91Ó°ÊÓ

The force field is$F_1=-\mathrm{yi}+\mathrm{xj}+{\mathrm{zk}}^{\text{'}}$

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

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

$$\begin{gathered} x=\cos t, \\ y=\sin t, \\ z=t \end{gathered}$$

Now write the other values.

$$\begin{gathered} dx=-\sin tdt, \\ dy=\cos tdt, \\ dz=dt \end{gathered}$$

Find the work.

$\begin{array}{rcl}{W}_1& =& ∫\mathrm{F}·d\mathrm{r}\\ & =& {∫}_0^{\mathrm{π}}si{n}^{2}tdt\end{array}$

Set the limits of integration from$0→\mathrm{π}$ because the position (1,0,0) corresponds to an angle of zero, while the position (-1,0,0) corresponds to an angle of $\mathrm{π}$.

$W_1=\mathrm{Ï€}/2$

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

$$\begin{gathered} x=\cos t, \\ y=\sin t, \\ z=t \end{gathered}$$

Now write the other values.

$$\begin{gathered} dx=-\sin tdt, \\ dy=\cos tdt, \\ dz=dt \end{gathered}$$

Find the work.

$$\begin{gathered} W_2=∫\mathrm{F}·d\mathrm{r} \\ {W}_2={∫}_0^{\mathrm{π}}tdt \\ {W}_2={\mathrm{π}}^2/2 \end{gathered}$$

The formula states the equation mentioned below.

$\begin{array}{rcl}W& =& W_1+W_2\\ & =& \frac{{\mathrm{Ï€}}^2+\mathrm{Ï€}}{2}\end{array}$

$W_1$is from(1,0,0) to (-1,0,0) .

y = 0

z = 0

dy = 0

dz = 0

Find the work.

$\begin{array}{rcl}{W}_1& =& ∫\mathrm{F}·d\mathrm{r}\\ & =& {∫}_1^{-1}0dx\end{array}$

$W_2$is from(-1,0,0) to$(-1,0,\mathrm{Ï€})$ .

y = 0

dx = 0

dy = 0

Find the work.

$W_2={\mathrm{Ï€}}^2/2$

The formula states the equation mentioned below.

$\begin{array}{rcl}W& =& W_1+W_2\\ & =& \frac{{\mathrm{Ï€}}^2}{2}\end{array}$

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

Putthe values given below in the above equation.

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

Therefore, the field is not conservative.