The angular velocity vector of a rigid object rotating about the z-axis is given by $\vec \omega = \omega \hat z$. At any point in the rotating object, the linear velocity vector is given by $\vec v = \vec \omega \times \vec r$, where $\vec r$ is the position vector to that point.
a) Assuming that $\omega$ is constant, evaluate $\vec v$ and $\vec \nabla \times \vec v$ in cylindrical coordinates.
b) Evaluate $\vec v$ in spherical coordinates.
c) Evaluate the curl of $\vec v$ in spherical coordinates and show that the resulting expression is equivalent to that given for $\vec \nabla \times \vec v$ in part a.
So for part a.) I get the following, \begin{align} \vec v = \vec \omega \times \vec r & = \begin{vmatrix} \hat \rho & \hat \phi & \hat z \\ 0 & 0 & \omega \\ \rho & 0 & z \\ \end{vmatrix} \\ & = \rho \omega \hat \phi \end{align}
Then \begin{align} \vec \nabla \times \vec v &=\frac{1}{\rho} \begin{vmatrix} \hat \rho & \rho\hat \phi & \hat z \\ \frac{\partial}{\partial \rho} & \frac{\partial}{\partial \phi} & \frac{\partial}{\partial z} \\ 0 & \rho^2 \omega & 0 \\ \end{vmatrix}\\ &= 2\omega \hat z \end{align}
for part b.) I use the relations $x = r\cos\theta \sin\phi$, $y = r \sin\theta \sin\phi$, and $ z = r \cos\theta$ to change $\vec \omega$ to spherical form
$\vec \omega = \omega \hat r$, the position vector in spherical form is
$\vec r = r \hat r$
so the velocity in spherical form is
\begin{align} \vec v = \omega \times \vec r & = \begin{vmatrix} \hat r & \hat \theta & \hat \phi \\ \omega & 0 & 0 \\ r & 0 & 0 \\ \end{vmatrix} \\ &= 0 \end{align}
I don't know what I'm doing wrong.
In spherical coordinates, $(r,\theta,\phi)$,the axial unit vector $\hat z$ is
$$\hat z=\hat r \cos(\theta)-\hat \theta \sin(\theta)$$
Then,
$$\begin{align} \vec v&=\vec \omega\times \vec r\\\\ &=\omega \hat \omega\times \vec r\\\\ &=\omega (\hat r \cos(\theta)-\hat \theta \sin(\theta))\times \vec r\\\\ &=\hat \phi\omega r \sin(\theta) \end{align}$$
Finally, we have
$$\begin{align} \nabla \times \vec v&= \omega r \nabla \times (\hat \phi r\sin(\theta))\\\\ &=\hat r\omega r\left(\frac{1}{r\sin(\theta)}\frac{\partial }{\partial \theta}(r\sin^2(\theta))\right)+\hat \theta \omega r\left(\frac1r \frac{\partial (r^2\sin(\theta))}{\partial r}\right)\\\\ &=\hat r 2\omega r\cos(\theta)+ \hat \theta 2\omega r\sin(\theta)\\\\ &=2\omega r\underbrace{\color{blue}{\left(\hat r \cos(\theta)+ \hat \theta \sin(\theta)\right)}}_{\color{blue}{=\hat z}}\\\\ \end{align}$$