To compute the curl of a vector field F = $\langle F_x, F_y, F_z \rangle$, where F = $\langle 2x^2 - y^2, 3xy, 2z \rangle$, we use the formula for the curl:

$\text{curl } \mathbf{F} = \nabla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right) \mathbf{i} + \left( \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) \mathbf{j} + \left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \mathbf{k}$

1. Compute $\frac{\partial F_z}{\partial y}$ and $\frac{\partial F_y}{\partial z}$:

   • $\frac{\partial F_z}{\partial y} = \frac{\partial (2z)}{\partial y} = 0$
   • $\frac{\partial F_y}{\partial z} = \frac{\partial (3xy)}{\partial z} = 0$

   So, the i-component is $0 - 0 = 0$.

2. Compute $\frac{\partial F_x}{\partial z}$ and $\frac{\partial F_z}{\partial x}$:

   • $\frac{\partial F_x}{\partial z} = \frac{\partial (2x^2 - y^2)}{\partial z} = 0$
   • $\frac{\partial F_z}{\partial x} = \frac{\partial (2z)}{\partial x} = 0$

   So, the j-component is $0 - 0 = 0$.

3. Compute $\frac{\partial F_y}{\partial x}$ and $\frac{\partial F_x}{\partial y}$:

   • $\frac{\partial F_y}{\partial x} = \frac{\partial (3xy)}{\partial x} = 3y$
   • $\frac{\partial F_x}{\partial y} = \frac{\partial (2x^2 - y^2)}{\partial y} = -2y$

   So, the k-component is $3y - (-2y) = 3y + 2y = 5y$.

Putting it all together, we get:

$\text{curl } \mathbf{F} = \langle 0, 0, 5y \rangle = 5y \mathbf{k}$

Answer:

$\text{curl } \mathbf{F} = 5y \mathbf{k}$

Would you like further details on any part of the solution?

Here are some questions that extend the topic:

1. How does the physical interpretation of curl apply to this vector field?
2. What is the divergence of this vector field?
3. How would the curl change if we modified one of the components of F?
4. What would be the curl of a constant vector field?
5. Can a vector field have both zero curl and zero divergence?

Tip: The curl represents the rotation or "twisting" tendency of a vector field in three dimensions.