Stokes’ Theorem

Stokes’ theorem says we can calculate the flux of curl ${\bf{F}}$ across surface $S$ by knowing information only about the values of ${\bf{F}}$ along the boundary of $S$. Conversely, we can calculate the line integral of vector field ${\bf{F}}$ along the boundary of surface $S$ by translating to a double integral of the curl of ${\bf{F}}$ over $S$.

Let $S$ be an oriented smooth surface with unit normal vector ${\bf{N}}$. Furthermore, suppose the boundary of $S$ is a simple closed curve $C$. The orientation of $S$ induces the positive orientation of $C$ if, as you walk in the positive direction around $C$ with your head pointing in the direction of ${\bf{N}}$, the surface is always on your left. With this definition in place, we can state Stokes’ theorem.

Proof

First, we look at an informal proof of the theorem. This proof is not rigorous, but it is meant to give a general feeling for why the theorem is true. Let $S$ be a surface and let $D$ be a small piece of the surface so that $D$ does not share any points with the boundary of $S$. We choose $D$ to be small enough so that it can be approximated by an oriented square $E$. Let $D$ inherit its orientation from $S$, and give $E$ the same orientation. This square has four sides; denote them $E_l$, $E_r$, $E_u$, and $E_d$ for the left, right, up, and down sides, respectively. On the square, we can use the flux form of Green’s theorem:

$\large{\displaystyle\int_{E_l+E_d+E_r+E_u}{\bf{F}}\cdot{d}{\bf{r}}=\displaystyle\iint_E\text{curl }{\bf{F}}\cdot{\bf{N}}{d}S=\displaystyle\iint_E\text{curl }{\bf{F}}\cdot{d}{\bf{S}}}$.

To approximate the flux over the entire surface, we add the values of the flux on the small squares approximating small pieces of the surface (Figure 2). By Green’s theorem, the flux across each approximating square is a line integral over its boundary. Let $F$ be an approximating square with an orientation inherited from $S$ and with a right side $E_l$ (so $F$ is to the left of $E$). Let $F_r$ denote the right side of $F$; then, $E_l=-F_r$. In other words, the right side of $F$ is the same curve as the left side of $E$, just oriented in the opposite direction. Therefore,

$\large{\displaystyle\int_{E_l}{\bf{F}}\cdot{d}{\bf{r}}=-\displaystyle\int_{F_r}{\bf{F}}\cdot{d}{\bf{r}}}$.

As we add up all the fluxes over all the squares approximating surface $S$, line integrals $\displaystyle\int_{E_l}{\bf{F}}\cdot{d}{\bf{r}}$ and $\displaystyle\int_{F_r}{\bf{F}}\cdot{d}{\bf{r}}$cancel each other out. The same goes for the line integrals over the other three sides of $E$. These three line integrals cancel out with the line integral of the lower side of the square above $E$, the line integral over the left side of the square to the right of $E$, and the line integral over the upper side of the square below $E$ (Figure 3). After all this cancelation occurs over all the approximating squares, the only line integrals that survive are the line integrals over sides approximating the boundary of $S$. Therefore, the sum of all the fluxes (which, by Green’s theorem, is the sum of all the line integrals around the boundaries of approximating squares) can be approximated by a line integral over the boundary of $S$. In the limit, as the areas of the approximating squares go to zero, this approximation gets arbitrarily close to the flux.

Figure 2. Chop the surface into small pieces. The pieces should be small enough that they can be approximated by a square.

Figure 3. (a) The line integral along ${E_l}$ cancels out the line integral along ${F_r}$ because ${E_l}=-{F_r}$. (b) The line integral along any of the sides of $E$ cancels out with the line integral along a side of an adjacent approximating square.

Let’s now look at a rigorous proof of the theorem in the special case that $S$ is the graph of function $z=f(x, y)$, where $x$ and $y$ vary over a bounded, simply connected region $D$ of finite area (Figure 4). Furthermore, assume that $f$ has continuous second-order partial derivatives. Let $C$ denote the boundary of $S$ and let $C'$ denote the boundary of $D$. Then, $D$ is the “shadow” of $S$ in the plane and $C'$ is the “shadow” of $C$. Suppose that $S$ is oriented upward. The counterclockwise orientation of $C$ is positive, as is the counterclockwise orientation of $C^\prime$. Let ${\bf{F}}(x,y,z)=\langle{P},Q,R\rangle$ be a vector field with component functions that have continuous partial derivatives.

Figure 4. $D$ is the “shadow,” or projection, of $S$ in the plane and $C'$ is the projection of $C$.

We take the standard parameterization of $S:x=x,y=y,z=g(x,y)$. The tangent vectors are ${\bf{t}}_x=\langle1,0,g_x\rangle$, and ${\bf{t}}_y=\langle0,1,g_y\rangle$, and therefore, ${\bf{t}}_x\times{\bf{t}}_y=\langle-g_x,-g_y,1\rangle$. By the equation to calculate scalar surface integrals,

$\large{\displaystyle\iint_S\text{curl }{\bf{F}}\cdot{d}{\bf{S}}=\displaystyle\iint_D[-(R_y-Q_z)z_x-(P_z-R_x)z_y+(Q_x-P_y)]dA}$,

where the partial derivatives are all evaluated at $(x,y,g(x,y))$, making the integrand depend on $x$ and $y$ only. Suppose $\langle{x}(t),y(t)\rangle$, $a\leq{t}\leq{b}$ is a parameterization of $C'$. Then, a parameterization of $C$ is $\langle{x}(t),y(t),g(x(t),y(t))\rangle$, $a\leq{t}\leq{b}$. Armed with these parameterizations, the Chain rule, and Green’s theorem, and keeping in mind that $P$, $Q$, and $R$ are all functions of $x$ and $y$, we can evaluate line integral $\displaystyle\int_C{\bf{F}}\cdot{d}{\bf{r}}$:

$\large{\begin{aligned} \displaystyle\int_C{\bf{F}}\cdot{d}{\bf{r}}&=\displaystyle\int_a^b(Px^\prime(t)+Qy^\prime(t)+Rz^\prime(t))dt \\ &=\displaystyle\int_a^b\left[Px^\prime(t)+Qy^\prime(t)+R\left(\frac{\partial{z}}{\partial{x}}\frac{dx}{dt}+\frac{\partial{z}}{\partial{y}}\frac{dy}{dt}\right)\right]dt \\ &=\displaystyle\int_a^b\left[\left((P+R\frac{\partial{z}}{\partial{x}}\right)x^\prime(t)+\left(Q+R\frac{\partial{z}}{\partial{y}}\right)y^\prime(t)\right]dt \\ &=\displaystyle\int_{C^\prime}\left(P+R\frac{\partial{z}}{\partial{x}}\right)dx+\left(Q+R\frac{\partial{z}}{\partial{y}}\right)dy \\ &=\displaystyle\iint_D\left[\frac{\partial}{\partial{x}}\left(Q+R\frac{\partial{z}}{\partial{y}}\right)-\frac{\partial}{\partial{y}}\left(P+R\frac{\partial{z}}{\partial{x}}\right)\right]dA \\ &=\displaystyle\iint_D\left(\frac{\partial{Q}}{\partial{x}}+\frac{\partial{Q}}{\partial{z}}\frac{\partial{z}}{\partial{x}}+\frac{\partial{R}}{\partial{x}}\frac{\partial{z}}{\partial{y}}+\frac{\partial{R}}{\partial{z}}\frac{\partial{z}}{\partial{x}}\frac{\partial{z}}{\partial{y}}+R\frac{\partial^2{z}}{\partial{x}\partial{y}}\right)-\left(\frac{\partial{P}}{\partial{y}}+\frac{\partial{P}}{\partial{z}}\frac{\partial{z}}{\partial{y}}+\frac{\partial{R}}{\partial{y}}\frac{\partial{z}}{\partial{x}}+\frac{\partial{R}}{\partial{z}}\frac{\partial{z}}{\partial{y}}\frac{\partial{z}}{\partial{x}}+R\frac{\partial^2{z}}{\partial{y}\partial{x}}\right)dA \end{aligned}}$.

By Clairaut’s theorem, $\frac{\partial^2{z}}{\partial{x}\partial{y}}=\frac{\partial^2{z}}{\partial{y}\partial{x}}$. Therefore, four of the terms disappear from this double integral, and we are left with

$\large{\displaystyle\iint_D[-R(R_y-Q_z)z_x-(P_z-R_x)z_y+(Q_x-P_y)]dA}$,

which equals $\displaystyle\iint_S\text{curl }{\bf{F}}\cdot{d}{\bf{S}}$.

$_\blacksquare$

We have shown that Stokes’ theorem is true in the case of a function with a domain that is a simply connected region of finite area. We can quickly confirm this theorem for another important case: when vector field ${\bf{F}}$ is conservative. If ${\bf{F}}$ is conservative, the curl of ${\bf{F}}$ is zero, so $\displaystyle\iint_S\text{curl }{\bf{F}}\cdot{d}{\bf{S}}=0$. Since the boundary of $S$ is a closed curve, $\displaystyle\int_C{\bf{F}}\cdot{d}{\bf{r}}$ is also zero.