Verify that the force field is conservative. Then find a scalar potential $\mathbf{\psi }$such that $\mathbf{F}\mathbf{=}\mathbf{\nabla }\mathbf{\psi F}\mathbf{=}\mathbf{y}\mathbf{}\mathbf{sin}\mathbf{2}\mathbf{xi}\mathbf{+}{\mathbf{sin}}^{\mathbf{2}}\mathbf{xj}$.

The force field is$F=y\sin 2xi+{\sin }^{2}xjandF=-\nabla \psi$.

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

The formula for the scalar potential is $\mathbf{w}\mathbf{=}\mathbf{\int }\mathbf{F}\mathbf{.}\mathbf{dr}$.

The force is said to be conservative if$\nabla \times F=0$. ……. (1)

Put equation (2) in equation (1).

$F=y\sin 2xi+{\sin }^{2}xj$ ……. (2)

The equation becomes as follows.

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

Hence the field is conservative

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

$$\begin{gathered} W=\int \left(-y\sin 2xi+{\sin }^{2}xj\right).\left(dxi+dyj+dzk\right) \\ W=\int \left(y-\sin 2x\right)dx+{\sin }^{2}xdy \end{gathered}$$

$W_{1}Isfrom\left(0,0,0\right)to\left(x,0\right).$

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

Substitute the above value in the equation mentioned below.

$$\begin{gathered} w=\int \left(-y\sin 2x\right)dx+{\sin }^{2}xdy \\ {w}_1={\int }_0^{x}0dx \\ {w}_1=0 \end{gathered}$$

$w_{2}Isfrom\left(x,0\right)to\left(x,y\right)$

x is constant.

dx=0

Substitute the above value in the equation mentioned below.

$$\begin{gathered} w=\int \left(y-\sin 2x\right)dx+{\sin }^{2}xdy \\ {w}_2={\int }_0^y{\sin }^{2}xdy \\ {w}_2={\left[y{\sin }^{2}x\right]}_0^y \\ {w}_2=y{\sin }^{2}x \end{gathered}$$

The formula states the equation mentioned below.

$$\begin{gathered} w=w_1+w_2 \\ w=0+y{\sin }^{2}x \\ w=y{\sin }^{2}x \end{gathered}$$

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 W in above equation.

Hence the scalar potential is $-y{\sin }^{2}x$.