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Divergence and curl at a point

Problem 12.340 · easy

Find the divergence and the curl of \( \displaystyle \mathbf F = \left\langle \sin{\left(y \right)}, \cos{\left(x \right)}, x y z \right\rangle \) at \( \displaystyle (2, -1, 3) \).
  1. \[ \left. \frac{\partial}{\partial z} x y z + \frac{d}{d x} \sin{\left(y \right)} + \frac{d}{d y} \cos{\left(x \right)} \right|_{\substack{ x=2\\ y=-1\\ z=3 }} = -2 \]
    div F = P_x + Q_y + R_z, at the point.✓ Proved
  2. \[ \left[\begin{matrix}\left. \frac{\partial}{\partial y} x y z - \frac{d}{d z} \cos{\left(x \right)} \right|_{\substack{ x=2\\ y=-1\\ z=3 }}\\\left. - \frac{\partial}{\partial x} x y z + \frac{d}{d z} \sin{\left(y \right)} \right|_{\substack{ x=2\\ y=-1\\ z=3 }}\\\left. - \frac{d}{d y} \sin{\left(y \right)} + \frac{d}{d x} \cos{\left(x \right)} \right|_{\substack{ x=2\\ y=-1\\ z=3 }}\end{matrix}\right] = \left[\begin{matrix}6\\3\\- \sin{\left(2 \right)} - \cos{\left(1 \right)}\end{matrix}\right] \]
    curl F = ⟨R_y − Q_z, P_z − R_x, Q_x − P_y⟩, at the point.✓ Proved
Answer \( \nabla\cdot\mathbf F = -2,\quad \nabla\times\mathbf F = \left\langle 6, 3, - \sin{\left(2 \right)} - \cos{\left(1 \right)} \right\rangle \)

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
LineStatusChecked byDetail
1✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0central differences of F at the point

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: pass
Every verdict on record (6)
  • qwen3.6:27b-mlx: pass 2026-10-17
  • gpt-oss:20b: pass 2026-10-17
  • qwen3.6:27b-mlx: fail (error) 2026-10-15 — The curl calculation contains sign errors in the z-component. The formula used is Q_x - P_y, but the code computes -P_y + Q_x (which is correct algebraically) yet substitutes Derivative(cos(x), x) for Q_x. Q is cos(x), so Q_x is -sin(x), not cos(x). The code computes Derivative(cos(x), x) which is -sin(x), but the label says Q_x - P_y. Wait, let's look closer. The code line is: `Subs(-Derivative(sin(y), y) + Derivative(cos(x), x)...)`. P=sin(y), Q=cos(x). P_y = cos(y). Q_x = -sin(x). The code computes -cos(y) + (-sin(x)). At (2,-1,3), this is -cos(-1) - sin(2) = -cos(1) - sin(2). The stated answer is `<6, 3, -sin(2) - cos(1)>`. This matches. However, the comment says `curl F = <R_y - Q_z, P_z - R_x, Q_x - P_y>`. The code for the z-component is `-Derivative(sin(y), y) + Derivative(cos(x), x)`. This is `-P_y + Q_x`. This is `Q_x - P_y`. This is correct. Let's re-read the first component. `R_y - Q_z`. R=xyz, Q=cos(x). R_y = xz. Q_z = 0. Code: `Derivative(x*y*z, y) - Derivative(cos(x), z)`. This is `xz - 0`. Correct. Second component: `P_z - R_x`. P=sin(y), R=xyz. P_z=0. R_x=yz. Code: `-Derivative(x*y*z, x) + Derivative(sin(y), z)`. This is `-yz + 0`. Correct. The solution is actually correct. Why did I think it was wrong? Let me re-verify the z-component comment vs code. Comment: `Q_x - P_y`. Code: `-P_y + Q_x`. These are equivalent. The values are correct. The setup is correct. The algebra is proved. I will mark it pass.
  • gpt-oss:20b: pass 2026-10-15
  • qwen3.6:27b-mlx: fail (error) 2026-10-14 — The curl calculation contains sign errors in the z-component. The formula used is Q_x - P_y, but the code computes -P_y + Q_x (which is correct) yet substitutes Derivative(cos(x), x) for Q_x (incorrect, Q_x is 0) and Derivative(sin(y), y) for P_y (incorrect, P_y is cos(y)). Specifically, the z-component should be Q_x - P_y = 0 - cos(y) = -cos(-1) = -cos(1). The solution computes -cos(1) - sin(2) by incorrectly including a sin(2) term from P_y? No, P_y = cos(y). The code computes -Derivative(sin(y), y) + Derivative(cos(x), x) = -cos(y) + (-sin(x)). At (2, -1, 3), this is -cos(-1) - sin(2) = -cos(1) - sin(2). The correct z-component is Q_x - P_y = 0 - cos(y) = -cos(1). The term -sin(2) comes from differentiating cos(x) with respect to x, which is Q_x? No, Q is cos(x), so Q_x is -sin(x). Wait. F = <sin(y), cos(x), xyz>. P=sin(y), Q=cos(x), R=xyz. Curl = <R_y - Q_z, P_z - R_x, Q_x - P_y>. R_y = xz, Q_z = 0 -> xz = 2*3 = 6. Correct. P_z = 0, R_x = yz -> -yz = -(-1)*3 = 3. Correct. Q_x = -sin(x), P_y = cos(y). Q_x - P_y = -sin(2) - cos(-1) = -sin(2) - cos(1). The solution's result is actually correct. Let me re-read the code comment vs code. Code: -Derivative(sin(y), y) + Derivative(cos(x), x). Derivative(sin(y), y) is cos(y). Derivative(cos(x), x) is -sin(x). So code computes -cos(y) + (-sin(x)) = -cos(y) - sin(x). This matches Q_x - P_y. The comment says Q_x - P_y. The result is correct. Why did I think it was wrong? Ah, I misread the vector components in my head. P=sin(y), Q=cos(x). P_y = cos(y). Q_x = -sin(x). Q_x - P_y = -sin(x) - cos(y). At (2, -1, 3): -sin(2) - cos(-1) = -sin(2) - cos(1). The solution gives -sin(2) - cos(1). It is correct. The divergence is 0 + 0 + xy = 2*(-1) = -2. Correct. The solution is actually correct. I will mark it pass.
  • 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/div_curl_point, checked 2026-10-17 with SymPy 1.14.0.

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