1. The problem is to determine if the vector function \( \mathbf{v} = \{6xy + z^3, 3x^3 - z, 3xz^2 - y\} \) is irrotational and solenoidal. 2. Recall the definitions: - A vector field is irrotational if its curl is zero: \( \nabla \times \mathbf{v} = \mathbf{0} \). - A vector field is solenoidal if its divergence is zero: \( \nabla \cdot \mathbf{v} = 0 \). 3. Compute the curl: \[ \nabla \times \mathbf{v} = \left( \frac{\partial v_3}{\partial y} - \frac{\partial v_2}{\partial z}, \frac{\partial v_1}{\partial z} - \frac{\partial v_3}{\partial x}, \frac{\partial v_2}{\partial x} - \frac{\partial v_1}{\partial y} \right) \] where \( v_1 = 6xy + z^3, v_2 = 3x^3 - z, v_3 = 3xz^2 - y \). Calculate each component: - \( \frac{\partial v_3}{\partial y} = \frac{\partial}{\partial y}(3xz^2 - y) = -1 \) - \( \frac{\partial v_2}{\partial z} = \frac{\partial}{\partial z}(3x^3 - z) = -1 \) - So first component: \( -1 - (-1) = 0 \) - \( \frac{\partial v_1}{\partial z} = \frac{\partial}{\partial z}(6xy + z^3) = 3z^2 \) - \( \frac{\partial v_3}{\partial x} = \frac{\partial}{\partial x}(3xz^2 - y) = 3z^2 \) - Second component: \( 3z^2 - 3z^2 = 0 \) - \( \frac{\partial v_2}{\partial x} = \frac{\partial}{\partial x}(3x^3 - z) = 9x^2 \) - \( \frac{\partial v_1}{\partial y} = \frac{\partial}{\partial y}(6xy + z^3) = 6x \) - Third component: \( 9x^2 - 6x \) 4. The curl is \( \{0, 0, 9x^2 - 6x\} \). For the vector field to be irrotational, this must be zero everywhere, which is not true unless \( 9x^2 - 6x = 0 \) for all \( x \), which is false. 5. Therefore, the vector field is not irrotational. The initial statement claiming it is irrotational is incorrect. 6. Next, compute the divergence: \[ \nabla \cdot \mathbf{v} = \frac{\partial v_1}{\partial x} + \frac{\partial v_2}{\partial y} + \frac{\partial v_3}{\partial z} \] Calculate each term: - \( \frac{\partial v_1}{\partial x} = \frac{\partial}{\partial x}(6xy + z^3) = 6y \) - \( \frac{\partial v_2}{\partial y} = \frac{\partial}{\partial y}(3x^3 - z) = 0 \) - \( \frac{\partial v_3}{\partial z} = \frac{\partial}{\partial z}(3xz^2 - y) = 6xz \) 7. So divergence is \( 6y + 0 + 6xz = 6y + 6xz \), which is not zero everywhere. 8. Therefore, the vector field is not solenoidal. The initial statement claiming it is solenoidal is incorrect. Summary: The mistake is in the evaluation of curl and divergence. The vector field is neither irrotational nor solenoidal because \( \nabla \times \mathbf{v} \neq \mathbf{0} \) and \( \nabla \cdot \mathbf{v} \neq 0 \).