Stokes’ Theorem for Line Integrals

$\newenvironment {prompt}{}{} \newcommand {\DeclareMathOperator }[2]{\@OldDeclareMathOperator {##1}{##2}\immediate \write \myfile {\unexpanded {\DeclareMathOperator }{\unexpanded {##1}}{\unexpanded {##2}}}} \newcommand {\reserved@a }[2]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\reserved@a }[1]{} \newcommand {\reserved@a }[2]{} \newcommand {\AppendGraphicsExtensions }[0]{\@ifundefined {Gin@extensions}{\let \Gin@extensions \@empty }{}\@ifstar {\grfext@Append \grfext@Check }{\grfext@Append \grfext@@Add }} \newcommand {\PrependGraphicsExtensions }[0]{\@ifundefined {Gin@extensions}{\let \Gin@extensions \@empty }{}\@ifstar {\grfext@Prepend \grfext@Check }{\grfext@Prepend \grfext@@Add }} \newcommand {\RemoveGraphicsExtensions }[1]{\@ifundefined {Gin@extensions}{\def \Gin@extensions {}}{\edef \grfext@tmp {\zap@space ##1 \@empty }\@for \grfext@ext :=\grfext@tmp \do {\def \grfext@next {\let \grfext@tmp \Gin@extensions \@expandtwoargs \@removeelement \grfext@ext \Gin@extensions \Gin@extensions \ifx \grfext@tmp \Gin@extensions \let \grfext@next \relax \fi \grfext@next }\grfext@next }}\grfext@Print \RemoveGraphicsExtensions } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} \newcommand {\epstopdfcall }[1]{\ifETE@InsideSetfile \expandafter \@firstoftwo \else \expandafter \@secondoftwo \fi {‘##1}{\Gin@base \Gin@ext }} \newcommand {\epstopdfDeclareGraphicsRule }[4]{\ifx \\##4\\\@PackageError {epstopdf-base}{Conversion command is missing}\@ehc \else \begingroup \@ifundefined {Gin@rule@##1}{}{\@PackageInfo {epstopdf-base}{Redefining graphics rule for ‘##1’}}\endgroup \@namedef {Gin@rule@##1}####1{{##2}{##3}{\epstopdfcall {##4}}}\fi } \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\GetTitleString }[0]{\ifGTS@expand \expandafter \GetTitleStringExpand \else \expandafter \GetTitleStringNonExpand \fi } \newcommand {\GetTitleStringExpand }[1]{\def \GetTitleStringResult {##1}\begingroup \GTS@DisablePredefinedCmds \GTS@DisableHook \edef \x {\endgroup \noexpand \def \noexpand \GetTitleStringResult {\GetTitleStringResult }}\x } \newcommand {\GetTitleStringNonExpand }[1]{\def \GetTitleStringResult {##1}\global \let \GTS@GlobalString \GetTitleStringResult \begingroup \GTS@RemoveLeft \GTS@RemoveRight \endgroup \let \GetTitleStringResult \GTS@GlobalString } \newcommand {\GTS@DisableHook }[0]{} \newcommand {\label@hook }[0]{} \newcommand {\@currentlabelname }[0]{} \newcommand {\reserved@a }[0]{} \newcommand {\@safe@activestrue }[0]{} \newcommand {\@safe@activesfalse }[0]{} \newcommand {\nameref }[0]{\@ifstar \T@nameref \T@nameref } \newcommand {\Sectionformat }[2]{##1} \newcommand {\vnameref }[1]{\unskip ~\nameref {##1}\@vpageref [\unskip ]{##1}} \newcommand {\ref }[0]{\@ifstar \@refstar \T@ref } \newcommand {\pageref }[0]{\@ifstar \@pagerefstar \T@pageref } \newcommand {\nameref }[0]{\@ifstar \@namerefstar \T@nameref } \newcommand {\reserved@a }[2]{} \newcommand {\reserved@a }[0]{\AtBeginDocument } \newcommand {\reserved@a }[1]{} \newcommand {\reserved@a }[2]{}$

Objectives:

1.: Know what the boundary of a surface is.
2.: Given an oriented surface $S$ , understand the positive orientation of the boundary $\partial S$ .
3.: Know the statement of Stokes’ theorem.
4.: Be able to use Stokes’ theorem to compute line integrals.

Recap Video

Here is a video highlights the main points of the section.

Two Other Videos

Here are two more videos:

This one gives a brief summary of the boundary orientation.
_
This second video gives the steps for using Stokes’ theorem to compute a line integral.
_

Example Video

The following video shows a worked example of using Stokes’ theorem to compute a line integral.

If $C$ is the intersection of $y^2+z^2=1$ with the plane $x+y+z=1$ , oriented counterclockwise when viewed from above, and $\textbf {F} = \left \langle x^3+y+z,yz+y,z^5 \right \rangle ,$ evaluate $\displaystyle \int _C \textbf {F} \cdot d\textbf {r}$ .

Problems

Boundary and Boundary Orientation

Consider the surface $S$ which is the portion of $z = 5 - x^2-y^2$ with $1 \leq z \leq 5$ . Does the surface $S$ have a nonempty boundary?

Yes No

The boundary of $S$ is the circle $x^2+y^2 = \answer {4}$ in the plane $z = \answer {1}$ .

The vertex of this paraboloid is at $z=5$ , and the boundary happens when $z=1$ . Plugging $z=1$ into the paraboloid gives $1 = 5-x^2-y^2$ , so $x^2+y^2 =4$ .

Consider the surface $S$ which is the portion of $x^2+y^2+z^2 = 1$ with $x \geq 0$ coupled with the disc $y^2+z^2 \leq 1$ in the plane $x=0$ . Does $S$ have a nonempty boundary?

Yes No

Notice that the disc $y^2+z^2 \leq 1$ in the plane $x=0$ closes off the surface, removing the one exposed edge to the original hemisphere.

If $S$ is an oriented surface, then the positive orientation for the boundary $C$ is as follows: stand on the boundary with your head in the direction of the normal. The positive direction is the one you should walk in so that the surface is to your left.

Let $S$ be the portion of $z = 5-x^2-y^2$ with $1 \leq z \leq 5$ . If $S$ has an outward normal orientation, then the positive orientation of the boundary $C$ is (choose one) clockwisecounterclockwise when viewed from above.

Here is how you want to imagine standing on the boundary.

Let $S$ be the portion of the plane $x+2y+3z=6$ in the first octant, oriented with downward normals. The boundary $\partial S$ is the triangle with vertices $(\answer {6},0,0)$ , $(0,\answer {3},0)$ , and $(0,0,\answer {2})$ . The positive orientation of $C$ is clockwisecounterclockwise when viewed from above.

Here is how you want to imagine standing on the boundary.

Stokes’ Theorem

Let $\textbf {F} = \left \langle e^{y-z},0,0 \right \rangle$ , and let $C$ be the square with vertices $(1,0,1)$ , $(1,1,1)$ , $(0,1,1)$ , and $(0,0,1)$ , oriented counterclockwise when viewed from above. Evaluate $\displaystyle \int _C \textbf {F} \cdot d\textbf {r}$ .

First, we should check whether $\textbf {F}$ is conservative. We compute $\text {curl}(\textbf {F}) = \left \langle \answer {0},\answer {-e^{y-z}},\answer {-e^{y-z}}\right \rangle .$ Thus, $\textbf {F}$ isis not conservative. However, the curve is closed, so we can try and use Stokes’ theorem. To do this, we have three things to examine:

Pick a surface $S$ which has a boundary of $C$ .
Make sure $\textbf {F}$ is has continuous partials on $S$ .
Decide on which orientation $S$ should have so that $C$ has positive orientation.

One possible surface is the portion of the plane $z =\answer {1}$ over the square $[0,1] \times [0,1]$ in the $xy$ -plane. The field $\textbf {F}$ has continuous partials on $S$ . Finally, $S$ should be oriented with normals so that $C$ has positive orientation.

Now, we compute $\displaystyle \iint _S \text {curl}(\textbf {F}) \cdot d\textbf {S}$ . We can parametrize the surface as $\textbf {r}(x,y) = \left \langle x,y,1 \right \rangle$ where $0 \leq x \leq 1$ and $0 \leq y \leq 1$ . Then $\textbf {r}_x = \left \langle \answer {1},\answer {0},\answer {0} \right \rangle ,$ $\textbf {r}_y = \left \langle \answer {0},\answer {1},\answer {0} \right \rangle ,$ and $\textbf {r}_x \times \textbf {r}_y = \left \langle \answer {0},\answer {0},\answer {1} \right \rangle .$ This is the normal vector.

We have $\text {curl}(\textbf {F})(\textbf {r}(x,y)) = \left \langle \answer {0},\answer {-e^{y-1}},\answer {-e^{y-1}} \right \rangle .$

$\begin{eqnarray*} \iint_S \text{curl}(\textbf{F}) \cdot d\textbf{S} &=& \int_0^1 \int_0^1 \answer{-e^{y-1}} dy \, dx \\ &=& \answer{e^{-1}-1}. \end{eqnarray*}$

Let $\textbf {F} = \left \langle -y^2,x,z^2 \right \rangle$ , and let $C$ be the intersection of $x^2+y^2=1$ and the plane $y+z=2$ , oriented counterclockwise when viewed from above. Evaluate $\displaystyle \int _C \textbf {F} \cdot d\textbf {r}$ .

Again, let’s check whether $\textbf {F}$ is conservative. We compute $\text {curl}(\textbf {F}) = \left \langle \answer {0},\answer {0},\answer {1+2y} \right \rangle .$ Therefore, $\textbf {F}$ isis not conservative.

However, the curve $C$ is closed so we can try and use Stokes’ theorem. Again, we need to pick a surface $S$ with $C$ as the boundary. In this case, a convenient surface $S$ to pick is the plane $y+z=2$ , and $x^2+y^2=1$ will give us the bounds. This is shown in the image below.

The plane should be oriented with upwarddownward normals.

Stokes’ theorem now says we can compute $\displaystyle \iint _S \text {curl}(\textbf {F}) \cdot d\textbf {S}$ . Parametrize the surface as $\textbf {r}(x,y) = \left \langle x,y,\answer {2-y} \right \rangle ,$ where the domain $D$ is $x^2+y^2 \leq \answer {1}$ . Then $\textbf {r}_x = \left \langle \answer {1},\answer {0},\answer {0} \right \rangle ,$ $\textbf {r}_y = \left \langle \answer {0},\answer {1},\answer {-1} \right \rangle ,$ and $\textbf {r}_x \times \textbf {r}_y = \left \langle \answer {0},\answer {1},\answer {1} \right \rangle .$ This normal is upward, which is the direction we decided on earlier. We compute $(\text {curl}(\textbf {F}))(\textbf {r}(x,y)) \cdot (\textbf {r}_x \times \textbf {r}_y) = \answer {1+2y}.$ Then

$\begin{eqnarray*} \iint_S \text{curl}(\textbf{F}) \cdot d\textbf{S} &=& \iint_D \answer{1+2y} dA \\ &=& \int_0^{2\pi} \int_{\answer{0}}^{\answer{1}} (1+2r\sin(\theta))rdr\,d\theta \\ &=& \answer{\pi}. \end{eqnarray*}$

If $\textbf {F} = \left \langle 2y,2z,2x \right \rangle$ , and $C$ is the triangle with vertices $(1,0,0)$ , $(0,1,0)$ , and $(0,0,1)$ , oriented counterclockwise when viewed from above, then $\displaystyle \int _C \textbf {F} \cdot d\textbf {r} = \answer {-3}$

Here $\text {curl}(\textbf {F}) = \left \langle -2,-2,-2 \right \rangle$ , and we can let $S$ be the surface $x+y+z=1$ oriented upward. Parametrize the surface as $\textbf {r}(x,y) = \left \langle x,y,1-x-y \right \rangle$ . Then $\textbf {r}_x \times \textbf {r}_y = \left \langle 1,1,1 \right \rangle$ , which is the correct normal. Notice $\text {curl}(\textbf {F}) \cdot (\textbf {r}_x \times \textbf {r}_y) = -6$ , so the flux is $\iint _S \textbf {F} \cdot d\textbf {S} = \int _0^1 \int _0^{1-x} -6 dy\, dx = -3.$

Let $\textbf {F} = \left \langle 2xy+x^2,x+y^{10},y+z+\sin (z^2) \right \rangle$ . Let $S$ be the portion of $z =1-x^2-y^2$ with $z \geq 0$ , oriented with outward normals, and let $\partial S$ be the boundary of $S$ with the positive orientation. Then $\displaystyle \int _{\partial S} \textbf {F} \cdot d\textbf {r} = \answer {\pi }$

The boundary of $S$ is the circle $x^2+y^2=1$ in the plane $z=0$ , oriented counterclockwise when viewed from above. We also have $\text {curl}(\textbf {F}) = \left \langle 1,0,1-2x \right \rangle$ . Now, to use Stokes’ theorem, we don’t need to use $S$ . All we need to do is pick a surface that has the circle $x^2+y^2=1$ , $z=0$ as its boundary. So we can actually take the disc $x^2+y^2\leq 1$ in the plane $z=0$ . (Of course, you could also take $S$ if you’d like to.) Call this surface $S'$ . Parametrize $S'$ as $\textbf {r}(x,y) = \left \langle x,y,0 \right \rangle$ , which has domain $x^2+y^2\leq 1$ (call this $D$ ). Then $\textbf {r}_x \times \textbf {r}_y = \left \langle 0,0,1 \right \rangle$ , which is the correct normal. We have $\text {curl}(\textbf {F}) \cdot (\textbf {r}_x \times \textbf {r}_y) = 1-2x,$ so the flux is $\iint _D (1-2x)dA = \pi .$

Let $\textbf {F} = \left \langle e^{x^2}+2y^2,\sin (y^2)+2xz,z^8+1 \right \rangle$ , and let $S$ be the intersection of the surface $z = xy$ with the cylinder $x^2+y^2=1$ , oriented counterclockwise when viewed from above. Then $\displaystyle \int _C \textbf {F} \cdot d\textbf {r} = \answer {0}$

The curl is $\text {curl}(\textbf {F}) = \left \langle -2x,0,2z-4y \right \rangle$ . We will take $S$ to be $z = xy$ , where $x^2+y^2=1$ will give the domain. We need $S$ to have the upward normal orientation. Parametrize $S$ as $\textbf {r}(x,y) = \left \langle x,y,xy \right \rangle$ with domain $x^2+y^2 \leq 1$ , call it $D$ . The cross product of the partials is $\textbf {r}_x \times \textbf {r}_y = \left \langle -y,-x,1 \right \rangle ,$ which is the right normal. We find $\text {curl}(\textbf {F})(\textbf {r}(x,y)) = \left \langle -2x,0,2xy-4y \right \rangle .$ Dot this with $\textbf {r}_x \times \textbf {r}_y$ to get $4xy-4y$ . Thus, the flux is $\iint _D (4xy-4y)dA = 0.$