Discussion
The divergence of the curl of a twice continuously differentiable vector function is __________.
Answer: A
The divergence of a curl is 0.
For a twice continuously differentiable vector field \(\mathbf{F} = (F_{1}, F_{2}, F_{3})\),
\[ \begin{aligned} \nabla \cdot (\nabla \times \mathbf{F}) &= \partial_{x}(\partial_{y}F_{3} - \partial_{z}F_{2}) \\ &\quad + \partial_{y}(\partial_{z}F_{1} - \partial_{x}F_{3}) \\ &\quad + \partial_{z}(\partial_{x}F_{2} - \partial_{y}F_{1}) \end{aligned} \]
The mixed partial derivatives cancel in pairs, since \(\partial_{x}\partial_{y}F_{3} = \partial_{y}\partial_{x}F_{3}\) and so on (equality of mixed partials, which is why the field must be twice continuously differentiable). The result is identically zero, so 1, 2 and infinity are all wrong. A curl field is always solenoidal.
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