Recall that one version of Green's Theorem (see equation 16.5.1) is $$\int_{\partial D} {\bf F}\cdot d{\bf r} =\dint{D}(\nabla\times {\bf F})\cdot{\bf k}\,dA. $$ 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:

Theorem 16.8.1 (Stokes's Theorem) Provided that the quantities involved are sufficiently nice, and in particular if $D$ is orientable, $$\int_{\partial D} {\bf F}\cdot d{\bf r} =\dint{D}(\nabla\times {\bf F})\cdot{\bf N}\,dS, $$ if $\partial D$ is oriented counter-clockwise relative to $\bf N$. $\square$

Note how little has changed: $\bf k$ becomes $\bf N$, a unit normal to the surface (just as $\bf k$ is a unit normal to the $x$-$y$-plane), 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 16.8.2 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 $x=0$. $\square$

Example 16.8.3 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 $$\dint{D} \nabla\times{\bf F}\cdot {\bf N}\,dS= \int_{\partial D}{\bf F}\cdot d{\bf r}$$ 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. $\square$

An interesting consequence of Stokes's Theorem is that if $D$ and $E$ are two orientable surfaces with the same boundary, then $$ \dint{D}(\nabla\times {\bf F})\cdot{\bf N}\,dS =\int_{\partial D} {\bf F}\cdot d{\bf r} =\int_{\partial E} {\bf F}\cdot d{\bf r} =\dint{E}(\nabla\times {\bf F})\cdot{\bf N}\,dS. $$ Sometimes both of the integrals $$\dint{D}(\nabla\times {\bf F})\cdot{\bf N}\,dS \qquad\hbox{and}\qquad\int_{\partial D} {\bf F}\cdot d{\bf r}$$ are difficult, but you may be able to find a second surface $E$ so that $$\dint{E}(\nabla\times {\bf F})\cdot{\bf N}\,dS $$ has the same value but is easier to compute.

Example 16.8.4 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 $y^2+z^2\le 1$, given by ${\bf r}=\langle 0,v\cos u,v\sin u\rangle$. The normal is ${\bf r_v}\times{\bf r_u}=\langle v,0,0\rangle$. 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, $$(\nabla\times{\bf F})\cdot {\bf N}= \langle 0,-\sin z-y,0\rangle\cdot\langle v,0,0\rangle=0,$$ so the surface integral is $$\dint{E}0\,dS=0,$$ as before. In this case, of course, it is still somewhat easier to compute the line integral, avoiding $\nabla\times{\bf F}$ entirely. $\square$

Example 16.8.5 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 $\ds\int_C {\bf F}\cdot d{\bf r}$ 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. $\square$

**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 $$\eqalign{ \dint{D} \nabla\times{\bf F}\cdot d{\bf S}&= \dint{E} \langle R_y-Q_z,P_z-R_x,Q_x-P_y\rangle\cdot \langle -g_x,-g_y,1\rangle\,dA\cr &=\dint{E} -R_yg_x+Q_zg_x-P_zg_y+R_xg_y+Q_x-P_y\,dA.\cr}$$ 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 $$\int_{\partial E} \left(P+R{\partial z\over\partial x}\right)\,dx+ \left(Q+R{\partial z\over\partial y}\right)\,dy= \dint{E} {\partial\over \partial x}\left(Q+R{\partial z\over\partial y}\right) -{\partial\over \partial y}\left(P+R{\partial z\over\partial x}\right)\,dA.$$ Now we can use the chain rule again to evaluate the derivatives inside this integral, and it becomes $$\eqalign{ \dint{E} &Q_x+Q_zg_x+R_xg_y+R_zg_xg_y+Rg_{yx}- \left(P_y+P_zg_y+R_yg_x+R_zg_yg_x+Rg_{xy}\right)\,dA\cr &=\dint{E} Q_x+Q_zg_x+R_xg_y-P_y-P_zg_y-R_yg_x\,dA,\cr }$$ which is the same as the expression we obtained for the surface integral. $\qed$

## Exercises 16.8

Here is a Sage cell if you'd like to use it.

**Ex 16.8.1**
Let ${\bf F}=\langle z,x,y\rangle$.
The plane $z=2x+2y-1$ and the paraboloid $z=x^2+y^2$ intersect in a
closed curve. Stokes's Theorem implies that
$$\dint{D_1} (\nabla\times{\bf F})\cdot {\bf N}\,dS=
\oint_C {\bf F}\cdot d{\bf r}=
\dint{D_2} (\nabla\times{\bf F})\cdot {\bf N}\,dS,
$$
where the line integral is computed over the intersection $C$ of the plane
and the paraboloid, and the two surface integrals are computed over
the portions of the two surfaces that have boundary $C$ (provided, of
course, that the orientations all match). Compute all three integrals.
(answer)

**Ex 16.8.2**
Let $D$ be the portion of $z=1-x^2-y^2$ above the $x$-$y$
plane, oriented up, and let ${\bf F}=\langle
xy^2,-x^2y,xyz\rangle$. Compute $\ds\dint{D} (\nabla\times{\bf
F})\cdot {\bf N}\,dS$.
(answer)

**Ex 16.8.3**
Let $D$ be the portion of $z=2x+5y$ inside $x^2+y^2=1$,
oriented up, and
let ${\bf F}=\langle y,z,-x\rangle$. Compute
$\ds\int_{\partial D} {\bf F}\cdot d{\bf r}$.
(answer)

**Ex 16.8.4**
Compute $\ds\oint_C x^2z\,dx + 3x\,dy - y^3\,dz$, where $C$
is the unit circle $\ds x^2+y^2=1$ oriented counter-clockwise.
(answer)

**Ex 16.8.5**
Let $D$ be the portion of $z=px+qy+r$ over a region in the
$x$-$y$ plane that has area $A$, oriented up, and
let ${\bf F}=\langle ax+by+cz,ax+by+cz,ax+by+cz\rangle$. Compute
$\ds\int_{\partial D} {\bf F}\cdot d{\bf r}$.
(answer)

**Ex 16.8.6**
Let $D$ be any surface and
let ${\bf F}=\langle P(x),Q(y),R(z)\rangle$ ($P$ depends only on $x$,
$Q$ only on $y$, and $R$ only on $z$). Show that
$\ds\int_{\partial D} {\bf F}\cdot d{\bf r}=0$.

**Ex 16.8.7**
Show that $\ds\int_C f\nabla g+g\nabla f\cdot d{\bf r}=0$, where
$\bf r$ describes a closed curve $C$ to which Stokes's Theorem
applies. (See theorems 12.4.2 and
16.5.2.)