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Mathematics LibreTexts

16.8: Stokes's Theorem

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

David Guichard (Whitman College)

  • Integrated by Justin Marshall.