the path gives the same work |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The parametrization is correct, the dot product calculation is correct, and the integration yields the correct result.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-27 — The parametrization is correct, the dot product calculation is correct, and the integration yields the correct result.gpt-oss:20b: pass 2026-09-27qwen3.6:27b-mlx: fail (error) 2026-09-27 — The dot product calculation in step 3 is incorrect. F(r(t)) = <-t, 2t> and r'(t) = <2, 1>, so F · r' = -2t + 2t = 0 is actually correct, wait. Let me re-evaluate. F = <-y, x>. r(t) = <2t, t>. So x=2t, y=t. F(r(t)) = <-t, 2t>. r'(t) = <2, 1>. Dot product: (-t)(2) + (2t)(1) = -2t + 2t = 0. The integral of 0 is 0. The answer 0 is correct. Why did I think it was wrong? Let me check the problem statement again. F = <-y, x>. Path (0,0) to (2,1). Parametrization r(t) = <2t, t>. F(r(t)) = <-t, 2t>. r'(t) = <2, 1>. Dot product is 0. The work is 0. The solution is correct. Wait, is the field conservative? curl F = d/dx(x) - d/dy(-y) = 1 - (-1) = 2. Not conservative. But the work along this specific line is 0. The solution is correct. I will mark it as pass.gpt-oss:20b: pass 2026-09-27
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_work, checked 2026-09-27 with SymPy 1.14.0.