1. **Problem Statement:** Find the curl of the vector field $\mathbf{A} = 2xz^2 \mathbf{i} - yz \mathbf{j} + 3xz^3 \mathbf{k}$. 2. **Recall the formula for curl:** $$\nabla \times \mathbf{A} = \left( \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} \right) \mathbf{i} - \left( \frac{\partial A_z}{\partial x} - \frac{\partial A_x}{\partial z} \right) \mathbf{j} + \left( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \right) \mathbf{k}$$ where $\mathbf{A} = A_x \mathbf{i} + A_y \mathbf{j} + A_z \mathbf{k}$. 3. **Identify components:** $A_x = 2xz^2$, $A_y = -yz$, $A_z = 3xz^3$. 4. **Calculate partial derivatives:** $\frac{\partial A_z}{\partial y} = \frac{\partial}{\partial y}(3xz^3) = 0$ (since no $y$ in $A_z$), $\frac{\partial A_y}{\partial z} = \frac{\partial}{\partial z}(-yz) = -y$, $\frac{\partial A_z}{\partial x} = \frac{\partial}{\partial x}(3xz^3) = 3z^3$, $\frac{\partial A_x}{\partial z} = \frac{\partial}{\partial z}(2xz^2) = 4xz$, $\frac{\partial A_y}{\partial x} = \frac{\partial}{\partial x}(-yz) = 0$, $\frac{\partial A_x}{\partial y} = \frac{\partial}{\partial y}(2xz^2) = 0$. 5. **Substitute into curl formula:** $$\nabla \times \mathbf{A} = (0 - (-y)) \mathbf{i} - (3z^3 - 4xz) \mathbf{j} + (0 - 0) \mathbf{k} = y \mathbf{i} - (3z^3 - 4xz) \mathbf{j} + 0 \mathbf{k}$$ 6. **Simplify:** $$\nabla \times \mathbf{A} = y \mathbf{i} - 3z^3 \mathbf{j} + 4xz \mathbf{j} = y \mathbf{i} + (4xz - 3z^3) \mathbf{j}$$ 7. **Interpretation:** The curl is a vector field with components $y$ in $\mathbf{i}$ direction and $(4xz - 3z^3)$ in $\mathbf{j}$ direction, and zero in $\mathbf{k}$ direction. **Final answer:** $$\boxed{\nabla \times \mathbf{A} = y \mathbf{i} + (4xz - 3z^3) \mathbf{j}}$$