91Ó°ÊÓ

To find a potential function \( f \), we must first verify that the vector field \( \mathbf{F} = 2xy z \mathbf{i} + x^2 z \mathbf{j} + x^2 y \mathbf{k} \) is conservative. A vector field is conservative if \( abla \times \mathbf{F} = \mathbf{0} \). Calculate curl of \( \mathbf{F} \): \[ abla \times \mathbf{F} = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ 2xyz & x^2z & x^2y \end{array} \right| \] Calculating the determinant: - The \( i \)-component: \( \frac{\partial}{\partial y}(x^2y) - \frac{\partial}{\partial z}(x^2z) = x^2 - x^2 = 0 \)- The \( j \)-component: \( \frac{\partial}{\partial z}(2xyz) - \frac{\partial}{\partial x}(x^2y) = 2xy - 2xy = 0 \)- The \( k \)-component: \( \frac{\partial}{\partial x}(x^2z) - \frac{\partial}{\partial y}(2xyz) = 2xz - 2xz = 0 \)Thus, \( abla \times \mathbf{F} = \mathbf{0} \), so \( \mathbf{F} \) is conservative.

Since \( \mathbf{F} \) is conservative, it has a potential function \( f \) such that \( abla f = \mathbf{F} \). Let's find \( f \) by integrating each component with respect to its respective variable.1. Integrate the \( x \)-component with respect to \( x \): \[ \frac{\partial f}{\partial x} = 2xy z \Rightarrow f = \int 2xy z \, dx = x^2 y z + g(y,z) \] where \( g(y,z) \) is an arbitrary function of \( y \) and \( z \).2. Integrate the \( y \)-component with respect to \( y \): \[ \frac{\partial f}{\partial y} = x^2 z \Rightarrow f = \int x^2 z \, dy = x^2 y z + h(x,z) \] Comparing with step 1, \( h(x,z) = c(z) \) is an arbitrary function of \( z \).3. Integrate the \( z \)-component with respect to \( z \): \[ \frac{\partial f}{\partial z} = x^2 y \Rightarrow f = \int x^2 y \, dz = x^2 y z + k(x,y) \] Again, comparing, \( k(x,y) = 0 \).Thus, the potential function is \( f(x,y,z) = x^2 y z + C \), where \( C \) is a constant.