Show that the divergence of the curl of a vector is zero.

Let \(\vec A=A_x\hat i+A_y\hat j+A_z\hat k\).

\[ \nabla\times\vec A=\Big(\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}\Big)\hat i+\Big(\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}\Big)\hat j+\Big(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}\Big)\hat k \]

Taking the divergence:

\[ \nabla\cdot(\nabla\times\vec A)=\frac{\partial^2 A_z}{\partial x\,\partial y}-\frac{\partial^2 A_y}{\partial x\,\partial z}+\frac{\partial^2 A_x}{\partial y\,\partial z}-\frac{\partial^2 A_z}{\partial y\,\partial x}+\frac{\partial^2 A_y}{\partial z\,\partial x}-\frac{\partial^2 A_x}{\partial z\,\partial y} \]

For well-behaved (continuous second derivative) functions the order of differentiation does not matter, so the terms cancel in pairs: \(\frac{\partial^2 A_z}{\partial x\partial y}=\frac{\partial^2 A_z}{\partial y\partial x}\), and likewise for the others.

\[ \therefore\ \boxed{\nabla\cdot(\nabla\times\vec A)=0} \]

Hence the divergence of the curl of any vector is zero. Physical meaning: a curl field has no sources or sinks (for example \(\nabla\cdot\vec B=0\) is consistent with \(\vec B=\nabla\times\vec A\)).