Question

Determine whether the vector field \(F\) is conservative. If it is, find a potential function for the vector field. $$ \mathbf{F}(x, y, z)=e^{z}(y \mathbf{i}+x \mathbf{j}+x y \mathbf{k}) $$

Short answer

The vector field \(F = (e^{z}y, e^{z}x, e^{z}xy)\) is conservative and its potential function is \(f(x, y, z) = e^{z}xy\).

Step-by-step solution

Check if the curl of \(F\) equals zero

Calculate the curl of the vector field \(F = (e^{z}y, e^{z}x, e^{z}xy)\). This is done by using the determinants of the matrix formed by the unit vectors \(i\), \(j\), and \(k\), the gradient operator \(\[ \nabla = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z})\], and the vector field itself. i.e.,\(\nabla \times F = \left| \begin{matrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ e^{z}y & e^{z}x & e^{z}xy \end{matrix} \right|\)

Calculate the determinants to get the curl of \(F\)

When we calculate the determinants, we get \(\nabla \times F = \left( \frac{\partial}{\partial y}(e^{z}xy) - \frac{\partial}{\partial z}(e^{z}x)\right)i - \left( \frac{\partial}{\partial x}(e^{z}xy) - \frac{\partial}{\partial z}(e^{z}y)\right)j + \left( \frac{\partial}{\partial x}(e^{z}x) - \frac{\partial}{\partial y}(e^{z}y)\right)k\). This simplifies to \(\nabla \times F = (e^{z} - e^{z})i - (e^{z} - e^{z})j + (e^{z} - e^{z})k = 0\)

Determine the potential function of \(F\)

Now find a potential function \(f\) such that the gradient of \(f\) is equal to \(F\). This is solved by integrating the first component of \(F\) regarding \(x\), the second component of \(F\) regarding \(y\), and the third component of \(F\) regarding \(z\) and then harmonising the results. Since \(F\) is conservative, the results will be harmonised. If we do that, we get \(f(x, y, z) = \int e^{z}y dx + \int e^{z}x dy + \int e^{z}xy dz = e^{z}xy\). So, the potential function for \(F\) is \(f(x, y, z) = e^{z}xy\).