91Ó°ÊÓ

The force field is$F_1=2xi-2yzj-y^{2}k\text{and}{F}_2=yi-xj$

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.

role="math" localid="1664271386097" $$\begin{gathered} ∇×{\mathrm{F}}_1=\left|\begin{array}{ccc}i& j& k\\ ∂/∂x& ∂/∂y& ∂/∂z\\ 2x& -2yz& -y^2\end{array}\right| \\ ∇×{\mathrm{F}}_1=(-2y+2y)\mathrm{i}-(0-0)\mathrm{j}+(0-0)\mathrm{k} \\ ∇×{\mathrm{F}}_1=0 \end{gathered}$$

$F_1$is conservative.

$\begin{array}{rcl}∇×{\mathrm{F}}_1& =& \left|\begin{array}{ccc}i& j& k\\ ∂/∂x& ∂/∂y& ∂/∂z\\ y& -x& 0\end{array}\right|\\ ∇×{\mathrm{F}}_2& =& (0-0)\mathrm{i}-(0-0)\mathrm{j}+(-1-1)\mathrm{k}\\ & ≠& 0\end{array}$

$F_2$is not conservative.

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

$\begin{array}{rcl}{W}_1& =& ∫{\mathrm{F}}_1·d\mathrm{r}\\ & =& ∫\left(2x\right)dx+(-2yz)dy+\left(-y^2\right)dz\end{array}$

$W_1$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.

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

$W_2$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.

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

$W_3$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.

$\begin{array}{rcl}{W}_{iii}& =& {∫}_0^y-2yzdya\\ & =& -y^{2}z\end{array}$

The formula states the equation mentioned below.

$$\begin{gathered} W=W_1+W_2+W_3 \\ W=x^2-y^{2}z \end{gathered}$$

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.

$\mathrm{φ}=y^{2}z-x^2$

Hence the scalar potential is .$\mathrm{φ}=y^{2}z-x^2$