Verify that the force field is conservative. Then find a scalar potential φ such that $\mathbf{F}\mathbf{=}\mathbf{2}\mathit{x}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{y}\mathbf{i}\mathbf{-}\left(x^2+1\right)\mathit{s}\mathit{i}\mathit{n}\mathbf{2}\mathit{y}\mathbf{j}$.

The force field is$\mathrm{F}=2xco{s}^{2}y\mathrm{i}-\left(x^2+1\right)sin2y\mathrm{j}$

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

The scalar potential is independent of the path. The scalar potential is the sum of potential in all the 3 dimensions calculated separately.

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 $\nabla \times F=0$.

Putthe values given below in the above equation.

$\begin{array}{rcl}\nabla \times \mathrm{F}& =& \left|\begin{array}{ccc}i& j& k\\ \partial /\partial x& \partial /\partial y& \partial /\partial z\\ 2xco{s}^{2}y& -\left(x^2+1\right)sin2y& 0\end{array}\right|\\ \nabla \times \mathrm{F}& =& (0-0)\mathrm{i}-(0-0)\mathrm{j}+(-2xsin02y0+0000020x00sin2y)\mathrm{k}\\ \nabla \times \mathrm{F}& =& 0\end{array}$

It must be noted that .${\cos }^{2}y=\frac{1}{2}+\frac{\left(\cos 2y\right)}{2}$${\cos }^{2}y=\frac{1}{2}+\frac{\left(\cos 2y\right)}{2}$

Therefore, the field is conservative.

The formula for the scalar potential is.$W=\int F.dr$$\begin{array}{rcl}W& =& \int \mathrm{F}·d\mathrm{r}\end{array}$

$\begin{array}{rcl}W& =& \int \mathrm{F}·d\mathrm{r}\\ & =& \int \left(2xco{s}^{2}y\right)dx+\left(-\left(x^2+1\right)sin2y\right)dy+\left(0\right)dz\end{array}$

$W_1$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.

$\begin{array}{rcl}{W}_1& =& {\int }_0^x\left(2x{\cos }^{2}0\right)dx\\ & =& x^2\end{array}$

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

x is constant.

dx = 0

y = 0

dy = 0

Substitute the above value in the equation mentioned below.

$\begin{array}{rcl}{W}_2& =& {\int }_0^z\left(0\right)dz\\ & =& 0\end{array}$

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

x is constant.

dx = 0

z = 0

dz = 0

Substitute the above value in the equation mentioned below.

$\begin{array}{rcl}{W}_3& =& {\int }_0^y\left(-\left(x^2+1\right)\sin 2y\right)dy\\ & =& \frac{\left(x^2+1\right)\cos 2y}{2}-\frac{x^2+1}{2}\end{array}$

The formula states the equation mentioned below.

$$\begin{gathered} W=W_1+W_2+W_3 \\ W=x^2+\frac{\left(x^2+1\right)\cos 2y}{2}-\frac{x^2+1}{2} \\ W=\frac{2\left(x^2+1\right){\cos }^{2}y-+2x^2-x^2-1-x^2-1}{2} \end{gathered}$$

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.

$\phi =1-{\cos }^{2}y\left(x^2+1\right)$

Hence the scalar potential is $\phi =1-{\cos }^{2}y\left(x^2+1\right)$.