Divergence and curl
Problem 12.79 · easy
Find the divergence and curl of \( \displaystyle \mathbf{F} = \langle - 2 \sin{\left(y \right)}, - y z, z \rangle \).
- \[ \frac{d}{d z} z + \frac{\partial}{\partial y} \left(- y z\right) + \frac{d}{d x} \left(- 2 \sin{\left(y \right)}\right) = 1 - z \]div F.✓ Proved
- \[ \left[\begin{matrix}\frac{d}{d y} z - \frac{\partial}{\partial z} \left(- y z\right)\\- \frac{d}{d x} z + \frac{d}{d z} \left(- 2 \sin{\left(y \right)}\right)\\\frac{\partial}{\partial x} \left(- y z\right) - \frac{d}{d y} \left(- 2 \sin{\left(y \right)}\right)\end{matrix}\right] = \left[\begin{matrix}y\\0\\2 \cos{\left(y \right)}\end{matrix}\right] \]curl F, component by component.✓ Proved
Answer \( \nabla\cdot\mathbf{F} = 1 - z,\quad \nabla\times\mathbf{F} = \langle y, 0, 2 \cos{\left(y \right)} \rangle \)
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 explanation has not been reviewed yet.
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 | difference quotients of each component at (0.3, 0.7, 1.1) agree |
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/divergence_curl, checked 2026-09-26 with SymPy 1.14.0.