91Ó°ÊÓ

The given equation is Q=2X .

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

The curve on the left part, from 3 to 4, is given by $x=x_l\left(y\right)$ and the right part, from 4 to 3, is given by $x=x_r\left(y\right)$where y takes the value c to d.

localid="1659150255472" $âˆ\neg \frac{∂Q}{∂z}dxdy={∫}_c^{d}dy{∫}_{X_l}^{X_r}\frac{∂Q}{∂z}dx$

Solve further.

$$\begin{gathered} {∫}_a^b\frac{d}{dt}f\left(t\right)dt=f\left(b\right)-f\left(a\right) \\ {∫}_c^d\left[Q\left(x_r,y\right)-Q\left(x_l,y\right)\right]dy \end{gathered}$$

localid="1659150154992" ${∫}_c^d\left[Q{x}_r\left(y\right),y\right]dy$, is the line integral of Q over the right part of the curve C in the counter clockwise direction.

Repeat the same process as done above.

${∫}_c^d\left[Q{x}_r\left(y\right),y\right]dy={∫}_c^d\left[Q{x}_l\left(y\right),y\right]dy$ is the line integral of Q over the left part of the curve C in the counter clockwise direction.

$$\begin{gathered} âˆ\neg \frac{∂Q}{∂z}dxdy={∫}_c^{d}Q\left(x_r\right(y),y)dy+{∫}_d^{c}Q\left(x_l\right(y),y)dy \\ âˆ\neg \frac{∂Q}{∂z}dxdy=âˆ\circledR Qdy \end{gathered}$$

Hence, the equation in 9.6 is derived.