In the given problem, we are asked to find the \(z\) component of the vorticity for a two-dimensional circular flow in Cartesian coordinates, where the velocity vector is given by:

\(\mathbf{u} = \frac{V(r)}{(x^{2}+y^{2})^{1/2}}(-y,x,0)\)

Here, \(r = (x^2 + y^2)^{1/2}\). The velocity components in terms of \(x\), \(y\), and \(z\) are:

• \(u_x = -\frac{V(r) y}{r}\)
• \(u_y = \frac{V(r) x}{r}\)
• \(u_z = 0\)

The vorticity vector \(\mathbf{\omega}\) in a three-dimensional flow is given by the curl of the velocity field:

\(\mathbf{\omega} = \nabla \times \mathbf{u}\)

Specifically, the \(z\)-component of the vorticity, \(\omega_z\), is given by:

\(\omega_z = \left( \frac{\partial u_y}{\partial x} - \frac{\partial u_x}{\partial y} \right)\)

Calculating these partial derivatives:

1. \(\frac{\partial u_y}{\partial x} = \frac{\partial}{\partial x} \left( \frac{V(r) x}{r} \right)\)
2. \(\frac{\partial u_x}{\partial y} = \frac{\partial}{\partial y} \left( -\frac{V(r) y}{r} \right)\)

Using the chain rule:

• \(\frac{\partial u_y}{\partial x} = \frac{x}{r} \cdot \frac{\partial V}{\partial r} \cdot \frac{\partial r}{\partial x} + \frac{V(r)}{r} - x \cdot \frac{V(r) \cdot x}{r^3}\)
• \(\frac{\partial u_x}{\partial y} = -\left( \frac{y}{r} \cdot \frac{\partial V}{\partial r} \cdot \frac{\partial r}{\partial y} + \frac{V(r)}{r} - y \cdot \frac{V(r) \cdot y}{r^3} \right)\)

We know that \(\frac{\partial r}{\partial x} = \frac{x}{r}\) and \(\frac{\partial r}{\partial y} = \frac{y}{r}\). Substitute these into the expressions:

Combine the results:

\(\omega_z = \frac{dV}{dr} + \frac{V}{r}\)

Therefore, the \(z\) component of the vorticity is:

\(\frac{dv}{dr} + \frac{V}{r}\)

Hence, the correct answer is:

\(\frac{dv}{dr} + \frac{V}{r}\)