1. **State the problem:** Determine if the vector field $\mathbf{v} = \{6xy + z^3, 3x^3 - z, 3xz^2 - y\}$ is irrotational or rotational by computing its curl. 2. **Recall the formula for curl:** For a vector field $\mathbf{v} = \{P, Q, R\}$, the curl is given by $$\nabla \times \mathbf{v} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)$$ 3. **Identify components:** - $P = 6xy + z^3$ - $Q = 3x^3 - z$ - $R = 3xz^2 - y$ 4. **Compute partial derivatives:** - $\frac{\partial R}{\partial y} = \frac{\partial}{\partial y}(3xz^2 - y) = -1$ - $\frac{\partial Q}{\partial z} = \frac{\partial}{\partial z}(3x^3 - z) = -1$ - $\frac{\partial P}{\partial z} = \frac{\partial}{\partial z}(6xy + z^3) = 3z^2$ - $\frac{\partial R}{\partial x} = \frac{\partial}{\partial x}(3xz^2 - y) = 3z^2$ - $\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(3x^3 - z) = 9x^2$ - $\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(6xy + z^3) = 6x$ 5. **Calculate each component of the curl:** - First component: $\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} = -1 - (-1) = 0$ - Second component: $\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} = 3z^2 - 3z^2 = 0$ - Third component: $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 9x^2 - 6x$ 6. **Curl vector:** $$\nabla \times \mathbf{v} = \{0, 0, 9x^2 - 6x\}$$ 7. **Interpretation:** The curl is zero if and only if $9x^2 - 6x = 0$ which factors as $$3x(3x - 2) = 0$$ so $x = 0$ or $x = \frac{2}{3}$. 8. **Conclusion:** Since the curl is not zero everywhere, the vector field is rotational in general. **Final answer:** The given vector function is rotational because its curl is not the zero vector everywhere.