
# 16.8: Stokes's Theorem

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Recall that one version of Green's Theorem (see equation 16.5.1) is

$$\int_{\partial D} {\bf F}\cdot d{\bf r} =\iint_\limits{D}(\nabla\times {\bf F})\cdot{\bf k}\,dA.$$

Here $$D$$ is a region in the $$x$$-$$y$$ plane and $$\bf k$$ is a unit normal to $$D$$ at every point. If $$D$$ is instead an orientable surface in space, there is an obvious way to alter this equation, and it turns out still to be true:

Stoke's Theorem

Provided that the quantities involved are sufficiently nice, and in particular if $$D$$ is orientable,

$$\int_{\partial D} {\bf F}\cdot d{\bf r}=\iint_\limits{D}(\nabla\times {\bf F})\cdot{\bf N}\,dS,$$

if $$\partial D$$ is oriented counter-clockwise relative to $$\bf N$$.

Note how little has changed: $$\bf k$$ becomes $$\bf N$$, a unit normal to the surface, and $$dA$$ becomes $$dS$$, since this is now a general surface integral. The phrase "counter-clockwise relative to $$\bf N$$" means roughly that if we take the direction of $$\bf N$$ to be "up", then we go around the boundary counter-clockwise when viewed from "above''. In many cases, this description is inadequate. A slightly more complicated but general description is this: imagine standing on the side of the surface considered positive; walk to the boundary and turn left. You are now following the boundary in the correct direction.

Example $$\PageIndex{2}$$:

Let $${\bf F}=\langle e^{xy}\cos z,x^2z,xy\rangle$$ and the surface $$D$$ be $$x=\sqrt{1-y^2-z^2}$$, oriented in the positive $$x$$ direction. It quickly becomes apparent that the surface integral in Stokes's Theorem is intractable, so we try the line integral. The boundary of $$D$$ is the unit circle in the $$y$$-$$z$$ plane, $${\bf r}=\langle 0,\cos u,\sin u\rangle$$, $$0\le u\le 2\pi$$. The integral is

$$\int_0^{2\pi} \langle e^{xy}\cos z,x^2z,xy\rangle\cdot\langle 0,-\sin u,\cos u\rangle\,du= \int_0^{2\pi} 0\,du = 0,$$

because $$x=0$$.

Example $$\PageIndex{3}$$:

Consider the cylinder $${\bf r}=\langle \cos u,\sin u, v\rangle$$, $$0\le u\le 2\pi$$, $$0\le v\le 2$$, oriented outward, and $${\bf F}=\langle y,zx,xy\rangle$$. We compute

$$\iint_\limits{D} \nabla\times{\bf F}\cdot {\bf N}\,dS= \int_{\partial D}{\bf F}\cdot d{\bf r}$$

in two ways.

First, the double integral is
$$\int_0^{2\pi}\int_0^2 \langle 0,-\sin u,v-1\rangle\cdot \langle \cos u, \sin u, 0\rangle\,dv\,du= \int_0^{2\pi}\int_0^2 -\sin^2 u\,dv\,du = -2\pi.$$

The boundary consists of two parts, the bottom circle $$\langle \cos t,\sin t, 0\rangle$$, with $$t$$ ranging from $$0$$ to $$2\pi$$, and $$\langle \cos t,\sin t, 2\rangle$$, with $$t$$ ranging from $$2\pi$$ to $$0$$. We compute the corresponding integrals and add the results:

$$\int_0^{2\pi} -\sin^2 t\,dt+\int_{2\pi}^0 -\sin^2t +2\cos^2t =-\pi-\pi=-2\pi,$$

as before.

An interesting consequence of Stokes's Theorem is that if $$D$$ and $$E$$ are two orientable surfaces with the same boundary, then

$$\iint_\limits{D}(\nabla\times {\bf F})\cdot{\bf N}\,dS =\int_{\partial D} {\bf F}\cdot d{\bf r} =\int_{\partial E} {\bf F}\cdot d{\bf r} =\iint_\limits{E}(\nabla\times {\bf F})\cdot{\bf N}\,dS.$$

Sometimes both of the integrals

$$\iint_\limits{D}(\nabla\times {\bf F})\cdot{\bf N}\,dS\qquad\hbox{and}\qquad\int_{\partial D} {\bf F}\cdot d{\bf r}$$

are difficult, but you may be able to find a second surface $$E$$ so that

$$\iint_\limits{E}(\nabla\times {\bf F})\cdot{\bf N}\,dS$$

has the same value but is easier to compute.

Example $$\PageIndex{4}$$:

In example 16.8.2 the line integral was easy to compute. But we might also notice that another surface $$E$$ with the same boundary is the flat disk $$y^2+z^2\le 1$$. The unit normal $$\bf N$$ for this surface is simply $${\bf i}=\langle 1,0,0\rangle$$. We compute the curl:

$$\nabla\times{\bf F}=\langle x-x^2,-e^{xy}\sin z-y,2xz-xe^{xy}\cos z\rangle.$$

Since $$x=0$$ everywhere on the surface,

$$(\nabla\times{\bf F})\cdot {\bf N}=\langle 0,-e^{xy}\sin z-y,2xz-xe^{xy}\cos z\rangle\cdot\langle 1,0,0\rangle=0,$$
so the surface integral is

$$\iint_\limits{E}0\,dS=0,$$

as before. In this case, of course, it is still somewhat easier to compute the line integral, avoiding \$$$nabla\times{\bf F}$$ entirely.

Example $$\PageIndex{5}$$:

Let $${\bf F}=\langle -y^2,x,z^2\rangle$$, and let the curve $$C$$ be the intersection of the cylinder $$x^2+y^2=1$$ with the plane $$y+z=2$$, oriented counter-clockwise when viewed from above. We compute $$\int_C {\bf F}\cdot d{\bf r}$$ in two ways.

First we do it directly: a vector function for $$C$$ is({\bf r}=\langle \cos u,\sin u, 2-\sin u\rangle\), so $${\bf r}'=\langle -\sin u,\cos u,-\cos u\rangle$$, and the integral is then

$$\int_0^{2\pi} y^2\sin u+x\cos u-z^2\cos u\,du =\int_0^{2\pi} \sin^3 u+\cos^2 u-(2-\sin u)^2\cos u\,du =\pi.$$

To use Stokes's Theorem, we pick a surface with $$C$$ as the boundary; the simplest such surface is that portion of the plane $$y+z=2$$ inside the cylinder. This has vector equation $${\bf r}=\langle v\cos u,v\sin u,2-v\sin u\rangle$$. We compute $${\bf r}_u= \langle -v\sin u,v\cos u,-v\cos u\rangle$$, $${\bf r}_v= \langle \cos u,\sin u, -\sin u\rangle$$, and $${\bf r}_u\times{\bf r}_v=\langle 0,-v,-v\rangle$$. To match the orientation of $$C$$ we need to use the normal $$\langle 0,v,v\rangle$$. The curl of $$\bf F$$ is $$\langle 0,0,1+2y\rangle= \langle 0,0,1+2v\sin u\rangle$$, and the surface integral from Stokes's Theorem is

$$\int_0^{2\pi}\int_0^1 (1+2v\sin u)v\,dv\,du=\pi.$$

In this case the surface integral was more work to set up, but the resulting integral is somewhat easier.

Proof of Stokes's Theorem

We can prove here a special case of Stokes's Theorem, which perhaps not too surprisingly uses Green's Theorem.

Suppose the surface $$D$$ of interest can be expressed in the form $$z=g(x,y)$$, and let $${\bf F}=\langle P,Q,R\rangle$$. Using the vector function $${\bf r}=\langle x,y,g(x,y)\rangle$$ for the surface we get the surface integral

\eqalign{\iint_\limits{D} \nabla\times{\bf F}\cdot d{\bf S}&= \iint_\limits{E} \langle R_y-Q_z,P_z-R_x,Q_x-P_y\rangle\cdot \langle -g_x,-g_y,1\rangle\,dA\cr &=\iint_\limits{E}-R_yg_x+Q_zg_x-P_zg_y+R_xg_y+Q_x-P_y\,dA.\cr}

Here $$E$$ is the region in the $$x$$-$$y$$ plane directly below the surface$$D$$.

For the line integral, we need a vector function for $$\partial D$$. If $$\langle x(t),y(t)\rangle$$ is a vector function for $$\partial E$$ then we may use $${\bf r}(t)=\langle x(t),y(t),g(x(t),y(t))\rangle$$ to represent $$\partial D$$. Then

$$\int_{\partial D}{\bf F}\cdot d{\bf r}=\int_a^b P{dx\over dt}+Q{dy\over dt}+R{dz\over dt}\,dt=\int_a^b P{dx\over dt}+Q{dy\over dt}+R\left({\partial z\over\partial x}{dx\over dt}+{\partial z\over\partial y}{dy\over dt}\right)\,dt.$$

using the chain rule for $$dz/dt$$. Now we continue to manipulate this:

\eqalign{\int_a^b P{dx\over dt}+Q{dy\over dt}+&R\left({\partial z\over\partial x}{dx\over dt}+{\partial z\over\partial y}{dy\over dt}\right)\,dt\cr &=\int_a^b \left[\left(P+R{\partial z\over\partial x}\right){dx\over dt}+ \left(Q+R{\partial z\over\partial y}\right){dy\over dt}\right]\,dt\cr &=\int_{\partial E} \left(P+R{\partial z\over\partial x}\right)\,dx+\left(Q+R{\partial z\over\partial y}\right)\,dy,\cr}

which now looks just like the line integral of Green's Theorem, except that the functions $$P$$ and $$Q$$ of Green's Theorem have been replaced by the more complicated $$P+R(\partial z/\partial x)$$ and $$Q+R(\partial z/\partial y)$$. We can apply Green's Theorem to get

$$\int_{\partial E} \left(P+R{\partial z\over\partial x}\right)\,dx+\left(Q+R{\partial z\over\partial y}\right)\,dy= \iint_\limits{E} {\partial\over \partial x}\left(Q+R{\partial z\over\partial y}\right)-{\partial\over \partial y}\left(P+R{\partial z\over\partial x}\right)\,dA.$$

Now we can use the chain rule again to evaluate the derivatives inside this integral, and it becomes

\eqalign{\iint_\limits{E} &Q_x+Q_zg_x+R_xg_y+R_zg_xg_y+Rg_{yx}-\left(P_y+P_zg_y+R_yg_x+R_zg_yg_x+Rg_{xy}\right)\,dA\cr&=\iint_\limits{E} Q_x+Q_zg_x+R_xg_y-P_y-P_zg_y-R_yg_x\,dA,\cr}

which is the same as the expression we obtained for the surface integral.

$$\square$$

## Contributors

• Integrated by Justin Marshall.