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The force field is$F=z^{2}sinhyj+2zcoshykandF=-鈭�蠁.$

A force is said to be conservative if $\mathbf{鈭�}\mathbf{脳}\mathit{F}\mathbf{=}\mathbf{0}\mathbf{.}$.

The formula for the scalar potential is .$\mathit{W}\mathbf{=}\mathbf{鈭�}\mathit{F}\mathbf{.}\mathit{d}\mathit{r}\mathbf{.}$

The force is said to be conservative if $鈭�脳F=0.$.

Putthe values given below in the above equation.

$F=z^{2}sinhyj+2zcoshyk$

The equation becomes as follows.

localid="1659148650409" $$\begin{gathered} 鈭�脳F=\left(\begin{array}{ccc}i& j& k\\ \frac{蠎}{蠎x}& \frac{蠎}{蠎y}& \frac{蠎}{蠎z}\\ 0& z^2\sin y& 2z\cos hy\end{array}\right) \\ 鈭�脳F=\left(\frac{蠎}{蠎y}\left(2z\cos hy\right)-\frac{蠎}{蠎z}\left(z^2\sin y\right)\right)i-\left(\frac{蠎}{蠎x}\left(2z\cos hy\right)-\frac{蠎}{蠎z}\left(0\right)\right)j+\left(\frac{蠎}{蠎x}\left(z^2\sin y\right)-\frac{蠎}{蠎z}\left(0\right)\right)k \\ 鈭�脳F=0 \end{gathered}$$

The field is conservative.

The formula for the scalar potential is$W=鈭�F.dr.$.

$$\begin{gathered} W=鈭�(z^{2}sinhyj+2zcoshyk).(dxi+dyj+dzk) \\ W=鈭�z^{2}sinhydy+2zcoshydz \end{gathered}$$

$W_{1}isfrom(0,0,0)to(x,0,0).$

$$\begin{gathered} y=0 \\ dy=0 \\ z=0 \\ dz=0 \end{gathered}$$

Substitute the above value in the equation mentioned below.

$$\begin{gathered} W=鈭�z^2\sin hydy+2z\cos hydz \\ {W}_1={鈭�}_0^{x}0脳dx \\ {W}_1=0 \end{gathered}$$

localid="1659148618945" $$\begin{gathered} W_{2}isfrom(x,0,0)to(x,0,z)andxisconstant \\ dx=0 \\ y=0 \\ dy=0 \end{gathered}$$

Substitute the above value in the equation mentioned below.

localid="1659148692346" $$\begin{gathered} W=鈭�z^2\sin hydy+2z\cos hydz \\ {W}_2={鈭�}_0^{z}2\cos hy\left(\frac{z^2}{2}\right)dz \\ {W}_2={\left[\left(\cos hy\right)z^2\right]}_0^z \\ {W}_2=z^2\cos hy \end{gathered}$$

localid="1659148631260" $$\begin{gathered} W_{3}isfrom(x,0,0)to(x,y,0)andxisconstant,zisconstant. \\ dx=0 \\ z=0 \\ dz=0 \end{gathered}$$

Substitute the above value in the equation mentioned below.

$$\begin{gathered} W=鈭�z^2\sin hydy+2z\cos hydz \\ {W}_3={鈭�}_0^{y}ody \\ {W}_3=0 \end{gathered}$$

The formula states the equation mentioned below.

$$\begin{gathered} W=W_1+W_2+W \\ W=0+z^2\cos hy+0 \\ W=z^2\cos hy \end{gathered}$$

The formula states the equation mentioned below.

$F=鈭�W$

It is given that$F=-鈭�蠁$.

By both the values of $F,-鈭�蠁=鈭�W.$, .

$蠁=-W.$

Put the value of in above equation.

$$\begin{gathered} 蠁=-(z^{2}coshy) \\ 蠁=-z^{2}coshy \end{gathered}$$

Hence the Scalar potential is $-z^{2}coshy.$.