1. **State the problem:** Determine if the vector field $ \mathbf{F}(x,y,z) = (4y + 3xyz)\mathbf{i} + (8xy + xz)\mathbf{j} + (11 - 2xyz)\mathbf{k} $ is conservative. 2. **Recall the definition:** A vector field $ \mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k} $ is conservative if there exists a scalar potential function $ \phi(x,y,z) $ such that $ \mathbf{F} = \nabla \phi $. 3. **Key test:** For $ \mathbf{F} $ to be conservative in a simply connected domain, the curl must be zero: $$ \nabla \times \mathbf{F} = \mathbf{0} $$ 4. **Identify components:** $$ P = 4y + 3xyz, \quad Q = 8xy + xz, \quad R = 11 - 2xyz $$ 5. **Compute partial derivatives:** $$ \frac{\partial R}{\partial y} = \frac{\partial}{\partial y}(11 - 2xyz) = -2xz $$ $$ \frac{\partial Q}{\partial z} = \frac{\partial}{\partial z}(8xy + xz) = x $$ $$ \frac{\partial P}{\partial z} = \frac{\partial}{\partial z}(4y + 3xyz) = 3xy $$ $$ \frac{\partial R}{\partial x} = \frac{\partial}{\partial x}(11 - 2xyz) = -2yz $$ $$ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(8xy + xz) = 8y + z $$ $$ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(4y + 3xyz) = 4 + 3xz $$ 6. **Calculate curl components:** $$ (\nabla \times \mathbf{F})_x = \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} = -2xz - x = -x(2z + 1) $$ $$ (\nabla \times \mathbf{F})_y = \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} = 3xy - (-2yz) = 3xy + 2yz = y(3x + 2z) $$ $$ (\nabla \times \mathbf{F})_z = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = (8y + z) - (4 + 3xz) = 8y + z - 4 - 3xz $$ 7. **Analyze if curl is zero:** - For $ (\nabla \times \mathbf{F})_x = 0 $, we need $ -x(2z + 1) = 0 $ which is not true for all $x,z$. - Since the curl is not zero everywhere, $ \mathbf{F} $ is **not conservative**. **Final answer:** The vector field $ \mathbf{F} $ is not conservative because its curl is not zero everywhere.