Divergence and curl at a point
Problem 12.488 · easy
Find the divergence and the curl of \( \displaystyle \mathbf F = \left\langle x^{2} y, - 2 y z, x z^{2} \right\rangle \) at \( \displaystyle (0, 1, 2) \).
- \[ \left. \frac{\partial}{\partial z} x z^{2} + \frac{\partial}{\partial x} x^{2} y + \frac{\partial}{\partial y} \left(- 2 y z\right) \right|_{\substack{ x=0\\ y=1\\ z=2 }} = -4 \]div F = P_x + Q_y + R_z, at the point.✓ Proved
- \[ \left[\begin{matrix}\left. \frac{\partial}{\partial y} x z^{2} - \frac{\partial}{\partial z} \left(- 2 y z\right) \right|_{\substack{ x=0\\ y=1\\ z=2 }}\\\left. - \frac{\partial}{\partial x} x z^{2} + \frac{\partial}{\partial z} x^{2} y \right|_{\substack{ x=0\\ y=1\\ z=2 }}\\\left. - \frac{\partial}{\partial y} x^{2} y + \frac{\partial}{\partial x} \left(- 2 y z\right) \right|_{\substack{ x=0\\ y=1\\ z=2 }}\end{matrix}\right] = \left[\begin{matrix}2\\-4\\0\end{matrix}\right] \]curl F = ⟨R_y − Q_z, P_z − R_x, Q_x − P_y⟩, at the point.✓ Proved
Answer \( \nabla\cdot\mathbf F = -4,\quad \nabla\times\mathbf F = \left\langle 2, -4, 0 \right\rangle \)
Every line of this solution was proved by the computer algebra system SymPy. 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 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | ✓ 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 | central differences of F at the point |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-21gpt-oss:20b: pass 2026-10-21qwen3.6:27b-mlx: pass 2026-10-20 — The solution correctly computes the divergence and curl of the vector field and evaluates them at the specified point. The algebraic steps are verified as correct.gpt-oss:20b: pass 2026-10-20qwen3.6:27b-mlx: fail (error) 2026-10-20 — The curl calculation in line 2 uses an incorrect formula. The standard definition is \nabla \times \mathbf{F} = \langle R_y - Q_z, P_z - R_x, Q_x - P_y \rangle. The solution computes the z-component as -P_y + Q_x (which is correct) but the y-component as -R_x + P_z (correct) and the x-component as R_y - Q_z (correct). Wait, let me re-read the matrix carefully. The matrix in line 2 is: Row 1: Subs(Derivative(x*z**2, y) - Derivative(-2*y*z, z), ...) -> R_y - Q_z. Correct for x-comp. Row 2: Subs(-Derivative(x*z**2, x) + Derivative(x**2*y, z), ...) -> -R_x + P_z. Correct for y-comp. Row 3: Subs(-Derivative(x**2*y, y) + Derivative(-2*y*z, x), ...) -> -P_y + Q_x. Correct for z-comp. Let's check the values: F = <x^2y, -2yz, xz^2> P = x^2y, Q = -2yz, R = xz^2 Div: P_x + Q_y + R_z = 2xy + (-2z) + 2xz. At (0,1,2): 0 - 4 + 0 = -4. Correct. Curl: x-comp: R_y - Q_z = 0 - (-2y) = 2y. At (0,1,2): 2(1) = 2. Correct. y-comp: P_z - R_x = 0 - z^2 = -z^2. At (0,1,2): -4. Correct. z-comp: Q_x - P_y = 0 - x^2 = -x^2. At (0,1,2): 0. Correct. The result <2, -4, 0> is correct. The formula stated in the comment is <R_y - Q_z, P_z - R_x, Q_x - P_y>. This is the standard formula. The code implements it correctly. Why did I think it was wrong? Let's look closer at the code vs the comment. Comment: curl F = <R_y - Q_z, P_z - R_x, Q_x - P_y> Code Row 1: R_y - Q_z. Correct. Code Row 2: -R_x + P_z. This is P_z - R_x. Correct. Code Row 3: -P_y + Q_x. This is Q_x - P_y. Correct. The solution is actually correct. I will mark it as pass.gpt-oss:20b: pass 2026-10-20
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/div_curl_point, checked 2026-10-21 with SymPy 1.14.0.