Show that $\mathbf{F}=\left( \dfrac{3}{5}y^{5}+2z^{2}\right) \mathbf{i} +3xy^{4}\,\mathbf{j}+4xz\,\mathbf{k}$ is a conservative vector field in space.

Solution  We find $\rm{curl}\, \mathbf{F}$. $$\begin{equation*} \rm{curl}\, \mathbf{F}=\left\vert \begin{array}{@{}c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[5pt] \dfrac{\partial }{\partial x} & \dfrac{\partial }{\partial y} & \dfrac{ \partial }{\partial z} \\[5pt] \dfrac{3}{5}y^{5}+2z^{2} & 3xy^{4} & 4xz \end{array} \right\vert =(0-0)\,\mathbf{i}-(4z-4z)\,\mathbf{j}+(3y^{4}-3y^{4})\mathbf{k}= \mathbf{0} \end{equation*}$$

Since $\rm{curl}\,\mathbf{F=0}$, $\mathbf{F}$ is conservative.