1. The problem is to 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. The curl of a vector field $ \mathbf{v} = \{P, Q, R\} $ is given by: $$\nabla \times \mathbf{v} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \n\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \n\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)$$ 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. Substitute into curl formula: $$\nabla \times \mathbf{v} = \left((-1) - (-1), 3z^2 - 3z^2, 9x^2 - 6x\right) = \left(0, 0, 9x^2 - 6x\right)$$ 6. The curl is $\{0, 0, 9x^2 - 6x\}$. For the vector field to be irrotational, curl must be $\{0,0,0\}$ everywhere. 7. Since $9x^2 - 6x = 3x(3x - 2)$ is not zero for all $x$, the vector field is rotational. 8. Therefore, the given vector function is rotational. Mathematica code correction: ```mathematica v = {6 x y + z^3, 3 x^3 - z, 3 x z^2 - y}; curlV = Curl[v, {x, y, z}]; If[curlV == {0, 0, 0}, Print["The given vector function is irrotational"], Print["The given vector function is rotational"] ] ```