Lecture 4 Stokes' Theorem

📌 Overview

This lecture introduces Stokes' Theorem, which relates a line integral around a closed boundary curve to a surface integral over any surface bounded by that curve. It can be viewed as a higher-dimensional generalization of Green's Theorem. The lecture covers the concept of induced orientation, the formal statement of the theorem, its application to conservative vector fields, and its role in Maxwell's equations for static fields.

🎯 Learning Objectives

• Determine the induced orientation of the boundary curve of an oriented surface using the right-hand rule.
• State and apply Stokes' Theorem to evaluate line integrals and surface integrals.
• Understand the relationship between Stokes' Theorem and Green's Theorem.
• Simplify surface integrals by applying Stokes' theorem twice (changing the surface while keeping the boundary).
• Relate the curl of a vector field to its path independence and conservativeness on simply connected domains.
• Apply Stokes' Theorem to physical contexts like Maxwell's equations for static electric and magnetic fields.

➗ Formulas & Definitions

💡 Intuitive stepwise derivation

From Green's to Stokes'

Stokes' Theorem is essentially "Green's Theorem in 3D".

1. Green's Theorem works in the $xy$-plane. It says the circulation around a boundary is the sum of "microscopic" circulations (curls) over the flat region $R$.
2. Stokes' Theorem lifts this to a curved surface $S$ in ${\mathbb{R}}^3$.
3. The boundary $C=\partial S$ is a closed loop in space.
4. The "microscopic" circulation at any point on the surface is the component of the curl $\mathrm{\nabla }\times \mathbf{F}$ in the direction of the surface normal $\mathbf{n}$.

The Right-Hand Rule for Orientation

To ensure the signs match, we use the Right-Hand Rule:

1. Point your thumb in the direction of the surface normal $\mathbf{n}$.
2. Your fingers curl in the direction of the positive (induced) orientation of the boundary curve $C$.
3. Alternatively: If you walk along the boundary with your head in the direction of $\mathbf{n}$, the surface $S$ must be on your left.

\usepackage{tikz-3dplot}
\begin{document}
\tdplotsetmaincoords{70}{110}
\begin{tikzpicture}[tdplot_main_coords, scale=2]
    % Draw surface (a cap)
    \draw[thick, fill=orange!50, opacity=0.6] plot[domain=0:360, samples=60] ({cos(\x)}, {sin(\x)}, {0.5*cos(2*\x) + 1});
    
    % Draw boundary curve
    \draw[blue, very thick, ->] plot[domain=0:180, samples=30] ({cos(\x)}, {sin(\x)}, {0.5*cos(2*\x) + 1});
    \draw[blue, very thick] plot[domain=180:360, samples=30] ({cos(\x)}, {sin(\x)}, {0.5*cos(2*\x) + 1}) node[below right] {$C = \partial S$};
    
    % Draw normal vector
    \draw[red, -stealth, very thick] (0,0,1) -- (0,0,1.8) node[above] {$\mathbf{n}$};
    
    % Label surface
    \node at (0.3, 0.3, 1.2) {$S$};
    
    % Right hand rule visualization (simplified)
    \draw[dashed, gray] (0,0,1) circle (0.3);
    \draw[->, gray] (0.3, 0, 1) arc (0:90:0.3);
\end{tikzpicture}
\end{document}

Notes / Example exercises

1. Oriented Boundary Curves

A surface $S$ is oriented by a unit normal vector field $\mathbf{n}$. This orientation induces an orientation on the boundary curve $C$.

Example: Bounded Cylinder (Slide 5-6)

Consider the cylinder $S=\{(x,y,z)\in {\mathbb{R}}^3\mid x^2+y^2=4,-1\le z\le 1\}$.
The boundary consists of two circles:

• $C_1$ at $z=1$
• $C_2$ at $z=-1$

If $S$ is oriented with the normal pointing towards the $z$-axis (inward):

• $C_1$ (top): Induced orientation is counterclockwise when viewed from above.
• $C_2$ (bottom): Induced orientation is clockwise when viewed from above.

2. Stokes' Theorem Statement

Let $S$ be a piecewise-smooth, oriented surface in ${\mathbb{R}}^3$ bounded by a simple closed, piecewise-smooth, positively oriented boundary curve $C$. If $\mathbf{F}$ is a vector field with continuous partial derivatives:

Comparison with Green's Theorem (Slide 7)

• Green's: ${\oint }_{\partial R}\mathbf{F}\cdot d\mathbf{r}={\iint }_R(\text{curl }\mathbf{F})\cdot \mathbf{k}dA$.
• Stokes': Generalizes this to any surface $S$ and any orientation $\mathbf{n}$.

3. Example: Evaluating a Line Integral (Slide 9)

Problem: Evaluate ${\oint }_C\mathbf{F}\cdot d\mathbf{r}$ where $\mathbf{F}=⟨-y^2,x,z^2⟩$ and $C$ is the intersection of $y+z=2$ and $x^2+y^2=1$, traversed CCW from above.

Step-by-Step Plan:

1. Find Curl: $\mathrm{\nabla }\times \mathbf{F}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ {\partial }_x& {\partial }_y& {\partial }_z\\ -y^2& x& z^2\end{array}\right|=⟨0-0,0-0,1-(-2y)⟩=⟨0,0,1+2y⟩$.
2. Choose Surface: The simplest surface is the disk $D$ in the plane $y+z=2$ bounded by the cylinder $x^2+y^2=1$.
3. Parametrize Surface: $z=2-y$. Normal vector $\mathbf{n}$ for $z=g(x,y)$ is $⟨-g_x,-g_y,1⟩=⟨0,1,1⟩$.
4. Setup Integral:$${\iint }_D(\mathrm{\nabla }\times \mathbf{F})\cdot d\mathbf{S}={\iint }_{x^2+y^2\le 1}⟨0,0,1+2y⟩\cdot ⟨0,1,1⟩dA={\iint }_{x^2+y^2\le 1}(1+2y)dA$$
5. Solve: By symmetry, $\iint 2ydA=0$. The remaining part is $\iint 1dA=\text{Area(Disk)}=\pi (1{)}^2=\pi$.
   Result: $\pi$.

4. One Boundary, Many Surfaces (Slide 12-15)

Corollary: If $S_1$ and $S_2$ share the same oriented boundary $C$, then:

This is extremely useful! If $S_1$ is a complex hemisphere or paraboloid, you can often replace it with a flat disk $S_2$ in the $xy$-plane.

Practice Exercise 1 (Slide 12)

Problem: Compute ${\iint }_S(\mathrm{\nabla }\times \mathbf{F})\cdot d\mathbf{S}$ where $\mathbf{F}=⟨xz,yz,xy⟩$ and $S$ is the part of the sphere $x^2+y^2+z^2=4$ inside the cylinder $x^2+y^2=1$ above the $xy$-plane.
Solution:

1. The boundary $C$ is the circle $x^2+y^2=1$ at height $z=\sqrt{4-1}=\sqrt{3}$.
2. Instead of the spherical cap, use the flat disk $D$ at $z=\sqrt{3}$.
3. On $D$, the normal is $\mathbf{k}=⟨0,0,1⟩$.
4. $\mathrm{\nabla }\times \mathbf{F}=⟨x-y,x-y,0⟩$.
5. $(\mathrm{\nabla }\times \mathbf{F})\cdot \mathbf{k}=0$.
   Result: 0.

5. Closed Surfaces (Slide 16-17)

If $S$ is a closed surface (like a full sphere), it has no boundary ($C=\mathrm{\varnothing }$).
By Stokes' Theorem:

Intuition: You can split the closed surface into two parts $S_1$ and $S_2$ sharing the same boundary $C$, but with opposite induced orientations relative to the outward normal. They cancel out.

6. Conservative Vector Fields (Slide 18)

On an open, simply connected domain $D$, the following are equivalent:
(a) $\mathbf{F}$ is conservative ($\mathbf{F}=\mathrm{\nabla }\varphi$).
(b) $\int \mathbf{F}\cdot d\mathbf{r}$ is path independent.
(c) ${\oint }_C\mathbf{F}\cdot d\mathbf{r}=0$ for any closed curve $C$.
(d) $\mathrm{\nabla }\times \mathbf{F}=\mathbf{0}$ on $D$.

Stokes' Theorem proves (d) $⟹$ (c).

7. Application: Maxwell's Equations (Slide 19-21)

For static fields:

• Electric Field $\mathbf{E}$: $\mathrm{\nabla }\times \mathbf{E}=0⟹{\oint }_C\mathbf{E}\cdot d\mathbf{r}=0$. The static electric field is conservative.
• Magnetic Field $\mathbf{B}$: $\mathrm{\nabla }\times \mathbf{B}={\mu }_0\mathbf{J}$. By Stokes: ${\oint }_C\mathbf{B}\cdot d\mathbf{r}={\mu }_0{\iint }_S\mathbf{J}\cdot d\mathbf{S}={\mu }_0{I}_{\text{enclosed}}$. This is Ampere's Law.

💡 Intuitive stepwise derivation: Amp�re's Law

1. Start with the differential form: $\mathrm{\nabla }\times \mathbf{B}={\mu }_0\mathbf{J}$.
2. Integrate both sides over a surface $S$: ${\iint }_S(\mathrm{\nabla }\times \mathbf{B})\cdot d\mathbf{S}={\iint }_S{\mu }_0\mathbf{J}\cdot d\mathbf{S}$.
3. Apply Stokes' Theorem to the left side: ${\oint }_{\partial S}\mathbf{B}\cdot d\mathbf{r}={\mu }_0{\iint }_S\mathbf{J}\cdot d\mathbf{S}$.
4. Recognize that ${\iint }_S\mathbf{J}\cdot d\mathbf{S}$ is the total current $I$ passing through the surface.
5. Conclusion: The circulation of $\mathbf{B}$ around a loop is proportional to the current passing through the loop.

✍️ Notes / Example exercises (Continued)

Practice Exercise 1 (Slide 28)

Problem: $\mathbf{F}=⟨\ln (z)\cos (xyz),5x{e}^{z^2}+\cos (e^{y^2}),{\arctan }^2(z)+\sin (y^2)⟩$.
$S:z=2-x^2-y^2$ with $z\ge 1$, normal has positive $z$-component.
Solution:

1. The boundary $C$ is at $z=1⟹1=2-x^2-y^2⟹x^2+y^2=1$.
2. Use the flat disk $D$ at $z=1$ instead of the paraboloid.
3. On $D$, $d\mathbf{S}=\mathbf{k}dA$.
4. We only need the $z$-component of $\mathrm{\nabla }\times \mathbf{F}$:
   $(\mathrm{\nabla }\times \mathbf{F}{)}_z=\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}=5e^{z^2}-(-\ln (z)\sin (xyz)\cdot xz)$.
5. At $z=1$, $\ln (1)=0$, so $(\mathrm{\nabla }\times \mathbf{F}{)}_z=5e^1=5e$.
6. ${\iint }_{D}5edA=5e\cdot \text{Area}(D)=5e\cdot \pi (1{)}^2=5\pi e$.
   Result: $5\pi e$.

Practice Exercise 2 (Slide 28)

Problem: $\mathbf{J}=⟨xy,3,xy{z}^2⟩$, $S:x^2+z^2\le 1,y=2$ oriented towards $y=0$ (negative $y$ direction). Evaluate ${\oint }_C\mathbf{B}\cdot d\mathbf{r}$.
Solution:

1. By Ampere's Law (Stokes + Maxwell): ${\oint }_C\mathbf{B}\cdot d\mathbf{r}={\iint }_S(\mathrm{\nabla }\times \mathbf{B})\cdot d\mathbf{S}={\iint }_S{\mu }_0\mathbf{J}\cdot d\mathbf{S}$.
2. $S$ is in the plane $y=2$, so the normal is $-\mathbf{j}=⟨0,-1,0⟩$.
3. $\mathbf{J}\cdot \mathbf{n}=⟨xy,3,xy{z}^2⟩\cdot ⟨0,-1,0⟩=-3$.
4. ${\iint }_S-3{\mu }_{0}dA=-3{\mu }_0\cdot \text{Area}(S)=-3{\mu }_0\cdot \pi (1{)}^2=-3\pi {\mu }_0$.
   Result: $-3\pi {\mu }_0$.