Chapter 4, Problem 4.25P

(a)

To determine

Show that the path independence of ${\displaystyle \underset{1}{\overset{2}{\int }}F\cdot dr}$ is equivalent to the statement that the integral ${\displaystyle \underset{\Gamma }{\oint }F\cdot dr}$.

Answer to Problem 4.25P

Showed that the path independence of ${\displaystyle \underset{1}{\overset{2}{\int }}F\cdot dr}$ is equivalent to the statement that the integral ${\displaystyle \underset{\Gamma }{\oint }F\cdot dr}$.

Explanation of Solution

Consider two paths A and B from point 1 to 2 as shown in the Figure below.

Write the expression for stoker’s theorem.

    ${\displaystyle \underset{\Gamma }{\oint }F\cdot dr}={\displaystyle \int \left(\nabla \times F\right)\cdot \widehat{n}dA}$        (I)

Let $F$ is the force acting on the particle and ${\displaystyle \underset{1}{\overset{2}{\int }}F\cdot dr}$ is path independent. Then it can be written as follows,

    ${\displaystyle \underset{1\text{path A}}{\overset{2}{\int }}F\cdot dr}={\displaystyle \underset{1\text{path B}}{\overset{2}{\int }}F\cdot dr}$        (II)

Consider the Figure 1, this closed surface is denoted by $\Gamma$, and it containing points 1 and 2 formed by the curve A and B.

    ${\displaystyle \underset{\Gamma }{\oint }F\cdot dr}={\displaystyle \underset{1\text{path A}}{\overset{2}{\int }}F\cdot dr}+{\displaystyle \underset{2\text{path B}}{\overset{1}{\int }}F\cdot dr}$        (III)

Use equation (II) in (III).

    $\begin{array}{c}{\displaystyle \underset{\Gamma }{\oint }F\cdot dr}={\displaystyle \underset{1\text{path A}}{\overset{2}{\int }}F\cdot dr}-{\displaystyle \underset{1\text{path B}}{\overset{2}{\int }}F\cdot dr}\\ ={\displaystyle \underset{1\text{path A}}{\overset{2}{\int }}F\cdot dr}-{\displaystyle \underset{1\text{path A}}{\overset{2}{\int }}F\cdot dr}\\ =0\end{array}$        (IV)

Conclusion:

Therefore, the path independence of ${\displaystyle \underset{1}{\overset{2}{\int }}F\cdot dr}$ is equivalent to the statement that the integral ${\displaystyle \underset{\Gamma }{\oint }F\cdot dr}$.

(b)

To determine

Show that Stokes’s theorem implies that $\nabla \times F=0$.

Answer to Problem 4.25P

Showed that Stokes’s theorem implies that $\nabla \times F=0$.

Explanation of Solution

Use equation (IV) in (I).

    $0={\displaystyle \oint \left(\nabla \times F\right)\cdot \widehat{n}dA}$        (V)

As $\widehat{n}$ and $dA$ are non-zero. So $\nabla \times F$ must be zero.

    $\nabla \times F=0$        (VI)

Conclusion:

Therefore, Stokes’s theorem implies that $\nabla \times F=0$.

(c)

To determine

Prove Stokes’s theorem.

Answer to Problem 4.25P

Proved Stokes’s theorem.

Explanation of Solution

Use $F_x\widehat{x}+F_y\widehat{y}+F_z\widehat{z}$ for $F$, and $dx\widehat{x}+dy\widehat{y}+dz\widehat{z}$ for $dr$ in the equation ${\displaystyle \underset{\Gamma }{\oint }F\cdot dr}$.

    $\begin{array}{c}{\displaystyle \underset{\Gamma }{\oint }F\cdot dr}=\left(F_x\widehat{x}+F_y\widehat{y}+F_z\widehat{z}\right)\cdot \left(dx\widehat{x}+dy\widehat{y}+dz\widehat{z}\right)\\ =F_{x}dx+F_{y}dy+F_{z}dz\end{array}$        (VII)

The value of $F_{z}dz$ is zero, since the curve is rectangle on xy plane.

Rewrite the equation (VII) in rectangle plane.

    ${\displaystyle \underset{\Gamma }{\oint }F\cdot dr}={\displaystyle \underset{x=B}{\overset{x=B+b}{\int }}{F}_{x}dx}+{\displaystyle \underset{y=C}{\overset{y=C+c}{\int }}{F}_{y}dy}+{\displaystyle \underset{x=B+b}{\overset{x=B}{\int }}{F}_{x}dx}+{\displaystyle \underset{y=C+c}{\overset{y=C}{\int }}{F}_{y}dy}$        (VIII)

When $x$ changes from $B$ to $B+b$, $y$ is constant at $C$, when $x$ changes from $B+b$ to $B$, $y$ is constant at $C+c$, when $y$ changes from $C$ to $C+c$, $x$ is constant at $B+b$, and when $y$ changes from $C+c$ to $C$, $y$ is constant at $B$.

Then rewrite the equation (VIII).

    $\begin{array}{c}{\displaystyle \underset{\Gamma }{\oint }F\cdot dr}={\displaystyle \underset{x=B}{\overset{x=B+b}{\int }}{F}_{x}dx}+{\displaystyle \underset{y=C}{\overset{y=C+c}{\int }}{F}_{y}dy}+{\displaystyle \underset{x=B+b}{\overset{x=B}{\int }}{F}_{x}dx}+{\displaystyle \underset{y=C+c}{\overset{y=C}{\int }}{F}_{y}dy}\\ =F_x\left(B+b,C,z\right)-F_x\left(B,C,z\right)+F_y\left(B+b,C+c,z\right)-F_y\left(B+b,C,z\right)+F_y\left(B+b,C,z\right)\\ +F_x\left(B,C+c,z\right)-F_x\left(B+b,C+c,z\right)+F_x\left(B,C+c,z\right)+F_y\left(B,C,z\right)+F_y\left(B,C+c,z\right)\end{array}$        (IX)

Write the expression for curl of a function.

    $\begin{array}{c}\nabla \times F=\left(\begin{array}{ccc}\widehat{x}& \widehat{y}& \widehat{z}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ F_x& F_y& F_z\end{array}\right)\\ =\widehat{x}\left[\frac{\partial F_z}{\partial y}-\frac{\partial F_y}{\partial z}\right]-\widehat{y}\left[\frac{\partial F_z}{\partial z}-\frac{\partial F_x}{\partial x}\right]+\widehat{z}\left[\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\right]\end{array}$        (X)

Use equation (X) in ${\displaystyle \oint \left(\nabla \times F\right)\cdot \widehat{n}dA}$ since $\widehat{n}=\widehat{z}$.

    $\begin{array}{c}{\displaystyle \int \left(\nabla \times F\right)\cdot \widehat{n}dA}={\displaystyle \int \left\{\widehat{x}\left[\frac{\partial F_z}{\partial y}-\frac{\partial F_y}{\partial z}\right]-\widehat{y}\left[\frac{\partial F_z}{\partial z}-\frac{\partial F_x}{\partial x}\right]+\widehat{z}\left[\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\right]\right\}\cdot \widehat{z}dA}\\ ={\displaystyle \int \left[\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\right]dxdy}\\ ={\displaystyle \int \frac{\partial F_y}{\partial x}dxdy}-{\displaystyle \int \frac{\partial F_x}{\partial y}dxdy}\\ ={\displaystyle \underset{x=B}{\overset{x=B+b}{\int }}{\displaystyle \underset{y=C}{\overset{y=C+c}{\int }}\frac{\partial F_y}{\partial x}dxdy}-{\displaystyle \underset{x=B}{\overset{x=B+b}{\int }}{\displaystyle \underset{y=C}{\overset{y=C+c}{\int }}\frac{\partial F_x}{\partial y}dxdy}}}\end{array}$        (XI)

Solve the equation (XI).

    $\begin{array}{c}{\displaystyle \int \left(\nabla \times F\right)\cdot \widehat{n}dA}={{\displaystyle \underset{y=C}{\overset{y=C+c}{\int }}\left[F_y\left(x,y,z\right)\right]}}_{x=B}^{x=B+b}dy-{{\displaystyle \underset{x=B}{\overset{x=B+b}{\int }}\left[F_x\left(x,y,z\right)\right]}}_{y=C}^{y=C+c}dx\\ ={\displaystyle \underset{y=C}{\overset{y=C+c}{\int }}\left[F_y\left(B+b,y,z\right)-F_y\left(B,y,z\right)\right]}dy-{\displaystyle \underset{x=B}{\overset{x=B+b}{\int }}\left[F_x\left(x,C+c,z\right)-F_x\left(x,C,z\right)\right]}dx\\ =\left[{\left[F_y\left(B+b,y,z\right)-F_y\left(B,y,z\right)\right]}_{y=C}^{y=C+c}-{\left[F_x\left(x,C+c,z\right)-F_x\left(x,C,z\right)\right]}_{x=B}^{x=B+b}\right]\\ =F_x\left(B+b,C,z\right)-F_x\left(B,C,z\right)+F_y\left(B+b,C+c,z\right)-F_y\left(B+b,C,z\right)+F_y\left(B+b,C,z\right)\\ +F_x\left(B,C+c,z\right)-F_x\left(B+b,C+c,z\right)+F_x\left(B,C+c,z\right)+F_y\left(B,C,z\right)+F_y\left(B,C+c,z\right)\end{array}$        (XII)

From the equation (IX) and (XII), it is proved that Stokes’s theorem that is ${\displaystyle \underset{\Gamma }{\oint }F\cdot dr}={\displaystyle \int \left(\nabla \times F\right)\cdot \widehat{n}dA}$.

Conclusion:

Therefore, Stokes’s theorem proved.