Problems for Chapter 3

If $\mathbf{F}$ is a central force, then for any point $P$ in space, the force is in the radial direction and its magnitude depends only on the distance $r$ from the origin:

$$ \mathbf{F} = F_r(r) \hat{\mathbf{e}}_r $$

Then we have:

$$ \begin{align*} F_x &= F_r(r) \sin \phi \cos \theta \\ F_y &= F_r(r) \sin \phi \sin \theta \\ F_z &= F_r(r) \cos \phi \end{align*} $$

and since:

$$ \begin{align*} x &= r \sin \phi \cos \theta \\ y &= r \sin \phi \sin \theta \\ z &= r \cos \phi \end{align*} $$

We have:

$$ \begin{align*} dx &= \sin \phi \cos \theta \, dr + r \cos \phi \cos \theta \, d\phi - r \sin \phi \sin \theta \, d\theta \\ dy &= \sin \phi \sin \theta \, dr + r \cos \phi \sin \theta \, d\phi + r \sin \phi \cos \theta \, d\theta \\ dz &= \cos \phi \, dr - r \sin \phi \, d\phi \end{align*} $$

Then:

$$ \begin{align*} F_x \, dx &+ F_y \, dy + F_z \, dz = \\ & F_r(r) (\sin^2 \phi \cos^2 \theta + \sin^2 \phi \sin^2 \theta + \cos^2 \phi) \, dr + \\ & F_r(r) (r \sin \phi \cos \phi \cos^2 \theta + r \sin \phi \cos \phi \sin^2 \theta - r \sin \phi \cos \phi) \, d\phi + \\ & F_r(r) (-r \sin^2 \phi \sin \theta \cos \theta + r \sin^2 \phi \sin \theta \cos \theta) \, d\theta \end{align*} $$

Which simplifies to:

$$ F_x \, dx + F_y \, dy + F_z \, dz = F_r(r) \, dr $$

On page 68 we saw that:

$$ \int_C \mathbf{F} \cdot \hat{\mathbf{t}} \, ds = \int_C (F_x \, dx + F_y \, dy + F_z \, dz) $$

Thus, we have:

$$ \begin{align*} \int_C \mathbf{F} \cdot \hat{\mathbf{t}} \, ds &= \int_C F_r(r) \, dr \\ &= \int_{r_1}^{r_2} F_r(r) \, dr \end{align*} $$

which is independent of the path taken and depends only on the initial and final radius.

Assume the force $\mathbf{F}$ is given by:

$$ \mathbf{F} = F_x \mathbf{i} + F_y \mathbf{j} + F_z \mathbf{k} $$

In the text on page 68 we saw that:

$$ \int_C \mathbf{F} \cdot \hat{\mathbf{t}} \, ds = \int_C (F_x \, dx + F_y \, dy + F_z \, dz) $$

Then for $C_1$, we have:

$$ \begin{align*} \int_{C_1} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds &= \int_{y + \Delta y / 2}^{y - \Delta y / 2} F_y\left(x - \frac{\Delta x}{2} , u\right) \, du \\[1em] & \approx -F_y\left(x - \frac{\Delta x}{2} , y\right) \Delta y \end{align*} $$

and for $C_2$, we have:

$$ \begin{align*} \int_{C_2} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds &= \int_{x - \Delta x / 2}^{x + \Delta x / 2} F_x\left(u, y - \frac{\Delta y}{2}\right) \, du \\[1em] & \approx F_x\left(x, y - \frac{\Delta y}{2}\right) \Delta x \end{align*} $$

and for $C_3$, we have:

$$ \begin{align*} \int_{C_3} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds &= \int_{C_3} \left( F_x \, dx + F_y \, dy \right) \\[1em] &= \int_{x + \Delta x / 2}^{x - \Delta x / 2} F_x\left(u, y(u)\right) \, du + \int_{y - \Delta y / 2}^{y + \Delta y / 2} F_y\left(x(u), u\right) \, du \\[1em] & \approx -F_x\left(x, y\right) \Delta x + F_y\left(x, y\right) \Delta y \end{align*} $$

Thus, we have:

$$ \begin{align*} \int_{C_1+C_2+C_3} \mathbf{F} &\cdot \hat{\mathbf{t}} \, ds \approx \\[0.5em] &\Delta y \left( F_y(x, y) - F_y\left(x - \frac{\Delta x}{2}, y\right) \right) + \\[0.5em] &\Delta x \left( F_x\left(x, y - \frac{\Delta y}{2}\right) - F_x(x, y) \right) \end{align*} $$

Since $\Delta A = \dfrac{\Delta x \Delta y}{2}$, we have:

$$ \begin{align*} \frac{1}{\Delta A} \int_{C_1+C_2+C_3} \mathbf{F} &\cdot \hat{\mathbf{t}} \, ds \approx \\[0.5em] & \frac{F_y(x, y) - F_y\left(x - \frac{\Delta x}{2}, y\right)} {\Delta x / 2} - \\[0.5em] &\frac{F_x(x, y) - F_x\left(x, y - \frac{\Delta y}{2}\right)}{\Delta y / 2} \end{align*} $$

So, we have:

$$ \lim_{\Delta A \to 0} \frac{1}{\Delta A} \int_{C_1+C_2+C_3} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds = \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} $$

$$ \begin{align*} \nabla \times \mathbf{F} &= \mathbf{i} \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right) \\ &+ \mathbf{j} \left( \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) \\ &+ \mathbf{k} \left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \end{align*} $$

Then:

$$ \begin{align*} \nabla \cdot (\nabla \times \mathbf{F}) &= \frac{\partial F_z}{\partial x \partial y} - \frac{\partial F_y}{\partial x \partial z} \\ &+ \frac{\partial F_x}{\partial y \partial z} - \frac{\partial F_z}{\partial x \partial y} \\ &+ \frac{\partial F_y}{\partial x \partial z} - \frac{\partial F_x}{\partial y \partial z} \\ &= 0 \end{align*} $$

Using the path shown in the figure, we’ll yield the $\theta$-component of $\nabla \times \mathbf{F}$ in cylindrical coordinates.

Viewed from above:

For $C_1$, we have $\hat{\mathbf{t}} = \hat{\mathbf{e}_r}$ and $ds = dr$. Then:

$$ \begin{align*} \int_{C_1} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds &= \int_{r - \Delta r / 2}^{r + \Delta r / 2} F_r(u, \theta, z - \frac{\Delta z}{2}) \, du \\[1em] &\approx F_r(r, \theta, z - \frac{\Delta z}{2}) \Delta r \end{align*} $$

For $C_2$, we have $\hat{\mathbf{t}} = \mathbf{\mathbf{e}_z}$ and $ds = dz$. Then:

$$ \begin{align*} \int_{C_2} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds &= \int_{z - \Delta z / 2}^{z + \Delta z / 2} F_z(r + \Delta r / 2, \theta, u) \, du \\[1em] &\approx F_z(r + \frac{\Delta r}{2}, \theta, z) \Delta z \end{align*} $$

For $C_3$, we have $\hat{\mathbf{t}} = -\hat{\mathbf{e}_r}$ and $ds = -dr$. Then:

$$ \begin{align*} \int_{C_3} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds &= \int_{r + \Delta r / 2}^{r - \Delta r / 2} (-F_r(u, \theta, z + \frac{\Delta z}{2})) \, (-du) \\[1em] &\approx -F_r(r, \theta, z + \frac{\Delta z}{2}) \Delta r \end{align*} $$

For $C_4$, we have $\hat{\mathbf{t}} = -\hat{\mathbf{e}_z}$ and $ds = -dz$. Then:

$$ \begin{align*} \int_{C_4} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds &= \int_{z + \Delta z / 2}^{z - \Delta z / 2} (-F_z(r - \Delta r / 2, \theta, u)) \, (-du) \\[1em] &\approx -F_z(r - \frac{\Delta r}{2}, \theta, z) \Delta z \end{align*} $$

Then, we have:

$$ \begin{align*} \int_{C_1 + C_2 + C_3 + C_4} \mathbf{F} &\cdot \hat{\mathbf{t}} \, ds \approx \\[0.5em] &\Delta r \left( F_r(r, \theta, z - \frac{\Delta z}{2}) - F_r(r, \theta, z + \frac{\Delta z}{2}) \right) + \\[0.5em] &\Delta z \left( F_z(r + \frac{\Delta r}{2}, \theta, z) - F_z(r - \frac{\Delta r}{2}, \theta, z) \right) \end{align*} $$

Since $\Delta A = \Delta r \Delta z$, we have:

$$ \begin{align*} \frac{1}{\Delta A} \int_{C_1 + C_2 + C_3 + C_4} \mathbf{F} &\cdot \hat{\mathbf{t}} \, ds \approx \\[0.5em] & \frac{F_r(r, \theta, z - \dfrac{\Delta z}{2}) - F_r(r, \theta, z + \dfrac{\Delta z}{2})} {\Delta z} - \\[0.5em] &\frac{F_z(r + \dfrac{\Delta r}{2}, \theta, z) - F_z(r - \dfrac{\Delta r}{2}, \theta, z)}{\Delta r} \end{align*} $$

So, $\theta$-component of $\nabla \times \mathbf{F}$ is:

$$ \lim_{\Delta A \to 0} \frac{1}{\Delta A} \int_{C_1 + C_2 + C_3 + C_4} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds = \frac{\partial F_r}{\partial z} - \frac{\partial F_z}{\partial r} $$

Using the path shown in the figure, we’ll yield the $r$-component of $\nabla \times \mathbf{F}$ in cylindrical coordinates.

For $C_1$, we have $\hat{\mathbf{t}} = \hat{\mathbf{e}_\theta}$ and $ds = r d\theta$. Then:

$$ \begin{align*} \int_{C_1} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds &= \int_{\theta - \Delta \theta / 2}^{\theta + \Delta \theta / 2} F_\theta(r, u, z - \frac{\Delta z}{2}) \, r \, du \\[1em] &\approx F_\theta(r, \theta, z - \frac{\Delta z}{2}) r \Delta \theta \end{align*} $$

For $C_2$, we have $\hat{\mathbf{t}} = \hat{\mathbf{e}_z}$ and $ds = dz$. Then:

$$ \begin{align*} \int_{C_2} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds &= \int_{z - \Delta z / 2}^{z + \Delta z / 2} F_z(r, \theta + \frac{\Delta \theta}{2}, u) \, du \\[1em] &\approx F_z(r, \theta + \frac{\Delta \theta}{2}, z) \Delta z \end{align*} $$

For $C_3$, we have $\hat{\mathbf{t}} = -\hat{\mathbf{e}_\theta}$ and $ds = -r d\theta$. Then:

$$ \begin{align*} \int_{C_3} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds &= \int_{\theta + \Delta \theta / 2}^{\theta - \Delta \theta / 2} (-F_\theta(r, u, z + \frac{\Delta z}{2})) \, (-r \, du) \\[1em] &\approx -F_\theta(r, \theta, z + \frac{\Delta z}{2}) r \Delta \theta \end{align*} $$

For $C_4$, we have $\hat{\mathbf{t}} = -\hat{\mathbf{e}_z}$ and $ds = -dz$. Then:

$$ \begin{align*} \int_{C_4} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds &= \int_{z + \Delta z / 2}^{z - \Delta z / 2} (-F_z(r, \theta - \frac{\Delta \theta}{2}, u)) \, (-du) \\[1em] &\approx -F_z(r, \theta - \frac{\Delta \theta}{2}, z) \Delta z \end{align*} $$

Then, we have:

$$ \begin{align*} \int_{C_1 + C_2 + C_3 + C_4} \mathbf{F} &\cdot \hat{\mathbf{t}} \, ds \approx \\[0.5em] &r \, \Delta \theta \left( F_\theta(r, \theta, z - \frac{\Delta z}{2}) - F_\theta(r, \theta, z + \frac{\Delta z}{2}) \right) + \\[0.5em] &\Delta z \left( F_z(r, \theta + \frac{\Delta \theta}{2}, z) - F_z(r, \theta - \frac{\Delta \theta}{2}, z) \right) \end{align*} $$

Since $\Delta A = r \Delta \theta \Delta z$, we have:

$$ \begin{align*} \frac{1}{\Delta A} \int_{C_1 + C_2 + C_3 + C_4} \mathbf{F} &\cdot \hat{\mathbf{t}} \, ds \approx \\[0.5em] & \frac{F_z(r, \theta + \dfrac{\Delta \theta}{2}, z) - F_z(r, \theta - \dfrac{\Delta \theta}{2}, z)} {r \, \Delta \theta} - \\[0.5em] &\frac{F_\theta(r, \theta, z + \dfrac{\Delta z}{2}) - F_\theta(r, \theta, z - \dfrac{\Delta z}{2})}{\Delta z} \end{align*} $$

So, $r$-component of $\nabla \times \mathbf{F}$ is:

$$ \lim_{\Delta A \to 0} \frac{1}{\Delta A} \int_{C_1 + C_2 + C_3 + C_4} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds = \frac{1}{r} \, \frac{\partial F_z}{\partial \theta} - \frac{\partial F_{\theta}}{\partial z} $$

Using $\nabla \times \mathbf{F} = \mathbf{i} \left( \frac{\partial{F_z}}{\partial y} - \frac{\partial{F_y}}{\partial z} \right) + \mathbf{j} \left( \frac{\partial{F_x}}{\partial z} - \frac{\partial{F_z}}{\partial x} \right) + \mathbf{k} \left( \frac{\partial{F_y}}{\partial x} - \frac{\partial{F_x}}{\partial y} \right)$, we get: $\nabla \times \mathbf{i}$ = $\nabla \times \mathbf{j}$ = $\nabla \times \mathbf{k} = 0$.

Then, $i$-component of $\oint_C \hat{\mathbf{t}} \\, ds$ is:

$$ \begin{align*} \mathbf{i} \cdot \oint_C \hat{\mathbf{t}} \, ds &= \oint_C \mathbf{i} \cdot \hat{\mathbf{t}} \, ds \\ &= \iint_S \nabla \times \mathbf{i} \cdot \hat{\mathbf{n}} \, dS \\ &= 0 \end{align*} $$

Similarly, we can show that $j$-component and $k$-component of $\oint_C \hat{\mathbf{t}} \\, ds$ are also zero.

Then, we have:

$$ \oint_C \hat{\mathbf{t}} \, ds = 0 $$

Form $C_3$ by joining $C_1$ and $C_2$ with a segment on $S$ and reversing the direction of $C_1$.

In figure below $a$ and $b$ segments completely overlap but in opposite directions. They connect inner and outer parts of the curve by moving along $S$.

Then $S$ is a capping surface for the region enclosed by $C_3$.

Then, we have:

$$ \oint_{C_3} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds = \iint_S \nabla \times \mathbf{F} \cdot \hat{\mathbf{n}} \, dS $$

Since $\nabla \times \mathbf{F} = 0$, we have:

$$ \oint_{C_3} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds = 0 \tag{1} $$

Since we have reversed the direction of $C_1$ and line integrals along $a$ and $b$ cancel each other, we have:

$$ \oint_{C_3} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds = -\oint_{C_1} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds + \oint_{C_2} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds \tag{2} $$

Putting (1) and (2) together, we have:

$$ -\oint_{C_1} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds + \oint_{C_2} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds = 0 $$

Thus, we have:

$$ \oint_{C_1} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds = \oint_{C_2} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds $$

$$ \begin{align*} (\nabla \times \mathbf{B})_r &= \frac{1}{r} \frac{\partial F_z}{\partial \theta} - \frac{\partial F_\theta}{\partial z} \\[0.8em] &= - \frac{\partial}{\partial z} \left( \frac{\mu_0 I}{2 \pi r} \right) \\[0.5em] &= 0 \\[1.5em] (\nabla \times \mathbf{B})_\theta &= \frac{\partial F_r}{\partial z} - \frac{\partial F_z}{\partial r} \\[0.5em] &= 0 \\[1.5em] (\nabla \times \mathbf{B})_z &= \frac{1}{r} \frac{\partial}{\partial r}(r F_\theta) - \frac{\partial F_r}{\partial \theta} \\[1em] &= \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\mu_0 I}{2 \pi r} \right)\\[1em] &= \frac{1}{r} \frac{\partial}{\partial r} \left( \frac{\mu_0 I}{2 \pi} \right) \\[0.5em] &= 0 \end{align*} $$

For a circle of radius $r$, $ds = r \\, d\theta$ and $\hat{\mathbf{t}} = \hat{\mathbf{e}_\theta}$. Then, we have:

$$ \oint_{circle} \mathbf{B} \cdot \hat{\mathbf{t}} \, ds = \int_{0}^{2 \pi} \frac{\mu_0 I}{2 \pi r} r \, d\theta = \mu_0 I $$

For any curve $C$ not passing through the wire, there exists a circle around the wire inside it. Putting together the results from part (a) and (b), circulation of $\mathbf{B}$ around $C$ is equal to the circulation of $\mathbf{B}$ around the circle, which is $\mu_0 I$.

So, for any curve $C$ not passing through the wire, we have:

$$ \oint_C \mathbf{B} \cdot \hat{\mathbf{t}} \, ds = \mu_0 I $$

Since $\mathcal{E} = \oint_C \mathbf{E} \cdot \hat{\mathbf{t}} \\, ds$, then for any capping surface $S$ of $C$, we have:

$$ \mathcal{E} = \iint_S \nabla \times \mathbf{E} \cdot \hat{\mathbf{n}} \, dS \tag{1} $$

Differentiating $\Phi = \iint_S \mathbf{B} \cdot \hat{\mathbf{n}} \\, dS$ with respect to $t$, we have:

$$ \frac{d \Phi}{d t} = \iint_S \frac{\partial \mathbf{B}}{\partial t} \cdot \hat{\mathbf{n}} \, dS $$

Where we used the fact that $S$ is stationary, so $\hat{\mathbf{n}}$ does not depend on $t$.

From the problem statement we know that $\mathcal{E} = - \dfrac{d \Phi}{dt}$. Then, we have:

$$ \mathcal{E} = - \iint_S \frac{\partial \mathbf{B}}{\partial t} \cdot \hat{\mathbf{n}} \, dS \tag{2} $$

Putting (1) and (2) together, we have:

$$ \iint_S \nabla \times \mathbf{E} \cdot \hat{\mathbf{n}} \, dS = - \iint_S \frac{\partial \mathbf{B}}{\partial t} \cdot \hat{\mathbf{n}} \, dS $$

Since this is true for any capping surface $S$ of $C$, we have:

$$ \nabla \times \mathbf{E} = - \frac{\partial \mathbf{B}}{\partial t} $$

Since $\nabla \times \mathbf{F} = 0$, the line integral is independent of the path taken.

Then we can use:

• $C_1$: straight segment from $(0, 0, 0)$ to $(1, 0, 0)$,
• $C_2$: straight segment from $(1, 0, 0)$ to $(1, 1, 0)$,
• $C_3$: straight segment from $(1, 1, 0)$ to $(1, 1, 1)$.

Then we have:

$$ \begin{align*} \int_{C_1} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds &= \int_{0}^{1} e^0 \, dx = 1 \\[1em] \int_{C_2} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds &= \int_{0}^{1} (1 - e^{-y}) \, dy = e^{-1} \\[1em] \int_{C_3} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds &= \int_{0}^{1} (e^{-1} - e^{-z}) \, dz = 2 e^{-1} - 1 \end{align*} $$

Then, we have:

$$ \int_C \mathbf{F} \cdot \hat{\mathbf{t}} \, ds = \int_{C_1 + C_2 + C_3} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds = 3 e^{-1} $$

Lemma. $\dfrac{\partial}{\partial t} \left( \nabla \cdot \mathbf{E} \right) = \nabla \cdot \left( \dfrac{\partial \mathbf{E}}{\partial t} \right)$.

Proof of Lemma. Let $\mathbf{E} = (E_x, E_y, E_z)$. Then

$$ \nabla \cdot \mathbf{E} = \dfrac{\partial E_x}{\partial x} + \dfrac{\partial E_y}{\partial y} + \dfrac{\partial E_z}{\partial z} $$

Differentiating with respect to $t$, we have:

$$ \begin{align*} \dfrac{\partial}{\partial t} \left( \nabla \cdot \mathbf{E} \right) &= \dfrac{\partial^2 E_x}{\partial x \partial t} + \dfrac{\partial^2 E_y}{\partial y \partial t} + \dfrac{\partial^2 E_z}{\partial z \partial t} \\[1em] &= \nabla \cdot \left( \frac{\partial E_x}{\partial t}, \frac{\partial E_y}{\partial t}, \frac{\partial E_z}{\partial t} \right) \\[1em] &= \nabla \cdot \left( \frac{\partial \mathbf{E}}{\partial t} \right) \end{align*} $$

Proof of Continuity Equation. From (M4) we have:

$$ \mu_0 \mathbf{J} = \nabla \times \mathbf{B} - \varepsilon_0 \mu_0 \frac{\partial \mathbf{E}}{\partial t} $$

Taking the divergence of both sides and using the linearity of the divergence operator, we have:

$$ \mu_0 \left( \nabla \cdot \mathbf{J} \right) = \nabla \cdot \left( \nabla \times \mathbf{B} \right) - \varepsilon_0 \mu_0 \nabla \cdot \frac{\partial \mathbf{E}}{\partial t} $$

In III.7 we proved that $\nabla \cdot \left( \nabla \times \mathbf{B} \right) = 0$. Then, we have:

$$ \mu_0 \left( \nabla \cdot \mathbf{J} \right) = - \varepsilon_0 \mu_0 \nabla \cdot \frac{\partial \mathbf{E}}{\partial t} $$

which means:

$$ \nabla \cdot \mathbf{J} = - \varepsilon_0 \nabla \cdot \frac{\partial \mathbf{E}}{\partial t} \tag{1} $$

Differentiating (M1) with respect to $t$, we have:

$$ \frac{\partial}{\partial t} \left( \nabla \cdot \mathbf{E} \right) = \frac{1}{\varepsilon_0} \frac{\partial \rho}{\partial t} $$

Using the lemma, we have:

$$ \nabla \cdot \frac{\partial \mathbf{E}}{\partial t} = \frac{1}{\varepsilon_0} \frac{\partial \rho}{\partial t} \tag{2} $$

Putting (1) and (2) together, we have:

$$ \nabla \cdot \mathbf{J} = - \frac{\partial \rho}{\partial t} $$

Rearranging, we have:

$$ \nabla \cdot \mathbf{J} + \frac{\partial \rho}{\partial t} = 0 $$

Lemma. $\nabla \cdot \left( \mathbf{E} \times \mathbf{B} \right) = \mathbf{B} \cdot \left( \nabla \times \mathbf{E} \right) - \mathbf{E} \cdot \left( \nabla \times \mathbf{B} \right)$.

Proof of Lemma. We have

$$ \mathbf{E} \times \mathbf{B} = (E_y B_z - E_z B_y, E_z B_x - E_x B_z, E_x B_y - E_y B_x) $$

Then

$$ \begin{align*} \nabla \cdot \left( \mathbf{E} \times \mathbf{B} \right) &= E_y \frac{\partial B_z}{\partial x} + B_z \frac{\partial E_y}{\partial x} - E_z \frac{\partial B_y}{\partial x} - B_y \frac{\partial E_z}{\partial x} + \\ &+ E_z \frac{\partial B_x}{\partial y} + B_x \frac{\partial E_z}{\partial y} - E_x \frac{\partial B_z}{\partial y} - B_z \frac{\partial E_x}{\partial y} + \\ &+ E_x \frac{\partial B_y}{\partial z} + B_y \frac{\partial E_x}{\partial z} - E_y \frac{\partial B_x}{\partial z} - B_x \frac{\partial E_y}{\partial z} \end{align*} $$

Rearranging, we have:

$$ \begin{align*} \nabla \cdot \left( \mathbf{E} \times \mathbf{B} \right) &= E_x \left(\frac{\partial B_y}{\partial z} - \frac{\partial B_z}{\partial y}\right) + E_y \left(\frac{\partial B_z}{\partial x} - \frac{\partial B_x}{\partial z}\right) + \\ &+ E_z \left(\frac{\partial B_x}{\partial y} - \frac{\partial B_y}{\partial x}\right) + B_x \left(\frac{\partial E_z}{\partial y} - \frac{\partial E_y}{\partial z}\right) + \\ &+ B_y \left(\frac{\partial E_x}{\partial z} - \frac{\partial E_z}{\partial x}\right) + B_z \left(\frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y}\right) \end{align*} $$

Which can be rewritten as $B \cdot \left( \nabla \times \mathbf{E} \right) - E \cdot \left( \nabla \times \mathbf{B} \right)$.

Proof of Energy Density Equation.

Inner product of sides of (M3) by $\mathbf{B}$ on the left gives:

$$ \mathbf{B} \cdot (\nabla \times \mathbf{E}) = - \mathbf{B} \cdot \frac{\partial \mathbf{B}}{\partial t} \tag{1} $$

Inner product of sides of (M4) by $E$ on the left gives:

$$ \mathbf{E} \cdot (\nabla \times \mathbf{B}) = \varepsilon_0 \mu_0 \mathbf{E} \cdot \frac{\partial \mathbf{E}}{\partial t} + \mu_0 \mathbf{E} \cdot \mathbf{J} \tag{2} $$

Putting the lemma with (1) and (2) gives:

$$ \nabla \left( \frac{\mathbf{E} \times \mathbf{B}}{\mu_0} \right) = -\frac{1}{\mu_0} \mathbf{B} \cdot \frac{\partial \mathbf{B}}{\partial t} - \varepsilon_0 \mathbf{E} \cdot \frac{\partial \mathbf{E}}{\partial t} - \mathbf{E} \cdot \mathbf{J} \tag{3} $$

On the other hand, differentiating sides of $\rho_E = \dfrac{1}{2}(\varepsilon_0 \mathbf{E} \cdot \mathbf{E} + \mathbf{B} \cdot \mathbf{B} / \mu_0)$ with respect to $t$ gives:

$$ \frac{\partial \rho_E}{\partial t} = \frac{1}{\mu_0} \frac{\partial \mathbf{B}}{\partial t} \mathbf{B} + \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t} \mathbf{E} \tag{4} $$

Putting (3) and (4) together we get:

$$ \frac{\partial \rho_E}{\partial t} + \nabla \cdot \left( \frac{\mathbf{E} \times \mathbf{B}}{\mu_0} \right) = - \mathbf{J} \cdot \mathbf{E} $$

We’ll use the divergence theorem:

$$ \iint_S \mathbf{G} \cdot \hat{\mathbf{n}} \, dS = \iiint_V \nabla \cdot \mathbf{G} \, dV \tag{1} $$

Calculating the surface integral.

First we calculate the surface integral $\iint_S \mathbf{G} \cdot \hat{\mathbf{n}} \\, dS$.

In upper and lower parts of the surface, the normal vector is $\pm \mathbf{k}$, and since $\mathbf{G}$ has $0$ as the $\mathbf{k}$ component, then the surface integral is zero.

To calculate the surface integral of the shaded part in the figure, since the value of $\mathbf{G}$ and the normal vector $\hat{\mathbf{n}}$ do not depend on the $z$-coordinate, we choose $dS$ as below. Then, $dS = h \\, ds$, where $h$ is the height of the surface.

Then surface integral of the shaded part in the figure will be equal to:

$$ h \, \oint_C \mathbf{G} \cdot \hat{\mathbf{n}} \, ds $$

Since $C$ is on the $xy$-plane, then its tangent vector is:

$$ \hat{\mathbf{t}} = \mathbf{i} \frac{dx}{ds} + \mathbf{j} \frac{dy}{ds} $$

Given that $C$ is oriented counterclockwise, then the normal vector is the tangent vector rotated by $90^\circ$ clockwise:

$$ \hat{\mathbf{n}} = \mathbf{i} \frac{dy}{ds} - \mathbf{j} \frac{dx}{ds} $$

Then, we have:

$$ \begin{align*} \oint_C \mathbf{G} \cdot \hat{\mathbf{n}} \, ds &= \oint_C \left( G_x \frac{dy}{ds} - G_y \frac{dx}{ds} \right) \, ds \\[0.8em] &= \oint_C G_x \, dy - G_y \, dx \end{align*} $$

and then the total surface integral is:

$$ \iint_S \mathbf{G} \cdot \hat{\mathbf{n}} \, dS = h \oint_C G_x \, dy - G_y \, dx \tag{2} $$

Calculating the volume integral.

Since value of $\mathbf{G}$ doesn’t depend on $z$, we can choose $dV$ as below. Then, $dV = h \\, dS$.

Then, we have:

$$ \begin{align*} \iiint_V \nabla \cdot \mathbf{G} \, dV &= h \iint_R \nabla \cdot \mathbf{G} \, dS \\[0.8em] &= h \iint_R \left( \frac{\partial G_x}{\partial x} + \frac{\partial G_y}{\partial y} \right) \, dx dy \tag{3} \end{align*} $$

---

Finally, putting (1), (2), and (3) together, we have:

$$ \oint_C G_x \, dy - G_y \, dx = \iint_R \left( \frac{\partial G_x}{\partial x} + \frac{\partial G_y}{\partial y} \right) \, dx dy $$

Since $S$ lies entirely in the $xy$-plane, then

• Assuming counterclockwise orientation, $\hat{\mathbf{n}} = \hat{\mathbf{k}}$,
• $dS = dx \\, dy$.

Since $\nabla \times \mathbf{F} = \mathbf{k} \left( \dfrac{\partial F_y}{\partial x} - \dfrac{\partial F_x}{\partial y} \right)$, we have:

$$ \nabla \times \mathbf{F} \cdot \hat{\mathbf{n}} = \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} $$

and then:

$$ \iint_S \nabla \times \mathbf{F} \cdot \hat{\mathbf{n}} \, dS = \iint_R \left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \, dx dy \tag{1} $$

---

Since $C$ is a closed curve lying entirely in the $xy$-plane, then:

$$ \oint_C \mathbf{F} \cdot \hat{\mathbf{t}} \, ds = \oint_C F_x \, dx + F_y \, dy \tag{2} $$

---

Then, by Stokes’ theorem we have:

$$ \oint_C \mathbf{F} \cdot \hat{\mathbf{t}} \, ds = \iint_S \nabla \times \mathbf{F} \cdot \hat{\mathbf{n}} \, dS \tag{3} $$

Putting (1), (2), and (3) together, we have:

$$ \oint_C F_x \, dx + F_y \, dy = \iint_R \left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \, dx dy $$

The divergence theorem for $\mathbf{i} F_x + \mathbf{j} F_y$ is the same as Stokes’ theorem for $-\mathbf{i} F_y + \mathbf{j} F_x$.

So they are identical in two dimensions.

Using the 2-dimensional Stokes’ theorem, we have:

$$ \oint_C F_x \, dx + F_y \, dy = \iint_R \left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \, dx \, dy $$

So, in order for the line integral to be equal to the area enclosed by $C$, we need:

$$ \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} = 1 $$

or equivalently:

$$ \nabla \times \mathbf{F} = \mathbf{k} $$

• $\mathbf{F} = \mathbf{i} y + 2 \mathbf{j} x$
• $\mathbf{F} = \hat{\mathbf{e}}_\theta \dfrac{r}{2}$ (cylendrical coordinates)

Let $\mathbf{F} = \mathbf{i} y + 2 \mathbf{j} x$ and consider the rectangle below:

Then, we have:

$$ \begin{align*} \int_{C_1} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds &= 0 \\[0.8em] \int_{C_2} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds &= \int_{0}^{a} 2 \cdot b \, dy = 2 a b \\[0.8em] \int_{C_3} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds &= \int_{b}^{0} a \, dx = - a b \\[0.8em] \int_{C_4} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds &= 0 \end{align*} $$

Putting together, we have:

$$ \oint_{C_1 + C_2 + C_3 + C_4} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds = 2 a b - a b = a b $$

Let $\mathbf{F} = \mathbf{i} y + 2 \mathbf{j} x$ and consider the right triangle below:

Then, we have:

$$ \begin{align*} \int_{C_1} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds &= 0 \\[0.8em] \int_{C_3} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds &= 0 \end{align*} $$

For $C_2$, note that $y = \dfrac{b}{a} (a - x)$, so $dy = -\dfrac{b}{a} \\, dx$.

Then, we have:

$$ \begin{align*} \int_{C_2} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds &= \int_{C_2} (y \, dx + 2 x \, dy) \\[0.8em] &= \int_{0}^{a} \left( \dfrac{b}{a} (a - x) \, dx - 2 x \dfrac{b}{a} \, dx \right) \\[0.8em] &= \int_{0}^{a} \left( \dfrac{b}{a} (a - 3 x) \right) \, dx \\[0.8em] &= \frac{ab}{2} \end{align*} $$

Let $\mathbf{F} = \hat{\mathbf{e}}_\theta \dfrac{r}{2}$. We have $ds = r \\, d\theta$ and $\hat{\mathbf{t}} = \hat{\mathbf{e}}_\theta$.

Then, we have:

$$ \begin{align*} \oint_C \mathbf{F} \cdot \hat{\mathbf{t}} \, ds &= \oint_C \left( \frac{r}{2} \right) r \, d\theta \\[0.8em] &= \frac{r^2}{2} \int_0^{2\pi} \, d\theta = \pi r^2 \end{align*} $$

We have $\hat{\mathbf{t}} = \mathbf{i} \dfrac{dx}{ds} + \mathbf{j} \dfrac{dy}{ds}$, then:

$$ \begin{align*} \mathbf{r} \times \hat{\mathbf{t}} &= \mathbf{k} \left( x \frac{dy}{ds} - y \frac{dx}{ds} \right) \\[0.8em] &= \mathbf{k} \left( (-y \mathbf{i} + x \mathbf{j}) \cdot \hat{\mathbf{t}} \right) \end{align*} $$

Then, we have:

$$ \frac{1}{2} \oint_C \mathbf{r} \times \hat{\mathbf{t}} \, ds = \frac{\mathbf{k}}{2} \oint_C (-y \mathbf{i} + x \mathbf{j}) \cdot \hat{\mathbf{t}} \, ds $$

Since $\nabla \times (-y \mathbf{i} + x \mathbf{j}) = 2 \mathbf{k}$, then by Stokes’ theorem we have:

$$ \oint_C (-y \mathbf{i} + x \mathbf{j}) \cdot \hat{\mathbf{t}} \, ds = 2 A $$

Then, we have:

$$ \frac{1}{2} \oint_C \mathbf{r} \times \hat{\mathbf{t}} \, ds = A \mathbf{k} $$

which has a magnitude of $A$.

Since $\nabla \cdot \mathbf{G} = 0$, then:

$$ \frac{\partial G_x}{\partial x} = - \frac{\partial G_y}{\partial y} - \frac{\partial G_z}{\partial z} \tag{1} $$

We have:

$$ \begin{align*} \nabla \times \mathbf{H} &= \mathbf{i} \left( \frac{\partial H_z}{\partial y} - \frac{\partial H_y}{\partial z} \right) + \mathbf{j} \left( \frac{\partial H_x}{\partial z} - \frac{\partial H_z}{\partial x} \right) + \mathbf{k} \left( \frac{\partial H_y}{\partial x} - \frac{\partial H_x}{\partial y} \right) \\[1em] &= \mathbf{i} \left( \frac{\partial H_z}{\partial y} - \frac{\partial H_y}{\partial z} \right) \+ \mathbf{j} \left( 0 - \frac{\partial H_z}{\partial x} \right) + \mathbf{k} \left( \frac{\partial H_y}{\partial x} - 0 \right) \end{align*} $$

We have:

$$ \begin{align*} \frac{\partial H_y}{\partial x} &= G_z(x, y, z) \\[1em] \frac{\partial H_y}{\partial z} &= \int_{x_0}^{x} \frac{\partial G_z(x', y, z)}{\partial z} \, dx' \\[1em] \frac{\partial H_z}{\partial x} &= - G_y(x, y, z) \\[1em] \frac{\partial H_z}{\partial y} &= - \int_{x_0}^{x} \frac{\partial G_y(x', y, z)}{\partial y} \, dx' + G_x(x_0, y, z) \end{align*} $$

As a direct consequence, we have:

$$ \begin{align*} (\nabla \times \mathbf{H})_y &= G_y(x, y, z) \\[1em] (\nabla \times \mathbf{H})_z &= G_z(x, y, z) \end{align*} $$

To calculate $(\nabla \times \mathbf{H})_x$, we have:

$$ (\nabla \times \mathbf{H})_x = \int_{x_0}^{x} \left( - \frac{\partial G_y(x', y, z)}{\partial y} - \frac{\partial G_z(x', y, z)}{\partial z} \right) \, dx' + G_x(x_0, y, z) $$

Using (1), we have:

$$ (\nabla \times \mathbf{H})_x = \int_{x_0}^{x} \frac{\partial G_x(x', y, z)}{\partial x} \, dx' + G_x(x_0, y, z) $$

Then using the fundamental theorem of calculus, we have:

$$ (\nabla \times \mathbf{H})_x = G_x(x, y, z) - G_x(x_0, y, z) + G_x(x_0, y, z) = G_x(x, y, z) $$

No, the vector function $\mathbf{H}$ is not unique. For example, adding constants to the components of $\mathbf{H}$ does not change the value of $\nabla \times \mathbf{H}$.

Possible, since $\nabla \cdot \mathbf{G} = 0$.

Using the formula from Problem III-24(a), we have:

$$ \mathbf{H} = \mathbf{j} \frac{x^2 - x_0^2}{2} + \mathbf{k} \left( -z (x - x_0) + \frac{y^2 - y_0^2}{2} \right) $$

For any $x_0$ and $y_0$.

Possible, since $\nabla \cdot \mathbf{G} = 0$.

Using the formula from Problem III-24(a), we have:

$$ \mathbf{H} = \mathbf{j} B_0 (x - x_0) $$

for any $x_0$.

Not possible, since $\nabla \cdot \mathbf{G} = 2x \neq 0$.

Possible, since $\nabla \cdot \mathbf{G} = 0$.

Using the formula from Problem III-24(a), we have:

$$ \mathbf{H} = -\mathbf{j} z (x - x_0) + \mathbf{k} \left( x y + x_0 y - 2 x_0 y_0 \right) $$

For any $x_0$ and $y_0$.

Not possible, since $\nabla \cdot \mathbf{G} = 2 \neq 0$.

$$ \begin{align*} \oint_C \mathbf{A} \cdot \hat{\mathbf{t}} \, ds &= \iint_S (\nabla \times \mathbf{A}) \cdot \hat{\mathbf{n}} \, dS \quad \text{(Stokes' Theorem)} \\[1em] &= \iint_S \mathbf{B} \cdot \hat{\mathbf{n}} \, dS \\[1em] \end{align*} $$

which is the flux of $\mathbf{B}$ through $S$.

Assume that $\mathbf{G} = \nabla \times \mathbf{H}$.

For any closed surface $S$ capping $C$ and $S'$ capping $C'$, let $V$ be the volume enclosed by $S$ and $S'$. Then, we have:

$$ \begin{align*} \iiint_V \nabla \cdot \mathbf{G} \, dV &= \iint_{S+S'} \mathbf{G} \cdot \hat{\mathbf{n}} \, dS &\text{(Divergence Theorem)} \\[1em] &= \iint_{S+S'} (\nabla \times \mathbf{H}) \cdot \hat{\mathbf{n}} \, dS \\[1em] &= \oint_{C + C'} \mathbf{H} \cdot \hat{\mathbf{t}} \, ds &\text{(Stokes' Theorem)} \end{align*} $$

Since $C$ and $C'$ follow the same path but in opposite directions, then the last integral is zero.

So, for volume bounded by any $S$ and $S'$ capping $C$ and $C'$, we have:

$$ \iiint_V \nabla \cdot \mathbf{G} \, dV = 0 $$

Since choice of $S$ and $S'$ is arbitrary, it implies that $\nabla \cdot \mathbf{G} = 0$.

Using the Stokes’ theorem, for a closed path $C$ and a surface $S$ capping $C$, we have:

$$ \iint_S \nabla \times \mathbf{H} \cdot \hat{\mathbf{n}} \, dS = \oint_C \mathbf{H} \cdot \hat{\mathbf{t}} \, ds $$

Since $\mathbf{G} = \nabla \times \mathbf{H}$, then we have:

$$ \iint_S \mathbf{G} \cdot \hat{\mathbf{n}} \, dS = \oint_C \mathbf{H} \cdot \hat{\mathbf{t}} \, ds $$

and this is the integral form of the equation $\mathbf{G} = \nabla \times \mathbf{H}$.

Consider the cuboid shown in the figure below:

Side S1.

• Normal: $\hat{\mathbf{j}}$,
• $\hat{\mathbf{n}} \times \mathbf{F} = \hat{\mathbf{i}} F_z - \hat{\mathbf{k}} F_x$,
• Area: $dS = \Delta x \\, \Delta z$.

Then, we have:

$$ \iint_{S_1} \hat{\mathbf{n}} \times \mathbf{F} \, dS \approx \Delta x \Delta z \left( \mathbf{i} F_z(x, y + \frac{\Delta y}{2}, z) - \mathbf{k} F_x(x, y + \frac{\Delta y}{2}, z) \right) $$

Then:

$$ \frac{1}{\Delta V} \iint_{S_1} \hat{\mathbf{n}} \times \mathbf{F} \, dS \approx \frac{\mathbf{i} F_z(x, y + \frac{\Delta y}{2}, z) - \mathbf{k} F_x(x, y + \frac{\Delta y}{2}, z)}{\Delta y} $$

Side S2.

• Normal: $-\hat{\mathbf{j}}$,
• $\hat{\mathbf{n}} \times \mathbf{F} = -\hat{\mathbf{i}} F_z + \hat{\mathbf{k}} F_x$,
• Area: $dS = \Delta x \\, \Delta z$.

Then, we have:

$$ \iint_{S_2} \hat{\mathbf{n}} \times \mathbf{F} \, dS \approx -\Delta x \Delta z \left( \mathbf{i} F_z(x, y - \frac{\Delta y}{2}, z) - \mathbf{k} F_x(x, y - \frac{\Delta y}{2}, z) \right) $$

Then:

$$ \frac{1}{\Delta V} \iint_{S_2} \hat{\mathbf{n}} \times \mathbf{F} \, dS \approx -\frac{\mathbf{i} F_z(x, y - \frac{\Delta y}{2}, z) - \mathbf{k} F_x(x, y - \frac{\Delta y}{2}, z)}{\Delta y} $$

Adding S1 and S2.

$$ \begin{align*} \frac{1}{\Delta V} \iint_{S_1+S_2} &\hat{\mathbf{n}} \times \mathbf{F} \, dS \approx \\ & \mathbf{i} \frac{F_z(x, y + \frac{\Delta y}{2}, z) - F_z(x, y - \frac{\Delta y}{2}, z)}{\Delta y} \\ \- &\mathbf{k} \frac{F_x(x, y + \frac{\Delta y}{2}, z) - F_x(x, y - \frac{\Delta y}{2}, z)}{\Delta y} \end{align*} $$

which means:

$$ \lim_{\Delta V \to 0} \frac{1}{\Delta V} \iint_{S_1+S_2} \hat{\mathbf{n}} \times \mathbf{F} \, dS = \mathbf{i} \frac{\partial F_z}{\partial y} - \mathbf{k} \frac{\partial F_x}{\partial y} $$

S3 and S4. Similarly, we can show that:

$$ \lim_{\Delta V \to 0} \frac{1}{\Delta V} \iint_{S_3+S_4} \hat{\mathbf{n}} \times \mathbf{F} \, dS = \mathbf{j} \frac{\partial F_x}{\partial z} - \mathbf{i} \frac{\partial F_y}{\partial z} $$

S5 and S6. Similarly, we can show that:

$$ \lim_{\Delta V \to 0} \frac{1}{\Delta V} \iint_{S_5+S_6} \hat{\mathbf{n}} \times \mathbf{F} \, dS = \mathbf{k} \frac{\partial F_y}{\partial x} - \mathbf{j} \frac{\partial F_z}{\partial x} $$

Putting these together, we have:

$$ \begin{align*} \nabla \times \mathbf{F} &= \lim_{\Delta V \to 0} \frac{1}{\Delta V} \iint_S \hat{\mathbf{n}} \times \mathbf{F} \, dS \\[1em] &= \mathbf{i} \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) + \mathbf{j} \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right) + \mathbf{k} \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right) \end{align*} $$

Assume we partition the volume $V$ into $N$ small cubes of volume $\Delta V$.

Since normal vector of overlapping sides of two adjacent cubes are equal and opposite, then $\hat{\mathbf{n}} \times \mathbf{F}$ of two overlapping sides of two adjacent cubes will cancel out.

Then, we have:

$$ \iint_S \hat{\mathbf{n}} \times \mathbf{F} \, dS = \iint_{S_1} \hat{\mathbf{n}} \times \mathbf{F} \, dS + \ldots + \iint_{S_N} \hat{\mathbf{n}} \times \mathbf{F} \, dS $$

where $S_i$ is the surface of the $i$-th cube.

Also, since we have:

$$ \nabla \times \mathbf{F}(x_i, y_i, z_i) \approx \frac{1}{\Delta V} \iint_{S_i} \hat{\mathbf{n}} \times \mathbf{F} \, dS $$

(where $(x_i, y_i, z_i)$ is the center of the $i$-th cube), then we have:

$$ (\nabla \times \mathbf{F}(x_i, y_i, z_i)) \Delta V \approx \iint_{S_i} \hat{\mathbf{n}} \times \mathbf{F} \, dS $$

Summing over all $N$ cubes and taking the limit as $\Delta V \to 0$, we have:

$$ \iiint_V \nabla \times \mathbf{F} \, dV = \iint_S \hat{\mathbf{n}} \times \mathbf{F} \, dS $$

Expanding $\nabla \times \mathbf{F}$ gives:

$$ \begin{align*} \iiint_V \nabla &\times \mathbf{F} \, dV = \\[1em] &\mathbf{i} \iiint_V \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right) dV + \\[1em] &\mathbf{j} \iiint_V \left( \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) dV + \\[1em] &\mathbf{k} \iiint_V \left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) dV \end{align*} $$

Which can be rewritten as:

$$ \begin{align*} \iiint_V \nabla \times \mathbf{F} \, dV &= \\[1em] &\mathbf{i} \iiint_V \nabla \cdot \left( 0, F_z, -F_y \right) \, dV + \\[1em] &\mathbf{j} \iiint_V \nabla \cdot \left(-F_z, 0, F_x \right) \, dV + \\[1em] &\mathbf{k} \iiint_V \nabla \cdot \left( F_y, -F_x, 0 \right) \, dV \end{align*} $$

Using the divergence theorem, we have:

$$ \begin{align*} \iiint_V \nabla \times \mathbf{F} \, dV &= \\[1em] &\mathbf{i} \iint_S \left( 0, F_z, -F_y \right) \cdot \hat{\mathbf{n}} \, dS + \\[1em] &\mathbf{j} \iint_S \left(-F_z, 0, F_x \right) \cdot \hat{\mathbf{n}} \, dS + \\[1em] &\mathbf{k} \iint_S \left( F_y, -F_x, 0 \right) \cdot \hat{\mathbf{n}} \, dS \end{align*} $$

which can be rewritten as:

$$ \begin{align*} \iiint_V &\nabla \times \mathbf{F} \, dV = \\[1em] &\iint_S \left( \mathbf{i} (F_z \hat{\mathbf{n}}_y - F_y \hat{\mathbf{n}}_z) + \\[1em] \mathbf{j} (-F_z \hat{\mathbf{n}}_x + F_x \hat{\mathbf{n}}_z) + \\[1em] \mathbf{k} (F_y \hat{\mathbf{n}}_x - F_x \hat{\mathbf{n}}_y) \right) \, dS \end{align*} $$

So:

$$ \iiint_V \nabla \times \mathbf{F} \, dV = \iint_S \hat{\mathbf{n}} \times \mathbf{F} \, dS $$

Step 1. Form right triangles taking each oblique segment of $P$ as the hypotenuse.

If right triangles overlap each other or the sides of $P$, then subdivide them into smaller right triangles:

Step 2. Now, we have only right triangles and polygons with sides parallel to the axes. Add vertical segments at concave 270° corners of the polygon to form rectangles:

Segments in patches that are not in the perimeter of $P$ are traced twice, and in opposite directions, so they cancel out. Thus, we can write:

$$ \oint_P \mathbf{F} \cdot \hat{\mathbf{t}} \, ds = \sum_{l=1}^N \oint_{C_l} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds \tag{1} $$

We can approximate the circulation around $C_l$ by:

$$ \oint_{C_l} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds \approx C(x_l, y_l) \, \Delta S_l \tag{2} $$

where $(x_l, y_l)$ is the center of the $l$-th patch.

To approximate $C(x_l, y_l)$, we can use the two dimensional Taylor series.

Taylor expansion of $C(x_l, y_l)$ gives:

$$ C(x_l, y_l) \approx C(x_0, y_0) + \left( \frac{\partial C}{\partial x} \right)_{x_0, y_0} (x_l - x_0) + \left( \frac{\partial C}{\partial y} \right)_{x_0, y_0} (y_l - y_0) + \cdots \tag{3} $$

Putting (1), (2), and (3) together, we have:

$$ \begin{align*} \oint_P \mathbf{F} &\cdot \hat{\mathbf{t}} \, ds \approx \\[1em] &C(x_0, y_0) \Delta A + \left( \frac{\partial C}{\partial x} \right)_{x_0, y_0} \sum_{l=1}^N (x_l - x_0) \Delta S_l \\[1em] &\quad + \left( \frac{\partial C}{\partial y} \right)_{x_0, y_0} \sum_{l=1}^N (y_l - y_0) \Delta S_l + \cdots \end{align*} $$

We have:

$$ \begin{align*} \lim_{N \to \infty} \sum_{l=1}^N (x_l - x_0) \Delta S_l &= \lim_{N \to \infty} \sum_{l=1}^N (x_l \Delta S_l) - x_0 \left( \lim_{N \to \infty} \sum_{l=1}^N \Delta S_l \right) \\[1em] &= \iint_R x \, dx \, dy - x_0 \Delta S \\[1em] &= \Delta S (\overline{x} - x_0) \end{align*} $$

Similarly, we can get:

$$ \lim_{N \to \infty} \sum_{l=1}^N (y_l - y_0) \Delta S_l = \Delta S (\overline{y} - y_0) $$

Putting this together, we have:

$$ \begin{align*} \oint_C \mathbf{F} \cdot \hat{\mathbf{t}} \, ds &= \lim_{N \to \infty} \sum_{l=1}^N \oint_{C_l} \mathbf{F} \cdot \hat{\mathbf{t}} \, ds \\[1em] &= \Bigl( C(x_0, y_0) + (\overline{x} - x_0) \left( \frac{\partial C}{\partial x} \right)_{x_0, y_0} \\[1em] &\+ (\overline{y} - y_0) \left( \frac{\partial C}{\partial y} \right)_{x_0, y_0} + \cdots \Bigr) \Delta S \end{align*} $$

As $\Delta S \to 0$, then $(\overline{x}, \overline{y}) \to (x, y)$, and we have:

$$ \begin{align*} \lim_{\Delta S \to 0} \frac{1}{\Delta S} \oint_P \mathbf{F} \cdot \hat{\mathbf{t}} \, ds &= \lim_{\Delta S \to 0} \Bigl( C(x_0, y_0) + (\overline{x} - x_0) \left( \frac{\partial C}{\partial x} \right)_{x_0, y_0} \\[1em] &\+ (\overline{y} - y_0) \left( \frac{\partial C}{\partial y} \right)_{x_0, y_0} + \cdots \Bigr) \\[1em] &= \lim_{\Delta S \to 0} C(x_l, y_l) \\[1em] &= \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \end{align*} $$