Surface integrals
Problem 12.357 · hard
Evaluate \( \displaystyle \iint_S 1 \, dS \), where \( \displaystyle S \) is the upper hemisphere \( \displaystyle x^2 + y^2 + z^2 = 1,\ z \ge 0 \).
- Parametrize S by r(u, v) = <sin(u)*cos(v), sin(u)*sin(v), cos(u)>; dS = ‖r_u × r_v‖ du dv.
- \[ \left(\sin{\left(u \right)} \sin^{2}{\left(v \right)} \cos{\left(u \right)} + \sin{\left(u \right)} \cos{\left(u \right)} \cos^{2}{\left(v \right)}\right)^{2} + \sin^{4}{\left(u \right)} \sin^{2}{\left(v \right)} + \sin^{4}{\left(u \right)} \cos^{2}{\left(v \right)} = \sin^{2}{\left(u \right)} \]‖r_u × r_v‖² = (sin(u))².✓ Proved
- \[ \int\limits_{0}^{2 \pi}\int\limits_{0}^{\frac{\pi}{2}} \sin{\left(u \right)}\, du\, dv = 2 \pi \]∬ f(r(u, v)) ‖r_u × r_v‖ du dv.✓ Proved
Answer \( 2 \pi \)
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: inconclusive — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The parametrization uses spherical coordinates where u is the polar angle. For the upper hemisphere (z >= 0), the polar angle u must range from 0 to
Every verdict on record (6)
qwen3.6:27b-mlx: inconclusive 2026-10-17 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The parametrization uses spherical coordinates where u is the polar angle. For the upper hemisphere (z >= 0), the polar angle u must range from 0 togpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: fail (style) 2026-10-15 — [domain objection, downgraded to style] The parameter bounds for u are incorrect; for the upper hemisphere, u ranges from 0 to pi/2, but the standard spherical parametrization usually defines u as the polar angle from the positive z-axis, so u should range from 0 to pi/2. Wait, let me re-evaluate. Standard spherical coordinates: x = sin(phi)cos(theta), y = sin(phi)sin(theta), z = cos(phi). Here u is phi. For the upper hemisphere (z >= 0), phi goes from 0 (north pole) to pi/2 (equator). The solution uses u in [0, pi/2]. This is correct for the upper hemisphere. Let me re-read carefully. Wait, the standard surface area of a unit hemisphere is 2*pi. The integral of sin(u) from 0 to pi/2 is [-cos(u)] from 0 to pi/2 = -(0 - 1) = 1. Then integrate 1 dv from 0 to 2pi gives 2pi. The calculation is correct. The bounds are correct. Why did I think it was wrong? Let's double check the parametrization. r(u,v) = <sin u cos v, sin u sin v, cos u>. r_u = <cos u cos v, cos u sin v, -sin u> r_v = <-sin u sin v, sin u cos v, 0> r_u x r_v = <sin^2 u cos v, sin^2 u sin v, sin u cos u>. Norm squared = sin^4 u (cos^2 v + sin^2 v) + sin^2 u cos^2 u = sin^4 u + sin^2 u cos^2 u = sin^2 u (sin^2 u + cos^2 u) = sin^2 u. Norm = sin u (since u in [0, pi/2], sin u >= 0). Integral is int_0^{2pi} int_0^{pi/2} sin u du dv = 2pi * 1 = 2pi. The solution is actually correct. I will mark it as pass.gpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: pass 2026-10-14 — The parametrization correctly covers the upper hemisphere with appropriate bounds, and the surface element calculation is correct.gpt-oss:20b: pass 2026-10-14
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 2026-10-17 with SymPy 1.14.0.