17.3 Stokes’ theorem

中文

Stokes’ theorem requires the concept of a submanifold with boundary. This is not an easy notion in general, but for most practical purposes it is sufficient to restrict ourselves to regions made up of ‘coordinate cubical regions’. Let ${\Gamma }_k$ be the standard unit $k$-cube,

$${\Gamma }_k=\{\mathbf{x}\in {\mathbb{R}}^k\mid 0\le x^i\le 1(i=1,\dots ,k)\}\subset {\mathbb{R}}^k.$$

The unit 0-cube is taken to be the singleton ${\Gamma }_0=\{0\}\subset \mathbb{R}$ . A $k$-cell in a manifold $M$ is a smooth map $\sigma :U\to M$ where $U$ is an open neighbourhood $U$ of ${\Gamma }_k$ in ${\mathbb{R}}^k$ (see Fig. 17.2), and its support is defined as the image of the standard $k$-cube, $\sigma ({\Gamma }_k)$ . A cubical $k$-chain in $M$ consists of a formal sum

$$C=c^1{\sigma }_1+c^2{\sigma }_2+\cdots +c^r{\sigma }_r\text{ where }{c}^i\in \mathbb{R},r\text{ a positive integer. }$$

The set of all cubical $k$-chains is denoted ${\mathcal{C}}_k.$ ; it forms an abelian group under addition of $k$-chains defined in the obvious way. It is also a vector space if we define scalar multiplication as $\begin{array}{r}aC={\sum }_{i}a{c}^i{\sigma }_i\end{array}$ where $a\in \mathbb{R}$

For each $i=1,\dots ,k$ and $ϵ=0$ , 1 define the maps ${\phi }_i^{ϵ}:{\Gamma }_{k-1}\to {\Gamma }_k$ by

$${\phi }_i^{ϵ}(y^1,\dots ,y^{k-1})=(y^1,\dots ,y^{i-1},ϵ,y^i,\dots ,y^{k-1}).$$

These maps can be thought of as the ($i$, 0)-face and ($i$, 1)-face respectively of ${\Gamma }_k$ . If the interior of the standard $k$-cube ${\Gamma }_k$ is oriented in the natural way, by assigning the $k$-form $\mathrm{d}{x}^1\wedge \cdots \wedge \mathrm{d}{x}^k$ to be positively oriented over ${\Gamma }_k$ , then for each face map $x^i=0$ or 1 the orientation on ${\Gamma }_{k-1}$ is assigned according to the following rule: set the $k$-form $\mathrm{d}{x}^i\wedge \mathrm{d}{x}^1\wedge \cdots \wedge \mathrm{d}{x}^{i-1}\wedge \mathrm{d}{x}^{i+1}\wedge \cdots \wedge \mathrm{d}{x}^k$ to be positively oriented $\mathrm{if}{x}^i$ is increasing outwards at the face, else it is negatively oriented. According to this rule the $(k-1)$-form $\mathrm{d}{\mathbf{y}}^1\wedge \mathrm{d}{y}^2\wedge \cdots \wedge \mathrm{d}{y}^{k-1}$ has orientation $(-1{)}^i$ on the ($i$, 0)-face, while on the ($i$, 1)-face it has orientation $(-1{)}^{i+1}$ . This is sometimes called the outward normal rule – the orientation on the boundary surface must be chosen such that at every point there exist local positively oriented coordinates such that the first coordinate $x^1$ points outwards from the surface.

Exercise

On the two-dimensional square, verify that the outward normal rule implies that the di rection of increasing $x$ or $y$ coordinate on each side proceeds in an anticlockwise fashion around the square.

Figure 17.2 $k$-Cell on a manifold $M$

This gives the rationale for the boundary map $\partial :{\mathcal{C}}_k\to (C{)}_{k-1}$ , defined by:

(i) for a $k$-cell $\sigma (k>0)$ , set

$$\partial \sigma =\sum _{i=1}^k(-1{)}^i\left(\sigma \circ {\phi }_i^0-\sigma \circ {\phi }_i^1\right),$$

(ii) for a cubical $k$-chain $C=\sum c^i{\sigma }_i$ set

$$\partial C=\sum _i{c}^i\partial {\sigma }_i.$$

An important identity is ${\partial }^2=0$ . For example if $\sigma$ is a 2-cell, its boundary is given by

$$\partial \sigma =-{\phi }_1^0+{\phi }_1^1+{\phi }_2^0-{\phi }_2^1.$$

The boundary of the face ${\phi }_1^0(z)=(0,z)\mathrm{is}\partial {\phi }_1^0=-{\rho }^{01}+{\rho }^{00}$ , where ${\rho }^{ab}:\{0\}\mathbb{R}$ is the map ${\rho }^{ab}(0)=(a,b)$ . Hence

$$\partial \circ \partial \sigma =-{\rho }^{01}+{\rho }^{00}+{\rho }^{11}-{\rho }^{10}+{\rho }^{10}-{\rho }^{00}-{\rho }^{11}+{\rho }^{01}=0.$$

For a $k$-cell,

$${\partial }^2\sigma =\sum _{ϵ=0}^1\sum _{{ϵ}^{\prime }=0}^1\left[\sum _i\sum _{j<i}(-1{)}^{i+ϵ+j+{ϵ}^{\prime }}\sigma \circ {\phi }_{ij}^{ϵ{ϵ}^{\prime }}+\sum _i\sum _{j\ge i}(-1{)}^{i+ϵ+j+{ϵ}^{\prime }+1}\sigma \circ {\phi }_{ij}^{ϵ{ϵ}^{\prime }}\right]$$

where

$${\phi }_{ij}^{ϵ{ϵ}^{\prime }}(z^1,\dots ,z^{k-2})=\{\begin{array}{ll}(z^1,\dots ,z^{j-1},{ϵ}^{\prime },z^j,\dots ,z^{i-1},ϵ,z^i,\dots ,z^{k-2})& \text{if}j<i,\\ (z^1,\dots ,z^{i-1},ϵ,z^i,\dots ,z^{j-1},{ϵ}^{\prime },z^j,\dots ,z^{k-2})& \text{if}j\ge i.\end{array}$$

It follows from this equation that all terms cancel in pairs, so that ${\partial }^2\sigma =0$ . The identity extends by linearity to all $k$-chains.

Exercise

Write out ${\partial }^2\sigma$ for a 3-cube, and verify the cancellation property.

For a $k$-form $\alpha$ on $M$ and a $k$-chain $C=\sum c^i{\sigma }_i$ we define the integral

$${\int }_C\alpha =\sum c^i{\int }_{{\sigma }_i}\alpha =\sum c^i{\int }_{{\Gamma }_k}{\sigma }_i^{\ast }\alpha .$$

Theorem 17.2 · Stokes’ theorem

(Stokes’ theorem) For any $(k+1)$ -chain $C$, and differential $k$-form on $M$,

$${\int }_C\mathrm{d}\alpha ={\int }_{\partial C}\alpha .\text{\hspace{1em}(17.2)}$$

Proof

By linearity, it is only necessary to prove the theorem for a $(k+1)$-cell $\sigma$ . The left-hand side of Eq. (17.2) can be written, using Theorem 16.2,

$${\int }_{\sigma }\mathrm{d}\alpha ={\int }_{{\Gamma }_{k+1}}{\sigma }^{\ast }\mathrm{d}\alpha ={\int }_{{\Gamma }_{k+1}}\mathrm{d}({\sigma }^{\ast }\alpha )$$

while the right-hand side is

$$\begin{array}{c}{\int }_{\partial \sigma }\alpha ={\sum }_{ϵ=0}^1{\sum }_{i=1}^{k+1}(-1{)}^{i+ϵ}{\int }_{\sigma \circ {\phi }_i^{ϵ}}\alpha \\ ={\sum }_{ϵ=0}^1{\sum }_{i=1}^{k+1}(-1{)}^{i+ϵ}{\int }_{{\Gamma }_k}{\left({\phi }_i^{ϵ}\right)}^{\ast }\circ {\sigma }^{\ast }\alpha .\end{array}$$

Since ${\sigma }^{\ast }\alpha$ is a differential $k$-form on ${\mathbb{R}}^{k+1}$ it can be written as

$${\sigma }^{\ast }\alpha =\sum _{i=1}^{k+1}{A}_i\mathrm{d}{x}^1\wedge \cdots \wedge \mathrm{d}{x}^{i-1}\wedge \mathrm{d}{x}^{i+1}\wedge \cdots \wedge \mathrm{d}{x}^{k+1}$$

where the $A_i$ are differentiable functions $A_i:V\to \mathbb{R}$ on an open neighbourhood $V$ of ${\Gamma }_{k+1}$ Hence

$$\begin{array}{l}\mathrm{d}({\sigma }^{\ast }\alpha )={\sum }_{i=1}^{k+1}{\sum }_{j=1}^{k+1}\frac{\partial A_i}{\partial x^j}\mathrm{d}{x}^j\wedge \mathrm{d}{x}^1\wedge \cdots \wedge \mathrm{d}{x}^{i-1}\wedge \mathrm{d}{x}^{i+1}\wedge \cdots \wedge \mathrm{d}{x}^{k+1}\\ ={\sum }_{i=1}^{k+1}(-1{)}^{i+1}\frac{\partial A_i}{\partial x^i}\mathrm{d}{x}^1\wedge \cdots \wedge \mathrm{d}{x}^j\wedge \cdots \wedge \mathrm{d}{x}^{k+1}.\end{array}$$

Substituting in the left-hand integral of Eq. (17.2) we have

$$\begin{array}{l}{\int }_{{\Gamma }_{k+1}}\mathrm{d}({\sigma }^{\ast }\alpha )={\sum }_{i=1}^{k+1}(-1{)}^{i+1}{\int }_0^1{\int }_0^1\cdots {\int }_0^1\frac{\partial A_i}{\partial x^i}\mathrm{d}{x}^1\mathrm{d}{x}^2\dots \mathrm{d}{x}^{k+1}\\ ={\sum }_{i=1}^{k+1}(-1{)}^{i+1}{\int }_0^1{\int }_0^1\cdots {\int }_0^1\mathrm{d}{x}^1\dots \mathrm{d}{x}^{i-1}\mathrm{d}{x}^{i+1}\dots \mathrm{d}{x}^{k+1}\\ \times [A_i(x^1,\dots ,x^{i-1},1,x^{i+1},\dots ,x^{k+1})\\ -A_i(x^1,\dots ,x^{i-1},0,x^{i+1},\dots ,x^{k+1})]\\ ={\sum }_{ϵ=0}^1{\sum }_{i=1}^{k+1}(-1{)}^{i+ϵ}{\int }_0^1\cdots {\int }_0^1\mathrm{d}{x}^1\dots \mathrm{d}{x}^{i-1}\mathrm{d}{x}^{i+1}\dots \\ \mathrm{d}{x}^{k+1}{A}_i(x^1,\dots ,x^{i-1},ϵ,x^{i+1},\dots ,x^{k+1})\\ ={\sum }_{ϵ=0}^1{\sum }_{i=1}^{k+1}(-1{)}^{i+ϵ}{\int }_{{\Gamma }_k}{\left({\phi }_i^{ϵ}\right)}^{\ast }\circ {\sigma }^{\ast }\alpha \\ ={\int }_{\partial \sigma }\alpha ,\end{array}$$

as required.

Example 17.2

In Example 15.9 we defined the integral of a differential 1-form $\omega$ over a curve with end points $\gamma :[t_1,t_2]M$ to be

$${\int }_{\gamma }\omega ={\int }_{t_1}^{t_2}⟨\omega ,\dot{\gamma }⟩\mathrm{d}t={\int }_{t_1}^{t_2}⟨{\gamma }^{\ast }\omega ,\frac{\mathrm{d}}{\mathrm{d}t}⟩\mathrm{d}t.$$

A 1-cell $\sigma :U\to M$ is a curve with parameter range $U=(-a,1+a)\supset {\Gamma }_1=[0,1].$ and can be made to cover an arbitrary range $[t_1,t_2]$ by a change of parameter $t{t}^{\prime }=t_1+(t_2-t_1)t$ . The integral of $\omega =w_i(x^j)\mathrm{d}{x}^i$ over the support of the 1-cell is

$${\int }_{\sigma }\omega ={\int }_{{\Gamma }_1}\sigma \ast \omega ={\int }_0^1{w}_i(\mathbf{x}(t))\frac{\mathrm{d}{x}^i}{\mathrm{d}t}\mathrm{d}t,$$

which agrees with the definition of the integral given in Example 15.9.

The boundary of the 1-cell $\sigma$ is

$$\partial \sigma =\sigma \circ {\phi }_1^1-\sigma \circ {\phi }_1^0$$

where the two terms on the right-hand side are the 0-cells ${\Gamma }_0=\{0\}\to \sigma (1)$ and ${\Gamma }_0=\{0\}\to \sigma (0)$ , respectively. Setting $\omega =df$ where $f$ is a differentiable function on $M$, Stokes theorem gives

$${\int }_{\sigma }\mathrm{d}f={\int }_{\partial \sigma }f=f(\sigma (1))-f(\sigma (0)).$$

If $M=\mathbb{R}$ and the 1-cell $\sigma$ is defined by $x=\sigma (t)=a+t(b-a)$ , Stokes’ theorem reduces to the fundamental theorem of calculus,

$${\int }_a^b\frac{\mathrm{d}f}{\mathrm{d}x}\mathrm{d}x={\int }_0^1\frac{\mathrm{d}f}{\mathrm{d}t}\mathrm{d}t={\int }_{\sigma }\mathrm{d}f=f(b)-f(a).$$

Regular domains

In the above discussion, the only requirement made concerning the cell maps $\sigma$ was that they be differentiable on a neighbourhood of the unit $k$-cube ${\Gamma }_k$ . For example, they could be completely degenerate and map the entire set ${\Gamma }_k$ into a single point of $M$. For this reason, the chains are sometimes called singular. A fundamental $n$-chain on an $n$-dimensional manifold $M$ has the form

$$C={\sigma }_1+{\sigma }_2+\cdots +{\sigma }_N$$

where each ${\sigma }_i:U\to {\sigma }_i(U)$ is a diffeomorphism and the interiors of the supports ${\sigma }_i({\Gamma }_k)$ of different cells are non-intersecting:

$${\sigma }_i(({\Gamma }_n{)}^o)\cap {\sigma }_j(({\Gamma }_n{)}^o)=\varnothing \text{ for }i\ne j.$$

A regular domain $D\subseteq M$ is a closed set of the form

$$D=\bigcup _{i=1}^n{\sigma }_i({\Gamma }_n)$$

where ${\sigma }_i$ are the $n$-cells of a fundamental chain $C$ (see Fig. 17.3). We may think of a regular domain as subdivided into cubical cells or a region with boundary – the ‘boundary consisting of boundary points of the chain that are not on the common faces of any pair of cells.

Theorem 17.3 · Independence of integration from cubulation

If $D$ is a regular domain ‘cubulated’ in two different ways byfundamental chains $C={\sigma }_i+\cdot \cdot \cdot +{\sigma }_N$ and $C^{\prime }={\tau }_1+\cdot \cdot \cdot +{\tau }_M$ then ${\int }_C\omega ={\int }_{C^{\prime }}\omega$for every differential $n$-form $\omega$.

Figure 17.3 A regular domain on a manifold

Proof

Let $A_{ij}={\sigma }_i({\Gamma }_n)\cap {\tau }_j({\Gamma }_n)$ and

$$B_{ij}=({\sigma }_i{)}^{-1}(A_{ij}),C_{ij}=({\tau }_j{)}^{-1}(A_{ij}).$$

The maps ${\tau }_j^{-1}\circ {\sigma }_i:B_{i}j{C}_{ij}$ are all diffeomorphisms and

$$\begin{array}{rl}& {\int }_{B_{ij}}{\sigma }_i^{\ast }\omega ={\int }_{B_{ij}}{\sigma }_i^{\ast }\circ ({\tau }_j^{-1}{)}^{\ast }\circ {\tau }_j^{\ast }\omega \\ & ={\int }_{B_{ij}}(({\tau }_j^{-1})\circ {\sigma }_i{)}^{\ast }\circ {\tau }_j^{\ast }\omega \\ & ={\int }_{C_{ij}}{\tau }_j^{\ast }\omega .\end{array}$$

Hence

$${\int }_C\omega =\sum _i{\int }_{{\sigma }_i}\omega =\sum _{i,j}{\int }_{B_{ij}}{\sigma }_i^{\ast }\omega =\sum _{i,j}{\int }_{C_{ij}}{\tau }_j^{\ast }\omega ={\int }_{C^{\prime }}\omega .$$

If $\alpha$ is a differential ($n$ 1)-form,

$${\int }_{\partial D}\alpha ={\int }_{\partial C}\alpha =\sum _{i=1}^N\sum _{j=1}^n\sum _{ϵ=0}^1{\int }_{{\Gamma }_{n-1}}{\left({\phi }_j^{ϵ}\right)}^{\ast }\circ {\sigma }_i^{\ast }\alpha .$$

Since the outward normals on common faces of adjoining cells are oppositely directed, the faces will be oppositely oriented and the integrals will cancel, leaving only an integra on the ‘free’ parts of the boundary of $D$ . This results in Stokes’ theorem for a regular domain

$${\int }_D\mathrm{d}\alpha ={\int }_{\partial D}\alpha .$$

A regular $k$-domain $D_k$ is defined as the image of a regular domain $D\subset K$ of a $k$-dimensional manifold $K$ under a regular embedding $\phi :K\to M$ , and for any $k$-form

$\beta$ and $(k-1)$-form $\alpha$ we set

$$\begin{array}{l}{\int }_{D_k}\beta ={\int }_{\phi (D)}\beta ={\int }_D\phi \ast \beta \\ {\int }_{\partial D_k}\alpha ={\int }_{\partial \phi (D)}\alpha ={\int }_{\partial D}\phi \ast \alpha .\end{array}$$

The general form of Stokes’ theorem asserts that for any $(k-1)$-form $\alpha$ and regular $k-$ domain $D_k$

$${\int }_{D_k}\mathrm{d}\alpha ={\int }_{\partial D_k}\alpha .\text{\hspace{1em}(17.3)}$$

Example 17.3

In low dimensions, Stokes’ theorem reduces to a variety offamiliar forms. For example, let $D$ be a regular 2-domain in ${\mathbb{R}}^2$ bounded by a circuit $C=\partial D$ having induced orientation according to the ‘right hand rule’. By this we mean that if the outward normal is taken locally in the direction of the first coordinate $x^1$ , and the tangent to $C$ in the direction $x^2$ , then $\mathrm{d}{x}^1\wedge \mathrm{d}{x}^2$ is positively oriented. If $\alpha =P\mathrm{d}x+Q\mathrm{d}y$ , then

$$\mathrm{d}\alpha =\frac{\partial P}{\partial y}\mathrm{d}y\wedge \mathrm{d}x+\frac{\partial Q}{\partial x}\mathrm{d}x\wedge \mathrm{d}y=\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\mathrm{d}x\wedge \mathrm{d}y,$$

and Stokes’ theorem is equivalent to Green’s theorem

$${\iint }_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy={\oint }_{C}Pdx+Qdy.$$

If $D$ is a bounded region in ${\mathbb{R}}^3$ whose boundary is a surface $S=\partial D$ , the induced orientation is such that if ($e$, $f$) are a correctly ordered pair of tangent vectors to $S$ and $n$ the outward normal to $S$, then ($n$, $e$, $f$) is a positively oriented basis of vectors in ${\mathbb{R}}^3$ . Let $\alpha =A_1\mathrm{d}y\wedge \mathrm{d}z+A_2\mathrm{d}z\wedge \mathrm{d}x+A_3\mathrm{d}x\wedge \mathrm{d}z$ be a 2-form, then

$$\mathrm{d}\alpha =\left(A_{1,1}+A_{2,2}+A_{3,3}\right)\mathrm{d}x\wedge \mathrm{d}y\wedge \mathrm{d}z=\nabla \cdot \mathbf{A}\mathrm{d}x\wedge \mathrm{d}y\wedge \mathrm{d}z$$

where $\mathbf{A}=(A_1,A_2,A_3)$ . Stokes’ theorem reads

$${\int }_D\mathrm{d}\alpha ={\iiint }_D\nabla \cdot \mathbf{A}\mathrm{d}x\mathrm{d}y\mathrm{d}z={\int }_{\partial D}\alpha ={\iint }_S{A}_1\mathrm{d}y\mathrm{d}z+A_2\mathrm{d}z\mathrm{d}x+A_3\mathrm{d}x\mathrm{d}z.$$

If the bounding surface is locally parametrized by two parameters, $x=x({\lambda }_1,{\lambda }_2),y=y({\lambda }_1,{\lambda }_2),z=z({\lambda }_1,{\lambda }_2)$ , then we can write

$${\iint }_S\alpha ={\iint }_S{ϵ}_{ijk}{A}_i\frac{\partial x^j}{\partial {\lambda }_1}\frac{\partial x^k}{\partial {\lambda }_2}d{\lambda }_1\wedge d{\lambda }_2$$

and it is common to write Stokes’ theorem in the standard Gauss theorem form

$${\iiint }_D\nabla \cdot \mathbf{A}\mathrm{d}x\mathrm{d}y\mathrm{d}z={\iint }_S\mathbf{A}\cdot \mathrm{d}\mathbf{S},$$

where $S$ is the vector area normal to $S$, having components

$$\mathrm{d}{S}_i={ϵ}_{ijk}\frac{\partial x^j}{\partial {\lambda }_1}\frac{\partial x^k}{\partial {\lambda }_2}\mathrm{d}{\lambda }_1\mathrm{d}{\lambda }_2.$$

Let $\Sigma$ be an oriented 2-surface in ${\mathbb{R}}^3$ , with boundary an appropriately oriented circui $C=\partial \Sigma$ , and $\alpha$ a differential 1-form $\alpha =A_1\mathrm{d}x+A_2\mathrm{d}y+A_3\mathrm{d}z=A_i\mathrm{d}{x}^i$ . Then

$$\begin{array}{l}{\int }_{\Sigma }\mathrm{d}\alpha ={\int }_{\Sigma }{A}_{i,j}\mathrm{d}{x}^j\wedge \mathrm{d}{x}^i\\ ={\iint }_{\Sigma }{A}_{k,j}\left(\frac{\partial x^j}{\partial {\lambda }_1}\frac{\partial x^k}{\partial {\lambda }_2}-\frac{\partial x^k}{\partial {\lambda }_1}\frac{\partial x^j}{\partial {\lambda }_2}\right)\mathrm{d}{\lambda }_1\wedge \mathrm{d}{\lambda }_2\\ ={\iint }_{\Sigma }{ϵ}_{ijk}{A}_{k,j}\mathrm{d}{S}_i\end{array}$$

and

$${\int }_{\partial \Sigma }\alpha ={\oint }_C{A}_i\mathrm{d}{x}^i={\oint }_C{A}_1\mathrm{d}x+A_2\mathrm{d}y+A_3\mathrm{d}z.$$

This can be expressed in the familiar form of Stokes’ theorem

$${\iint }_{\Sigma }(\nabla \times \mathbf{A})\cdot \mathrm{d}\mathbf{S}={\oint }_C\mathbf{A}\cdot \mathrm{d}\mathbf{r}.$$

Exercise

Show that $\mathrm{d}{S}_i$ is ‘normal’ to the surface $S$ in the sense tha $\mathrm{d}{S}_i\partial x^i/\partial {\lambda }_a=0\mathrm{for}a=1,2$

Problems

Problem 17.4

Let $\alpha =y^2\mathrm{d}x+x^2\mathrm{d}y$ . If ${\gamma }_1$ is the stretch of $y$-axis from $(x=0,y=-1)\mathrm{to}(x=0,y=1)$ , and ${\gamma }_2$ the unit right semicircle connecting these points, evaluate

$${\int }_{{\gamma }_1}\alpha ,{\int }_{{\gamma }_2}\alpha \text{ and }{\int }_{S^1}\alpha .$$

Verify Stokes’ theorem for the unit circle and the unit right semicircular region encompassed by ${\gamma }_1$ and ${\gamma }_2$ .

Problem 17.5

If $\alpha =x\mathrm{d}y\wedge \mathrm{d}z+y\mathrm{d}z\wedge \mathrm{d}x+z\mathrm{d}x\wedge \mathrm{d}y$ compute ${\int }_{\partial \Omega }\alpha$ where $\Omega$ is (i) the uni cube, (ii) the unit ball in ${\mathbb{R}}^3$ . In each case verify Stokes’ theorem,

$${\int }_{\partial \Omega }\alpha ={\int }_{\Omega }\mathrm{d}\alpha .$$

Problem 17.6

Let $S$ be the surface of a cylinder of elliptical cross-section and height 2$h$ given by

$$x=a\cos \theta ,y=b\sin \theta (0\le \theta <2\pi ),-h\le z\le h.$$

(a) Compute ${\int }_S\alpha$ where $\alpha$ $x$ $\mathrm{d}y$ $\mathrm{d}z$ $y$ $\mathrm{d}z$ $dx$ 2$z$ $dx$ $\mathrm{d}y$.

(b) Show $\mathrm{d}\alpha =0$, and find a 1-form $\omega$ such that $\alpha =\mathrm{d}\omega$

(c) Verify Stokes’ theorem $\begin{array}{r}{\int }_S\alpha ={\int }_{\partial S}\omega .\end{array}$

Problem 17.7

A torus in ${\mathbb{R}}^3$ may be represented parametrically by

$$x=\cos \varphi (a+b\cos \psi ),y=\sin \varphi (a+b\cos \psi ),z=b\sin \psi$$

where $0\le \varphi <2\pi ,0\le \psi <2\pi$ . If $b$ is replaced by a variable $\rho$ that ranges from 0 to ${}^{b,}$ show that

$$\mathrm{d}x\wedge \mathrm{d}y\wedge \mathrm{d}z=\rho (a+\rho \cos \psi )\mathrm{d}\varphi \wedge \mathrm{d}\psi \wedge \mathrm{d}\rho .$$

By integrating this 3-form over the region enclosed by the torus, show that the volume of the solid torus is $2{\pi }^{2}a{b}^2$ . Can you see this by a simple geometrical argument?

Evaluate the volume by performing the integral of the 2-form $\alpha =x$ $\mathrm{d}y$ $\mathrm{d}z$ over the surface of the torus and using Stokes’ theorem.

Problem 17.8

Show that in $n$ dimensions, if $V$ is a regular $n$-domain with boundary $S=\partial V$ , and we set $\alpha$ to be an ($n$ 1)-form with components

$$\alpha =\sum _{i=1}^n(-1{)}^{i+1}{A}^i\mathrm{d}{x}^1\wedge \cdots \wedge \mathrm{d}{x}^{i-1}\wedge \mathrm{d}{x}^{i+1}\wedge \cdots \wedge \mathrm{d}{x}^n,$$

Stokes’ theorem can be reduced to the $n$-dimensional Gauss theorem

$${\int }_V\cdots \int A_{,i}^i\mathrm{d}{x}^1\dots \mathrm{d}{x}^n={\int }_S\cdots \int A^i\mathrm{d}{S}_i$$

where $\mathrm{d}{S}_i=\mathrm{d}{x}^1\dots \mathrm{d}{x}^{i-1}\mathrm{d}{x}^{i+1}\dots \mathrm{d}{x}^n$ is a ‘vector volume element’ normal to $S$.