Divergence and curl
Problem 12.80 · easy
Find the divergence and curl of \( \displaystyle \mathbf{F} = \langle - 2 \sin{\left(y \right)}, 2 x y z, 2 \cos{\left(z \right)} \rangle \).
- \[ \frac{\partial}{\partial y} 2 x y z + \frac{d}{d x} \left(- 2 \sin{\left(y \right)}\right) + \frac{d}{d z} 2 \cos{\left(z \right)} = 2 x z - 2 \sin{\left(z \right)} \]div F.✓ Proved
- \[ \left[\begin{matrix}- \frac{\partial}{\partial z} 2 x y z + \frac{d}{d y} 2 \cos{\left(z \right)}\\\frac{d}{d z} \left(- 2 \sin{\left(y \right)}\right) - \frac{d}{d x} 2 \cos{\left(z \right)}\\\frac{\partial}{\partial x} 2 x y z - \frac{d}{d y} \left(- 2 \sin{\left(y \right)}\right)\end{matrix}\right] = \left[\begin{matrix}- 2 x y\\0\\2 y z + 2 \cos{\left(y \right)}\end{matrix}\right] \]curl F, component by component.✓ Proved
Answer \( \nabla\cdot\mathbf{F} = 2 x z - 2 \sin{\left(z \right)},\quad \nabla\times\mathbf{F} = \langle - 2 x y, 0, 2 y z + 2 \cos{\left(y \right)} \rangle \)
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.