If it is known that $$\begin{equation*} \mathbf{F}=\mathbf{F}(x,y)=(6xy+y^{3})\mathbf{i}+(3x^{2}+3xy^{2})\mathbf{j} \end{equation*}$$

is a conservative vector field, find a potential function $f$ for $\mathbf{F}$.

Solution We seek a function $f(x,y)$ for which $$\begin{equation*} \nabla\! \ f=\mathbf{F}=(6xy+y^{3})\mathbf{i}+(3x^{2}+3xy^{2})\mathbf{j} \end{equation*}$$

Let $$P(x,y)=6xy+y^{3}\qquad \hbox{and}\qquad Q(x,y)=3x^{2}+3xy^{2}$$

Since $$\nabla\! \ f=\dfrac{\partial f}{\partial x}\mathbf{i}+\dfrac{\partial f}{ \partial y}\mathbf{j}=P(x,y)\mathbf{i}+Q(x,y)\mathbf{j}$$

we have $$\begin{eqnarray*} \dfrac{\partial f}{\partial x}=6xy+y^{3}\qquad \hbox{and}\qquad \dfrac{\partial f}{\partial y}=3x^{2}+3xy^{2}\tag{1} \end{eqnarray*}$$

Now integrate the left equation partially with respect to $x$. $$f(x,y) =\int ( 6xy+y^{3})\, dx=3x^{2}y+xy^{3}+h(y)$$

where the “constant of integration,” $h(y)$, is a function of $y$.

Next differentiate $f$ with respect to $y$: $$\begin{equation*} \dfrac{\partial f}{\partial y}=3x^{2}+3xy^{2}+h' (y) \end{equation*}$$

From (1), we have $$\begin{equation*} \dfrac{\partial f}{\partial y}=3x^{2}+3xy^{2} \end{equation*}$$

So, $h' (y)=0$ or $h(y) =K,$ where $K$ is a constant. A potential function $f$ for $\mathbf{F}$ is $$f(x,y)=3x^{2}y+xy^{3}+K$$