6.7 Stokes’ theorem

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

which equals ${\displaystyle \underset{S}{\iint }\text{curl}\text{F}·d\text{S}.}$

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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 F is conservative. If F is conservative, the curl of F is zero, so ${\displaystyle \underset{S}{\iint }\text{curl}\text{F}·d\text{S}=0.}$ Since the boundary of S is a closed curve, ${\displaystyle {\int }_C\text{F}·d\text{r}}$ is also zero.

Verifying stokes’ theorem for a specific case

Verify that Stokes’ theorem is true for vector field $\text{F}\left(x,y\right)=⟨\text{−}z,x,0⟩$ and surface S , where S is the hemisphere, oriented outward, with parameterization

$\text{r}(\varphi ,\theta )=⟨\text{sin}\varphi \text{cos}\theta ,\text{sin}\varphi \text{sin}\theta ,\text{cos}\varphi ⟩,0\le \theta \le \pi ,0\le \varphi \le \pi$ as shown in the following figure.

Let C be the boundary of S. Note that C is a circle of radius 1, centered at the origin, sitting in plane $y=0.$ This circle has parameterization $⟨\text{cos}t,0,\text{sin}t⟩,0\le t\le 2\pi .$ By [link] ,

$\begin{array}{cc}\hfill {\displaystyle {\int }_C\text{F}·d\text{r}}& ={\displaystyle {\int }_0^{2\pi }⟨\text{−}\text{sin}t,\text{cos}t,0⟩·⟨\text{−}\text{sin}t,0,\text{cos}t⟩}dt\hfill \\ & ={\displaystyle {\int }_0^{2\pi }{\text{sin}}^{2}t}dt=\pi .\hfill \end{array}$

By [link] ,

$\begin{array}{cc}\hfill {\displaystyle {\iint }_S\text{curl}\text{F}·d\text{S}}& ={\displaystyle {\iint }_D\text{curl}\text{F}\left(\text{r}\left(\varphi ,\theta \right)\right)}·\left({\text{t}}_{\varphi }\times {\text{t}}_{\theta }\right)dA\hfill \\ & ={\displaystyle {\iint }_D⟨0,\mathrm{-1},1⟩·⟨\text{cos}\theta {\text{sin}}^2\varphi ,\text{sin}\theta {\text{sin}}^2\varphi ,\text{sin}\varphi \text{cos}\varphi ⟩dA}\hfill \\ & ={\displaystyle {\int }_0^{\pi }{\displaystyle {\int }_0^{\pi }\left(\text{sin}\varphi \text{cos}\varphi -\text{sin}\theta {\text{sin}}^2\varphi \right)}}d\varphi d\theta \hfill \\ & =\frac{\pi }{2}{\displaystyle {\int }_0^{\pi }\text{sin}\theta d\theta }=\pi .\hfill \end{array}$

Therefore, we have verified Stokes’ theorem for this example.

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Applying stokes’ theorem

Stokes’ theorem translates between the flux integral of surface S to a line integral around the boundary of S . Therefore, the theorem allows us to compute surface integrals or line integrals that would ordinarily be quite difficult by translating the line integral into a surface integral or vice versa. We now study some examples of each kind of translation.

Calculating a surface integral

Calculate surface integral ${\displaystyle {\iint }_S\text{curl}\text{F}·d\text{S}},$ where S is the surface, oriented outward, in [link] and $\text{F}=⟨z,2xy,x+y⟩.$

Note that to calculate ${\displaystyle {\iint }_S\text{curl}\text{F}·d\text{S}}$ without using Stokes’ theorem, we would need to use [link] . Use of this equation requires a parameterization of S . Surface S is complicated enough that it would be extremely difficult to find a parameterization. Therefore, the methods we have learned in previous sections are not useful for this problem. Instead, we use Stokes’ theorem, noting that the boundary C of the surface is merely a single circle with radius 1.

The curl of F is $⟨1,1,2y⟩.$ By Stokes’ theorem,

${\displaystyle {\iint }_S\text{curl}\text{F}·d\text{S}}={\displaystyle {\int }_C\text{F}·d\text{r},}$

where C has parameterization $⟨\text{cos}t,\text{sin}t,1⟩,0\le t<2\pi .$ By [link] ,

$\begin{array}{cc}\hfill {\displaystyle {\iint }_S\text{curl}\text{F}}·d\text{S}& ={\displaystyle {\int }_C\text{F}·d\text{r}}\hfill \\ & ={\displaystyle {\int }_0^{2\pi }⟨1,2\text{sin}t\text{cos}t,\text{cos}t+\text{sin}t⟩·⟨\text{−}\text{sin}t,\text{cos}t,0⟩dt}\hfill \\ & ={\displaystyle {\int }_0^{2\pi }\left(\text{−}\text{sin}t+2\text{sin}t{\text{cos}}^{2}t\right)dt}\hfill \\ & ={\left[\text{cos}t-\frac{2{\text{cos}}^{3}t}{3}\right]}_0^{2\pi }\hfill \\ & =\text{cos}(2\pi )-\frac{2{\text{cos}}^3(2\pi )}{3}-\left(\text{cos}(0)-\frac{2{\text{cos}}^3(0)}{3}\right)\hfill \\ & =0.\hfill \end{array}$

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