Surface integrals
Problem 12.451 · hard
Evaluate \( \displaystyle \iint_S x + y + z \, dS \), where \( \displaystyle S \) is the part of the plane \( \displaystyle x + y + z = 1 \) in the first octant.
- Parametrize S by r(u, v) = <u, v, -u - v + 1>; dS = ‖r_u × r_v‖ du dv.
- \[ \left(\sqrt{\left(0 \left(-1\right) + 1\right)^{2} + \left(1 - 1 \cdot 0\right)^{2} + \left(0 \cdot 0 + 1 \cdot 1\right)^{2}}\right)^{2} = 3 \]‖r_u × r_v‖² = (sqrt(3))².✓ Proved
- \[ \int\limits_{0}^{1}\int\limits_{0}^{1 - u} \sqrt{3}\, dv\, du = \frac{\sqrt{3}}{2} \]∬ f(r(u, v)) ‖r_u × r_v‖ du dv.✓ Proved
Answer \( \frac{\sqrt{3}}{2} \)
Lines: 2 proved, 1 not checked. 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 | Not checked | — | 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 times area summed over a triangulated mesh of the surface |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The parametrization r(u, v) = <u, v, 1 - u - v> is incorrect for the first octant because it allows z to be negative (when u+v > 1). The correct bounds for u and v must ensure z >= 0, i.e., u >= 0, v >= 0, and u + v <= 1. The solution uses bounds 0 to 1 for both, integrating over a square that includes regions where z < 0, which is not part of the surface S.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-19 — The parametrization r(u, v) = <u, v, 1 - u - v> is incorrect for the first octant because it allows z to be negative (when u+v > 1). The correct bounds for u and v must ensure z >= 0, i.e., u >= 0, v >= 0, and u + v <= 1. The solution uses bounds 0 to 1 for both, integrating over a square that includes regions where z < 0, which is not part of the surface S.gpt-oss:20b: pass 2026-10-19qwen3.6:27b-mlx: fail (error) 2026-10-16 — The parametrization r(u, v) = <u, v, -u - v + 1> is incorrect for the plane x + y + z = 1; it should be r(u, v) = <u, v, 1 - u - v>. Consequently, the normal vector and the integrand substitution are wrong.gpt-oss:20b: fail (error) 2026-10-16 — The solution omits the factor x+y+z in the integrand; it integrates only the surface element √3 over the triangular region, which is not the requested surface integral.qwen3.6:27b-mlx: fail (error) 2026-10-15 — The parametrization r(u, v) = <u, v, -u - v + 1> is incorrect for the plane x + y + z = 1; it should be r(u, v) = <u, v, 1 - u - v>. The given parametrization corresponds to x + y - z = 1, leading to the wrong normal vector and incorrect integral evaluation.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/surface_integral, checked 2026-10-19 with SymPy 1.14.0.