Question

Write out the equations corresponding to (9.3) and (9.4) for Q=2Xbetween points 3 and 4 in Figure 9.2, and add them to get (9.6).

Short answer

The solution is derived to be as mentioned below.

$\iint \frac{\partial Q}{\partial z}dxdy=\oint Qdy〗$

Step-by-step solution

Given Information

The given equation is Q=2X .

Definition of conservative force and scalar potential

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

Calculate 2D version of Stokes theorem and divide the curve into two parts

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" $\iint \frac{\partial Q}{\partial z}dxdy={\int }_c^{d}dy{\int }_{X_l}^{X_r}\frac{\partial Q}{\partial z}dx$

Use fundamental theorem of calculus in 2nd part of the integral.

Solve further.

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

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

Repeat the same process

Repeat the same process as done above.

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

$$\begin{gathered} \iint \frac{\partial Q}{\partial z}dxdy={\int }_c^{d}Q\left(x_r\right(y),y)dy+{\int }_d^{c}Q\left(x_l\right(y),y)dy \\ \iint \frac{\partial Q}{\partial z}dxdy=\oint Qdy \end{gathered}$$

Hence, the equation in 9.6 is derived.