Verify that the force field is conservative. Then find a scalar potential φ such that

$\mathit{F}\mathbf{=}\mathbf{-}\mathbf{\nabla }\mathit{\phi }\mathbf{,}\mathit{F}\mathbf{=}\frac{\mathbf{y}}{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}}}\mathit{i}\mathbf{+}\frac{\mathbf{x}}{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}}}\mathit{j}$

The force field is$F=\frac{y}{\sqrt{1-x^2{y}^2}}i+\frac{x}{\sqrt{1-x^2{y}^2}}j$

and.$F=-\nabla \phi$

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

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

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

Put the values given below in the above equation.

$F=\frac{y}{\sqrt{1-x^2{y}^2}}i+\frac{y}{\sqrt{1-x^2{y}^2}}j$

The equation becomes as follows.

$$\begin{gathered} \nabla \times F=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ \frac{y}{\sqrt{1-x^2{y}^2}}& \frac{x}{\sqrt{1-x^2{y}^2}}& 0\end{array}\right| \\ \nabla \times F=\left(\frac{\partial }{\partial y}\left(0\right)-\frac{\partial }{\partial z}\left(\frac{x}{\sqrt{1-x^2{y}^2}}\right)\right)i-\left(\frac{\partial }{\partial x}\left(0\right)-\frac{\partial }{\partial z}\left(\frac{y}{\sqrt{1-x^2{y}^2}}\right)\right)j+\left(\frac{\partial }{\partial x}\left(\frac{x}{\sqrt{1-x^2{y}^2}}\right)-\frac{\partial }{\partial y}\left(\frac{y}{\sqrt{1-x^2{y}^2}}\right)\right)k \\ \nabla \times F=0 \end{gathered}$$

The field is conservative.

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

$$\begin{gathered} W=\int (\frac{y}{\sqrt{1-x^2{y}^2}}i+\frac{x}{\sqrt{1-x^2{y}^2}}j).(dxi+dyj+dzk) \\ W=\int \frac{y}{\sqrt{1-x^2{y}^2}}dx+\frac{x}{\sqrt{1-x^2{y}^2}}dy \end{gathered}$$

$W_1$is from (0,0,0) to (x,0,0).

y = 0

dy = 0

z = 0

dz = 0

Substitute the above values in the equation mentioned below.

$$\begin{gathered} W=\int \frac{y}{\sqrt{1-x^2{y}^2}}dx+\frac{x}{\sqrt{1-x^2{y}^2}}dy \\ {W}_1=\underset{0}{\overset{x}{\int }}0\times dx \\ {W}_1=0 \end{gathered}$$

$W_2$is from (x,0,0) to (x,0,z) and x is constant.

dx = 0

y = 0

dy = 0

Substitute the above value in the equation mentioned below.

$$\begin{gathered} W=\int \frac{y}{\sqrt{1-x^2{y}^2}}dx+\frac{x}{\sqrt{1-x^2{y}^2}}dy \\ {W}_2=\underset{0}{\overset{z}{\int }}0dz \\ {W}_2=0 \end{gathered}$$

$W_3$is from (x,0,0) to (x,y,0) and x, z is constant.

dx = 0

z = 0

dz = 0

Substitute the above value in the equation mentioned below.

$$\begin{gathered} W=\int \frac{y}{\sqrt{1-x^2{y}^2}}dx+\frac{x}{\sqrt{1-x^2{y}^2}}dy \\ {W}_3=\underset{0}{\overset{y}{\int }}\frac{x}{\sqrt{1-x^2{y}^2}}dy \\ {W}_3={\left[\frac{1}{x}\left(x{\sin }^{-1}y\right)\right]}_0^y \\ {W}_3={\sin }^{-1}y \end{gathered}$$

The formula states the equation mentioned below.

$$\begin{gathered} W=W_1+W_2+W_3 \\ W=0+0+{\sin }^{-1}y \\ W={\sin }^{-1}y \end{gathered}$$

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.

$\phi =-{\sin }^{-1}y$

Hence the Scalar potential is $-{\sin }^{-1}y$.