Solution (a) Let $P(x,y)=y\cos x+2xe^{y}$ and $Q(x,y)=\sin x+x^{2}e^{y}+4$. Then $$\begin{equation*} \dfrac{\partial P}{\partial y}=\cos x+2xe^{y}=\dfrac{\partial Q}{\partial x} \end{equation*}$$

Since $P,$ $\ Q,$ $\ \dfrac{\partial P}{\partial y}$, and $\dfrac{\partial Q }{\partial x}$ are continuous everywhere in the $xy$-plane, and since $\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$, the vector field $\mathbf{F}=\mathbf{F}(x,y)=(y\cos x+2xe^{y})\mathbf{i}+(\sin x+x^{2}e^{y}+4)\mathbf{j}$ is conservative and $\int_{C}\mathbf{F}\,{\cdot}\, d\mathbf{r}$ is independent of the path.

(b) Since $\mathbf{F}$ is a conservative vector field, there is a potential function $f$ for $\mathbf{F}$ for which $$\begin{equation*} \nabla\! \ f=\mathbf{F}=(y\cos x+2xe^{y})\mathbf{i}+(\sin x+x^{2}e^{y}+4)\mathbf{j} \end{equation*}$$

Since $\nabla \! \ f=\dfrac{\partial f}{\partial x}\mathbf{i}+\dfrac{\partial f}{\partial y}\mathbf{j}$, $$\begin{equation*} \dfrac{\partial f}{\partial x}=y\cos x+2xe^{y}\qquad \hbox{and}\qquad \dfrac{\partial f}{\partial y}=\sin x+x^{2}e^{y}+4 \end{equation*}$$

We integrate the function on the right partially with respect to $y.$ Then $$\begin{equation*} f(x,y)=\int ( \sin x+x^{2}e^{y}+4)\, dy=y\sin x+x^{2}e^{y}+4y+k(x) \end{equation*}$$

in which the “constant of integration,” $k(x)$, is a function of $x$. Now we differentiate $f$ with respect to $x$ to get $$\begin{equation*} \dfrac{\partial f}{\partial x}=y\cos x+2xe^{y}+k' (x) \end{equation*}$$

But $\dfrac{\partial f}{\partial x}=y\cos x+2xe^{y}.$ So, $$\begin{eqnarray*} y\cos x+2xe^{y}&=& y\cos x+2xe^{y}+k' (x) \\[4pt] k' (x)& =&0 \end{eqnarray*}$$

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Then $$k(x)=K\qquad \hbox{where \(K\) is a constant}$$

Therefore, $$f(x,y)=y\sin x+x^{2}e^{y}+4y+K$$

is a potential function for $\mathbf{F}$.