Stokes'theorem

Stokes' theorem

A generalization of Green's theorem by Irish mathematician and physicist George Gabriel Stokes (1819-1903) is called Stokes' theorem．First, To study Stokes' theorem, we need to orient the surface.．

Definition 4..1  

At each point $(x,y,z)$ on the curved surface $S$, select the normal vector $\boldsymbol{n}(x,y,z)$ appropriately, and $\boldsymbol{n}(x,y,z)$ can be made continuous on $S$. Then the surface $S$ is said to be orientable Also, the $\boldsymbol{n}$ selected at each point at this time is called the unit normal vector determined by the orientation of the curved surface.

Stokes' theorem gives an equation that rewrites the line integral along the bounding curve $\partial S$ of the oriented surface $S$ to the surface integral on $S$．

Theorem 4..4  

[Stokes' theorem] Let $S:z = f(x,y), (x,y) \in \Omega$ be an oriented curved surface bounded by some piecewise smooth closed curves．Also, the vector field $\boldsymbol{F}$ is of the $C ^{1}$ class on $S$. then,

$$\oint_{\partial S}\boldsymbol{F}\cdot d\boldsymbol{r} = \iint_{S}(\nabla \times \boldsymbol{F})\cdot \boldsymbol{n} dS$$

However, $\partial S$ represents the boundary of $S$, and the direction of the line integral on the curve $\partial S$ shall go around $\partial S$ so that the region $S$ is on the left. In other words, it follows the right-hand rule for the unit normal vector $\boldsymbol{n}$.

Figure 4.2: Stokes' theorem

Proof Firsr consider $\iint_{S} [\nabla \times F_{1}\boldsymbol{i}] \cdot\boldsymbol{n}dS$．

$$\nabla \times F_{1}\boldsymbol{i} = \left\vert\begin{array}{ccc} ... ...}}{\partial z}\boldsymbol{j} - \frac{\partial F_{1}}{\partial y}\boldsymbol{k}$$

implies

$$[\nabla \times F_{1}\boldsymbol{i}] \cdot\boldsymbol{n}dS = \left... ...- \frac{\partial F_{1}}{\partial y}\boldsymbol{n} \cdot\boldsymbol{k}\right )dS$$ (4.1)

Let $\boldsymbol{r} = {}^t[x,y,z]$ be a position vector. Then $\boldsymbol{r}_{y} = (0,1,z_{y})$ is a tangent vector of $S$. Thus it is orthogonal to the normal vecto $\boldsymbol{n}$.

$$\boldsymbol{n} \cdot\boldsymbol{r}_{y} = \boldsymbol{n} \cdot\boldsymbol{j} + z_{y}\boldsymbol{n} \cdot\boldsymbol{k} = 0 \$$   or$$\ \boldsymbol{n} \cdot\boldsymbol{j} = - z_{y}\boldsymbol{n} \cdot\boldsymbol{k}$$

Putting this into the equation 4.1，

$$[\nabla \times F_{1}\boldsymbol{i}] \cdot\boldsymbol{n}dS = - \le... ...ac{\partial F_{1}}{\partial y} \right ) \boldsymbol{n} \cdot \boldsymbol{k} dS$$

Here we can set $F_{1}(x,y,z) = F_{1}(x,y,f(x,y)) = H(x,y)$ on $S$. Thus by the differentiation of composite functions, we have

$$\frac{\partial H}{\partial y} = \frac{\partial F_{1}}{\partial y} + \frac{\partial F_{1}}{\partial z}\frac{\partial z}{\partial y}$$

Thus,

$$[\nabla \times F_{1}\boldsymbol{i}] \cdot\boldsymbol{n}dS = - \fr... ...}\boldsymbol{n} \cdot \boldsymbol{k} dS = - \frac{\partial H}{\partial y}dx dy$$

Therefore,

$$\iint_{S}[\nabla \times F_{1}\boldsymbol{i}] \cdot\boldsymbol{n}dS = \iint_{\Omega}- \frac{\partial H}{\partial y}dx dy$$

Here, $\Omega$ is a normal projection of $S$ on the $xy$ plane. The integral on the right is the integral on the plane, which is $\oint_{\partial \Omega} {\bf H}dx$ according to Green's theorem. The value of $H(x,y)$ at the point $(x, y)$ on the boundary of $\Omega$ and at the point $(x,y,z)$ on the boundary of $S$ The values of $F_{1}(x,y,z)$ are equal, and $dx$ is the same for both curves.

$$\oint_{\partial \Omega} {\bf H}dx = \oint_{\partial S}{F_{1}}dx$$

or

$$\iint_{S}[\nabla \times F_{1}\boldsymbol{i}] \cdot\boldsymbol{n}dS = \oint_{\partial S}{F_{1}}dx$$

Similarly, if you take a normal projection onto another plane,

$$\iint_{S}[\nabla \times F_{2}\boldsymbol{j}] \cdot\boldsymbol{n}d... ...times F_{3}\boldsymbol{k}] \cdot\boldsymbol{n}dS = \oint_{\partial S}{F_{3}}dz$$

Adding these

$$\iint_{S}(\nabla \times \boldsymbol{F}) \cdot\boldsymbol{n}dS = \oint_{\partial S}\boldsymbol{F}\cdot d\boldsymbol{r}$$

Example 4..4  

For $${\boldsymbol{F} = -y\boldsymbol{i} + x\boldsymbol{j} + \boldsymbol{k}, \ S: z = (4 - x^2 - y^2)^{1/2}}$$，show that Stokes' theorem holds.

Answer The boundary of $S$, $\partial S$ is a circle $x^2 + y^2 = 4$．Thus the position vector is $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k} = 2\cos{t}\boldsymbol{i} + 2\sin{t}\boldsymbol{j}$. Then we find the line integral

$$\oint_{\partial S}\boldsymbol{F}\cdot d\boldsymbol{r} = \oint_{\partial S}(-2\sin{t}\boldsymbol{i} + 2\cos{t}\boldsymbol{j} + \boldsymbol{k})\cdot(-2\sin{t}\boldsymbol{i} + 2\cos{t}\boldsymbol{j})dt$$

$$= \int_{0}^{2\pi}[4\sin^{2}{t} + 4\cos^{2}{t}]dt = 8\pi$$

Next we find the surface integral.．

$$\nabla \times \boldsymbol{F} = \left\vert\begin{array}{ccc} \bol... ...c{\partial}{\partial z}\\ -y & x & 1 \end{array}\right\vert = 2\boldsymbol{k}$$

We find the unit normal vector. Then

$$\boldsymbol{r}_{x} \times \boldsymbol{r}_{y} = \left\vert\begin{a... ...t\vert = \frac{x}{z}\boldsymbol{i} + \frac{y}{z}\boldsymbol{j} + \boldsymbol{k}$$

Thus,

$$\boldsymbol{n} = \frac{ \frac{x}{z}\boldsymbol{i} + \frac{y}{z}\b... ... \frac{x}{z}\boldsymbol{i} + \frac{y}{z}\boldsymbol{j} + \boldsymbol{k}\Vert }$$

Therefore,，

$$\iint_{S}(\nabla \times \boldsymbol{F}) \cdot\boldsymbol{n} dS = 2\iint_{S}\boldsymbol{k} \cdot\frac{\boldsymbol{r}_{x} \times \bo... ...bol{r}_{y} \Vert} \Vert\boldsymbol{r}_{x} \times \boldsymbol{r}_{y} \Vert dx dy$$

$$= 2\iint_{\Omega}\boldsymbol{k} \cdot\boldsymbol{r}_{x} \times \boldsymbol{r}_{y} dx dy$$

$$= \iint_{\Omega}2 dx dy = 2(4\pi) = 8 \pi$$

This shows that Stokes' theorem hols.

So far we have already learned that in a conservative field, the vector field is equal to the gradient of the scalar field, and the rotation of the vector field is zero. Now let's investigate what holds true in relation to line integrals.

Theorem 4..5  

In the vector field $\boldsymbol{F}(x,y,z)$, the following 3 conditions are equivalent.

(1) There exists a scalar function $\phi(x,y,z)$ so that $\boldsymbol{F} = \nabla \phi$．( $\boldsymbol{F}$ is conservative)

(2) $\nabla \times \boldsymbol{F} = {\bf0}$ holds everywhwer．(no votex)

(3) For any close curve $C$, $\oint_{C} \boldsymbol{F}\cdot d \boldsymbol{r} = 0$ holds (inepenent of path)．

Proof Show (1) $\Rightarrow$ (2) $\Rightarrow$ (3) $\Rightarrow$ (1)．

(1) $\Rightarrow$ (2) $\nabla \times \boldsymbol{F} = \nabla \times \nabla \phi = \left \vert \begin{a... ...artial z}\\ \phi_{x} & \phi_{y} & \phi_{z} \end{array} \right \vert = {\bf0}$

(2) $\Rightarrow$ (3) Consider the surface $S$ bounded by $C$ and apply Stokes' theorem.

$$\oint_{C} \boldsymbol{F} \cdot d \boldsymbol{r} = \iint_{S} (\nab... ...symbol{F}) \cdot\boldsymbol{n} dS = \iint_{S} {\bf0} \cdot\boldsymbol{n}dS = 0$$

(3) $\Rightarrow$ (1) Two curves $C_{1},C_{2}$ connecting the fixed point P $(x_{0},y_{0},z_{0})$ and the moving point Q $(x,y,z)$. Starting from P passing $C_{1}$ to Q and from Q passing $C_{2}$ to P. Set this whole path $C$.

$$\int_{{\rm P}(C_{1}){\rm Q}}\boldsymbol{F} \cdot d\boldsymbol{r} ... ...ol{F} \cdot d\boldsymbol{r} = \oint_{C}\boldsymbol{F} \cdot d\boldsymbol{r} = 0$$

Thus，

$$\int_{{\rm P}(C_{1}){\rm Q}}\boldsymbol{F} \cdot d\boldsymbol{r} = \int_{{\rm P}(C_{2}){\rm Q}}\boldsymbol{F} \cdot d\boldsymbol{r}$$

That is, the line integral from P to Q, $\int_{{\rm PQ}}\boldsymbol{F} \cdot d\boldsymbol{r}$ is given by the function of the coordinate $(x,y,z)$ of the end point Q regardless of the route in the middle. So if you set this to $\phi(x,y,z)$, then we have

$$\int_{\rm PQ}\boldsymbol{F} \cdot d\boldsymbol{r} = \phi(x,y,z)$$

We let vector equation for any curve from P to Q as $\boldsymbol{r} = (x(s),y(s),z(s))$. Then

$$\int_{{\rm PQ}}\boldsymbol{F} \cdot d\boldsymbol{r} = \int_{s_{0}... ... \left (F_{1}\frac{dx}{ds} + F_{2}\frac{dy}{ds} + F_{3}\frac{dz}{ds}\right )ds$$

Thus

$$F_{1}\frac{dx}{ds} + F_{2}\frac{dy}{ds} + F_{3}\frac{dz}{ds} = \f... ...hi}{\partial y} \frac{dy}{ds} + \frac{\partial \phi}{\partial z} \frac{dz}{ds}$$

The curve PQ is arbitrary. Thus, $x(s),y(s),z(s)$ is also an arbitrary function.，

$$F_{1} = \frac{\partial \phi}{\partial x}, F_{2} = \frac{\partial \phi}{\partial y}, F_{3} = \frac{\partial \phi}{\partial z}$$

Therefore,

$$\boldsymbol{F} = F_{1}\boldsymbol{i} + F_{2}\boldsymbol{j} + F_{3... ...}\boldsymbol{j} + \frac{\partial \phi}{\partial z}\boldsymbol{k} = \nabla \phi$$

From this theorem, we can see that the answer is 0 even without integration if the vector field has a scalar potential when performing line integrals

Example 4..5  

Find $${\int_{C}((2x+yz)\boldsymbol{i} + zx\boldsymbol{j} + xy\boldsymbol{k}) \cdot d\boldsymbol{r}}$$．where $C$ is a curve connecting from the point $(1,0,-1)$ to $(2,-1,3)$ and back to the starting point.．

Answer

$$\nabla \times ((2x+yz)\boldsymbol{i} + zx\boldsymbol{j} + xy\bold... ...\frac{\partial}{\partial z}\\ 2x+yz & zx & xy \end{array}\right\vert = {\bf0}$$

Then，${\bf F}$ has a scalar potential．So, we find $f$ so that ${\bf F} = \nabla f$． $f_{x} = 2x+yz, f_{y} = zx, f_{z} = xy$ implies

$$f = ^int f_{x}dx = x^2 + xyz + g(y,z)$$

Here， $f_{y} = zx$ imlies $f_{y} = xz + g_{y}(y,z)$. Thus，$g_{y} = 0$. in other words， $g(y,z) = h(z)$. Next， $f_{z} = xy$ implies $f_{z} = xy + h'(z)$. Thus $h(z) = 0$．Here，let $h(z) = 0$. Then

$$f = x^2 + xyz$$

Using this, we have

$$\int_{C}((2x+yz)\boldsymbol{i} + zx\boldsymbol{j} + xy\boldsymbol{k}) \cdot d\boldsymbol{r} = f(2,-1,3) - f(1,0,-1) = -2 - 1 = -3$$

Example 4..6  

For any surface $S$ and its boundary $C$ within the common domain of $\phi, \psi$, prove the following equation holds.

$$\iint_{S}\{(\nabla \phi) \times (\nabla \psi)\}\cdot\boldsymbol{n}dS = \int_{C}\phi(\nabla \psi)\cdot d\boldsymbol{r}$$

Answer We first calculate $\nabla \times (\phi \nabla \psi)$.

$$\nabla \times (\phi \nabla \psi) = (\nabla \phi) \times (\nabla \psi) + \phi \nabla \times (\nabla \psi)$$

Here， $\nabla \times (\nabla \psi) = {\bf0}$ implies

$$\nabla \times (\phi \nabla \psi) = (\nabla \phi) \times (\nabla \psi)$$

Therefore, by Stokes' theorem，

$$\iint_{S}\{(\nabla \phi) \times (\nabla \psi)\}\cdot\boldsymbol{n}dS = \int_{C}\phi(\nabla \psi)\cdot d\boldsymbol{r}$$

Question 4..1  

Let $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z \boldsymbol{k}$．Prove for any surface $S$ and its boundary $C$, the following is true.

$$(1)\ \int_{C}d\boldsymbol{r} = {\bf0}\hskip 1cm (2)\ \int_{C}\boldsymbol{r} \cdot d\boldsymbol{r} = 0$$

Exercise4.2

1.

Prove the following equation holds for any surface $S$ and its boundary $C$ in the common domain of the scalar fields $\phi, \psi$.．

$$\int_{C}\phi(\nabla \psi)\cdot d\boldsymbol{r} = -\int_{C}\psi(\nabla \phi)\cdot d\boldsymbol{r}$$

2.

Let $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k},\ r = \vert\boldsymbol{r}\vert$ and $\phi$ be a scalar field. Then prove the following equation holds for any surface $S$ and its boundary $C$．

(1) $\int_{C}\boldsymbol{r} \times d\boldsymbol{r} = 2\int_{S}\boldsymbol{n}dS$

(2) $\int_{C}r^{k}\boldsymbol{r} \cdot d\boldsymbol{r} = 0$

(3) $\int_{C}\boldsymbol{r} (\nabla \phi) \cdot d\boldsymbol{r} = \int_{S}(\nabla \phi) \times \boldsymbol{n}dS$

3.

Prove the following holds for any surface $S$ and its boundary $C$ in the common domain of the scalar field $\phi$ and the vector field ${\bf A}$.

$$\int_{S}\phi(\nabla \times {\bf A})\cdot\boldsymbol{n} dS = \int_... ...d\boldsymbol{r} - \int_{S}\{(\nabla \phi)\times {\bf A}\} \cdot\boldsymbol{n}dS$$

4.

Let $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k},\ r = \vert\boldsymbol{r}\vert$ and $\phi$ be a scalar field. Prove that for any surface $S$ and its boundary $C$, the following equation holds.

$$\int_{C}\frac{\boldsymbol{r} \times d\boldsymbol{r}}{r^3} = -\int... ...}\cdot\boldsymbol{n}}{r^5}\boldsymbol{r} - \frac{\boldsymbol{n}}{r^3} \right)dS$$

5.

It is assumed that the vector field ${\bf A}$ is defined in the whole space.．About the border $C$ of any curved surface if

$$\int_{C}{\bf A} \cdot d\boldsymbol{r} = 0$$

${\bf A}$ has a scalar potential. Prove this.