Line integrals and work
Problem 12.83 · medium
Find the work done by \( \displaystyle \mathbf{F} = \langle x + y, x - y \rangle \) along the line segment from (0, 0) to (1, 1).
- Parametrize: r(t) = ⟨t, t⟩ for 0 ≤ t ≤ 1.Reviewed
- \[ \left[\begin{matrix}\frac{d}{d t} t\\\frac{d}{d t} t\end{matrix}\right] = \left[\begin{matrix}1\\1\end{matrix}\right] \]r'(t).✓ Proved
- \[ 2 t \]F(r(t)) · r'(t).Reviewed
- \[ \int\limits_{0}^{1} 2 t\, dt = 1 \]W = ∫ F · dr.✓ Proved
Answer \( 1 \)
Lines: 2 proved, 2 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. 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 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | both sides are the same expression: the line states a value, and the working that produced it is not on the line |
| 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 for the line segment from (0,0) to (1,1). The computation of F(r(t)) and the dot product with r'(t) are correct, and the integral evaluates correctly to 1.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-09-28 — The parametrization is correct for the line segment from (0,0) to (1,1). The computation of F(r(t)) and the dot product with r'(t) are correct, and the integral evaluates correctly to 1.gpt-oss:20b: pass 2026-09-28qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly parametrizes the line segment, computes the dot product, and evaluates the integral to find the work done.gpt-oss:20b: fail (error) 2026-09-26 — The dot product in step 3 is incorrect: F(r(t)) = ⟨2t,0⟩, so F·r' = 2t, not 2t,2t. This mis‑computes the integrand and leads to an incorrect work value.qwen3.6:27b-mlx: pass 2026-09-26 — The parametrization, substitution, and integration are all correct and lead to the right answer.gpt-oss:20b: fail (error) 2026-09-26 — The dot product in step 3 is incorrect: F(r(t)) = ⟨2t,0⟩, so F(r(t))·r'(t)=2t, not “2*t, 2*t”. This mis‑computes the integrand and leads to an incorrect work value.
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 2026-10-13 with SymPy 1.14.0.