91Ó°ÊÓ

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$\mathbf{∇}\mathbf{×}\mathit{F}\mathbf{=}\mathbf{0}$ .

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 $\mathit{W}\mathbf{=}{\mathbf{∫}}_{}^{}\mathit{F}\mathbf{.}\mathit{d}\mathit{r}$.

The force is said to be conservative if $∇×F=0$.

Putthe values given below in the above equation.

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

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=∫F.dr$$\begin{array}{rcl}W& =& ∫\mathrm{F}·d\mathrm{r}\end{array}$

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

$W_1$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.

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

$W_2$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.

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

$W_3$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.

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

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.

$F=∇W$

It is given that $F=-∇\mathrm{φ}$.

By both the values of F,$-∇\mathrm{φ}=∇W$ .

$\mathrm{φ}=-W$

Put the value of W in above equation.

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

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