Verify that the force field is conservative. Then find a scalar potential $\mathit{\theta }$ such that $\mathit{F}\mathbf{=}\mathbf{-}\mathbf{\nabla }\mathit{\phi }\mathbf{,}\mathit{F}\mathbf{=}{\mathit{z}}^{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathbf{}\mathit{y}\mathbf{}\mathit{j}\mathbf{+}\mathbf{2}\mathit{z}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathbf{}\mathit{y}\mathbf{}\mathit{k}\mathbf{,}\mathbf{}\mathit{k}\mathbf{}\mathbf{=}\mathbf{}\mathit{c}\mathit{o}\mathit{n}\mathit{s}\mathit{t}\mathit{a}\mathit{n}\mathit{t}\mathbf{.}$

The force field is$F=z^{2}sinhyj+2zcoshykandF=-\nabla \phi .$

A force is said to be conservative if $\mathbf{\nabla }\mathbf{\times }\mathit{F}\mathbf{=}\mathbf{0}\mathbf{.}$.

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

The force is said to be conservative if $\nabla \times F=0.$.

Putthe values given below in the above equation.

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

The equation becomes as follows.

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

The field is conservative.

The formula for the scalar potential is$W=\int F.dr.$.

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

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

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

Substitute the above value in the equation mentioned below.

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

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

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.

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

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.

$F=\nabla W$

It is given that$F=-\nabla \phi$.

By both the values of $F,-\nabla \phi =\nabla W.$, .

$\phi =-W.$

Put the value of in above equation.

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

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