Curl of a Gradient: Calculating in (x,y) Plane

[yungman]

Let $v(x,y)$ be function of (x,y) and not z.
$$\nabla v=\hat x \frac{\partial v}{\partial x}+\hat y \frac{\partial v}{\partial y}$$
$$\nabla \times \nabla v=\left|\begin{array} \;\hat x & \hat y & \hat z \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} & 0 \end{array}\right|=\hat x\left(-\frac{\partial^2 v}{\partial y\partial z}\right)-\hat y\left(-\frac{\partial^2 v}{\partial x\partial z}\right) +\hat z\left(\frac{\partial^2 v}{\partial y\partial x}-\frac{\partial^2 v}{\partial x\partial y} \right) =0$$

What did I do wrong?

 

[WannabeNewton]

What makes you think that's wrong? The curl of the gradient of a smooth scalar field $f:\mathbb{R}^{3}\rightarrow \mathbb{R}$ always vanishes: $(\nabla \times \nabla f)^{i} = \epsilon^{ijk}\partial_{j}\partial_{k}f = 0$.

 

[yungman]

What makes you think that's wrong? The curl of the gradient of a smooth scalar field $f:\mathbb{R}^{3}\rightarrow \mathbb{R}$ always vanishes: $(\nabla \times \nabla f)^{i} = \epsilon^{ijk}\partial_{j}\partial_{k}f = 0$.

I did not know that!

Thanks

 

[LCKurtz]

What is going on here is the use of Clairaut's theorem, which says that with enough continuity the mixed partial derivatives in any order give the same value. That's why, for example, $v_{xy} = v_{yx}$.