Stokes_Theorem

Introduction

In my second semester of High School Junior year, when we were studying vector calculus @Calculus 3, our teacher gave us an assignment: proof of Stokes formula. CALCULUS EARLY TRANSCENDENTALS EIGHTH EDITION JAMES STEWART (our Calculus3 textbook) only has a special form of stokes' theorem proof. Thus, I looked up the prood at old Chinese Text Book. It is a pity that our high school calculus education with American calculus education are quite different compared to Chinese's. I searched on zhihu, Bilibili and other public platforms, but failed to find any suitable results. Luckily, I finally figure it out when I went back home during the epidemic.

Stoke's Theorem

For surface S in \(R^3\) space is an oriented surface in \(x,y,z\) coordinate with boundary \(\partial S\). Let \(R\) be abounded, open region in \(s, t\) plane with smooth boundary \(\partial R\) . Suppose that \(F\) is a continuously differentiable vector field. \(\vec{r}\) is a smooth parametrization that maps \(R\) to \(S\), and \(\partial R\) to \(\partial S\). Then

\[ \iint_S{curl\,\,\vec{F}\,\,·\,\,d\vec{S}}=\iint_S{\left( \nabla \times \vec{F} \right) \,\,·\,\,\vec{n}\,\,dS}=\int_{\partial S}{\vec{F}\,\,·\,\,\vec{T}\,\,ds}=\,\,\int_{\partial S}{\vec{F}\,\,·\,\,d\vec{r}} \]

stks

Proof

Let

\[ \vec{r}\left( s,t \right) =\left[ \begin{array}{c} x\left( s,t \right)\\ y\left( s,t \right)\\ z\left( s,t \right)\\ \end{array} \right] \,\,: R\rightarrow S. \]

Then.

\[ d\vec{r}\mid_{\partial R}^{}\,\,=\left[ \begin{matrix} \frac{\partial x}{\partial s}& \frac{\partial x}{\partial t}\\ \frac{\partial y}{\partial s}& \frac{\partial y}{\partial t}\\ \frac{\partial z}{\partial s}& \frac{\partial z}{\partial t}\\ \end{matrix} \right] \left[ \begin{array}{c} ds\\ dt\\ \end{array} \right] \,\,=\left[ \begin{array}{c} \frac{\partial x}{\partial s}ds+\frac{\partial x}{\partial t}dt\\ \frac{\partial y}{\partial s}ds+\frac{\partial y}{\partial t}dt\\ \frac{\partial z}{\partial s}ds+\frac{\partial z}{\partial t}dt\\ \end{array} \right] \,\,=\left[ \begin{array}{c} \frac{\partial x}{\partial s}\\ \frac{\partial y}{\partial s}\\ \frac{\partial z}{\partial s}\\ \end{array} \right] ds+\left[ \begin{array}{c} \frac{\partial x}{\partial t}\\ \frac{\partial y}{\partial t}\\ \frac{\partial z}{\partial t}\\ \end{array} \right] dt\,\,=\frac{\partial \vec{r}}{\partial s}ds+\frac{\partial \vec{r}}{\partial t}dt \]

Hence.

\[ \int_{\partial S}{\vec{F}·}d\vec{r}=\int_{\partial S}{\left( \vec{F}·\frac{\partial \vec{r}}{\partial s}ds+\vec{F}·\frac{\partial \vec{r}}{\partial s} \right)} \]

We define a 2-dimensional vector field \(G=(G1,G2)\) on the s,t, plane by

\[ G_1=\vec{F}\,\,·\,\,\frac{\partial \vec{r}}{\partial s}\,\, and\,\, G_2=\vec{F}\,\,·\,\,\frac{\partial \vec{r}}{\partial t} \]

Therefore, we put G into the original line integral

\[ \int_{\partial S}{\vec{F}\,\,·\,\,d\vec{r}=}\int_{\partial R}{\left( G_1ds+G_2dt \right) =\int_R{\begin{array}{c} \left( \frac{\partial G_2}{\partial s}-\frac{\partial G_1}{\partial t} \right) dsdt\,\,,\\ \end{array}}\,\,} \]

\[ \int_{\partial S}{\vec{F}\,\,·\,\,d\vec{r}=}\int_{\partial R}{\left( G_1ds+G_2dt \right) \,\,} \]

On the other hand.

\[ \iint_S{curl\,\,\vec{F}\,\,·\,\,d\vec{S}\,\,=\,\,\iint_R{curl\,\,\vec{F}\mid_{\vec{r}}^{}·\frac{\partial \vec{r}}{\partial s}\times \frac{\partial \vec{r}}{\partial t}dsdt}} \]

We expand it, get

\[ curl\,\,\vec{F}\mid_{\vec{r}}^{}\cdot \frac{\partial \vec{r}}{\partial s}\times \frac{\partial \vec{r}}{\partial t}\,\,=\,\,\left| \begin{matrix} \frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}& \frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x}& \frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\\ \frac{\partial x}{\partial s}& \frac{\partial y}{\partial s}& \frac{\partial z}{\partial s}\\ \frac{\partial x}{\partial t}& \frac{\partial y}{\partial t}& \frac{\partial z}{\partial t}\\ \end{matrix} \right| \]

\[ =\frac{\partial \vec{F}}{\partial s}·\frac{\partial \vec{r}}{\partial t}-\frac{\partial \vec{F}}{\partial t}·\frac{\partial \vec{r}}{\partial s}\,\,=\frac{\partial G_2}{\partial s}-\frac{\partial G_1}{\partial t} \]

Hence.

\[ \int_R{curl\,\,F\mid_{\vec{r}}^{}\cdot \frac{\partial \vec{r}}{\partial s}}\times \frac{\partial \vec{r}}{\partial t}\,\,dsdt\,\,=\,\,\int_R{\begin{array}{c} \left( \frac{\partial G_2}{\partial s}-\frac{\partial G_1}{\partial t} \right) dsdt\,\,\\ \end{array}} \]

For the line integral part,we have

\[ \int_{\partial S}{\vec{F}\,\,·\,\,d\vec{r}=}\int_{\partial R}{\begin{array}{c} \left( G_1ds+G_2dt \right) \,\,\\ \end{array}} \]

Use Green's Theorem, we know

\[ \int_{\partial R}{\begin{array}{c} \left( G_1ds+G_2dt \right)\\ \end{array}}=\int_R{\begin{array}{c} \left( \frac{\partial G_2}{\partial s}-\frac{\partial G_1}{\partial t} \right) dsdt\,\,,\\ \end{array}} \]

So, we can conclude that

\[ \int_{\partial S}{\vec{F}·}d\vec{r}=\int_R{\begin{array}{c} \left( \frac{\partial G_2}{\partial s}-\frac{\partial G_1}{\partial t} \right) dsdt=\int_R{curl\,\,F\mid_{\vec{r}}^{}\cdot \frac{\partial \vec{r}}{\partial s}}\times \frac{\partial \vec{r}}{\partial t}\\ \end{array}}dsdt=\iint_S{curl\,\,F\,\,·\,\,d\vec{S}} \]

Thus, we finished our proof of Stokes' Theorem.

Reference

[1] Fichtengoltz, G. M. (1959). Course on the Differential and Integral.

[2] Stewart, J., Clegg, D. K., & Watson, S. (2020).Calculus: early transcendentals. Cengage Learning.

After

You can browse my other articles at my zhihu account : https://zhuanlan.zhihu.com/p/543973521