Divergence of a Curl Proof Not coming out to 0

Question

In all physics courses we are taught that the divergence of a curl is always zero: $\nabla \cdot(\nabla \times\vec{V}) = 0$

So to prove this to myself I simply solve it to get $0$, but I am not coming up to zero. Can someone please point out where my mistake is?

\begin{align*} \nabla \times \vec{V}&= \begin{vmatrix}\hat{x} &\hat{y} &\hat{z} \\\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ V_x & V_y & V_z \end{vmatrix}\\ &=\left( \frac{\partial}{\partial y}V_z - \frac{\partial}{\partial z}V_y\right)\hat{x} + \left( \frac{\partial}{\partial x}V_z - \frac{\partial}{\partial z}V_x\right)\hat{y} + \left( \frac{\partial}{\partial x}V_y - \frac{\partial}{\partial y}V_x\right)\hat{z}\\ &=\frac{\partial}{\partial x}\left(\frac{\partial}{\partial y}V_z - \frac{\partial}{\partial z}V_y\right)\hat{x} + \frac{\partial}{\partial y}\left( \frac{\partial}{\partial x}V_z - \frac{\partial}{\partial z}V_x\right)\hat{y} + \frac{\partial}{\partial z}\left( \frac{\partial}{\partial x}V_y - \frac{\partial}{\partial y}V_x\right)\hat{z}\\ &= \frac{\partial^2 V_z}{\partial x \partial y} - \frac{\partial^2 V_y}{\partial x \partial z} + \frac{\partial^2 V_z}{\partial y \partial x} - \frac{\partial^2 V_x}{\partial y \partial z} + \frac{\partial^2 V_y}{\partial x \partial z} - \frac{\partial^2 V_x}{\partial z \partial y} \end{align*}

Only one term cancels out the $\frac{\partial^2 V_y}{\partial x \partial z}$? Leaving me with:

$$ \nabla\cdot(\nabla \times\vec{V}) = 2\left (\frac{\partial^2 V_z}{\partial x \partial y} - \frac{\partial^2 V_x}{\partial z \partial y}\right ). $$

I know I have screwed up somewhere, but I have been staring at my work for a while now and I can't seem to figure out where I went wrong. All terms should cancel out...

Thank you.

Answer 1

The correct form for curl is $$ \nabla \times \vec{V}=\left( \frac{\partial}{\partial y}V_z - \frac{\partial}{\partial z}V_y\right)\hat{x} + \underbrace{\left(\frac{\partial}{\partial z}V_x - \frac{\partial}{\partial x}V_z\right)}_{\text{note the switch}}\hat{y} + \left( \frac{\partial}{\partial x}V_y - \frac{\partial}{\partial y}V_x\right)\hat{z} $$ Another version of curl (which is convenient for this proof) is $$ \nabla \times \vec{V} = \hat x \times \frac{\partial V}{\partial x} + \hat y \times \frac{\partial V}{\partial y} + \hat z \times \frac{\partial V}{\partial z} $$ Similarly, divergence is given as $$ \nabla \cdot \vec{V} = \hat x \cdot \frac{\partial V}{\partial x} + \hat y \cdot \frac{\partial V}{\partial y} + \hat z \cdot \frac{\partial V}{\partial z} $$

Answer 2

You have a sign error in your middle term in $\nabla \times \vec{V}$ (the coefficient of $\hat{y}$). If you're calculating the determinant by expansion by minors, remember that the signs of the terms need to alternate; see eq. (2) in the above link.