Surface integrals
Problem 12.445 · hard
Evaluate \( \displaystyle \iint_S 1 \, dS \), where \( \displaystyle S \) is the part of the plane \( \displaystyle x + y + z = 3 \) in the first octant.
- Parametrize S by r(u, v) = <u, v, -u - v + 3>; dS = ‖r_u × r_v‖ du dv.Reviewed
- \[ \left(\sqrt{\left(0 \left(-1\right) + 1\right)^{2} + \left(1 - 1 \cdot 0\right)^{2} + \left(0 \cdot 0 + 1 \cdot 1\right)^{2}}\right)^{2} = 3 \]‖r_u × r_v‖² = (sqrt(3))².✓ Proved
- \[ \int\limits_{0}^{3}\int\limits_{0}^{3 - u} \sqrt{3}\, dv\, du = \frac{9 \sqrt{3}}{2} \]∬ f(r(u, v)) ‖r_u × r_v‖ du dv.✓ Proved
Answer \( \frac{9 \sqrt{3}}{2} \)
Lines: 2 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. 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 times area summed over a triangulated mesh of the surface |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The parametrization, surface element calculation, and integration bounds are all correct. The solution is sound.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-18 — The parametrization, surface element calculation, and integration bounds are all correct. The solution is sound.gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: fail (error) 2026-10-16 — The parametrization r(u, v) = <u, v, -u - v + 3> is incorrect because it yields z = 3 - u - v, which does not satisfy the plane equation x + y + z = 3 (it satisfies x + y + z = 3 - 2z, or rather, substituting gives u + v + (3 - u - v) = 3, wait. Let's re-check. x=u, y=v, z=3-u-v. Sum = u+v+3-u-v = 3. The parametrization is actually correct. Let me re-read carefully. r(u,v) = <u, v, 3-u-v>. The text says <u, v, -u - v + 3>. This is the same. So the parametrization is correct. The normal vector calculation: r_u = <1, 0, -1>, r_v = <0, 1, -1>. Cross product = <1, 1, 1>. Norm = sqrt(3). This is correct. The bounds: u from 0 to 3, v from 0 to 3-u. This is correct for the first octant. The integral is sqrt(3) * Area(Triangle). Area = 1/2 * 3 * 3 = 9/2. Result 9*sqrt(3)/2. The solution seems correct. Why would it fail? Let me look closer at the sentence. "Parametrize S by r(u, v) = <u, v, -u - v + 3>". This is fine. "dS = ‖r_u × r_v‖ du dv". This is fine. The equations are marked proved. Is there a subtle error? Ah, look at the cross product calculation in line 2. It calculates `(sqrt((0*(-1) + 1)**2 + (1 - 1*0)**2 + (0*0 + 1*1)**2))**2`. This looks like a garbled attempt to calculate the norm squared. The components of the cross product are 1, 1, 1. The norm squared is 3. The expression inside the sqrt in line 2 is `(0*(-1) + 1)**2` which is `1^2=1`. `(1 - 1*0)**2` which is `1^2=1`. `(0*0 + 1*1)**2` which is `1^2=1`. Sum is 3. Sqrt(3) squared is 3. The equation says `(sqrt(3))**2 = (sqrt(3))**2`. This is a tautology and doesn't actually show the derivation from the cross product components, but it is marked 'proved' by a CAS, so I shouldn't check the algebra. However, the prompt says "Check that... each conclusion follows from the lines before it". The sentence in line 1 sets up the method. Line 2 is an equation. Line 3 is an equation. The solution is technically sound. Wait, is there a sign error? No. Is the region correct? First octant means x>=0, y>=0, z>=0. u>=0, v>=0, 3-u-v>=0 => u+v<=3. Correct. Why would this be a fail? Maybe the parametrization is not one-to-one? It is. Maybe the orientation? dS is scalar, so orientation doesn't matter. Let me look really closely at the parametrization string: `<u, v, -u - v + 3>`. This is correct. Let me look at the bounds in line 3: `(v, 0, 3 - u), (u, 0, 3)`. This is correct. Is it possible the problem implies a specific orientation for a vector surface integral? No, it's `1 dS`, a scalar surface integral. The answer is correct. The steps are correct. Is there a style issue? The sentence in line 1 is clear. The equations are verified. I will mark it pass. Wait, let me double check the "adversarial" nature. Is there a trap? `r(u, v) = <u, v, -u - v + 3>`. If u=0, v=0, z=3. If u=3, v=0, z=0. If u=0, v=3, z=0. This traces the triangle with vertices (0,0,3), (3,0,0), (0,3,0). This is the part of the plane in the first octant. Everything seems correct. I will output pass.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: pass 2026-10-15 — The parametrization, surface element calculation, and integration bounds are all correct. The solution correctly models the region in the first octant and computes the area.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 2026-10-18 with SymPy 1.14.0.