# Stokes' Theorem

- Page ID
- 21061

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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.

## Example \(\PageIndex{2}\): Calculating a Surface Integral

Calculate surface integral

\[\iint_S curl \, \vecs{F} \cdot d\vecs S, \nonumber \]

where \(S\) is the surface, oriented outward, in Figure \(\PageIndex{6}\) and \(\vecs{F} = \langle z,\, 2xy, \, x + y \rangle\).

## Solution

Note that to calculate

\[ \iint_S curl \, \vecs F \cdot d\vecs S \nonumber \]

without using Stokes’ theorem, we would need the equation for scalar surface integrals. 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 \(\vecs{F}\) is \(\langle 1,1,2y \rangle\). By Stokes’ theorem,

\[\iint_S curl \, \vecs F \cdot d\vecs S = \int_C \vecs F \cdot d\vecs r, \nonumber \]

where \(C\) has parameterization \(\langle \cos t, \, \sin t, \, 1 \rangle, 0 \leq t \leq 2\pi\). By the equation for vector line integrals,

\[ \begin{align*} \iint_S curl \, F \cdot d\vecs S &= \int_C \vecs{F} \cdot d \vecs{r} \\[4pt] &= \int_0^2 \langle 1, \, \sin t \, \cos t, \, \cos t + \sin t \rangle \cdot \langle - \sin t, \, \cos t, \, 0 \rangle \, dt \\[4pt] &= \int_0^{2\pi} ( - \sin t + 2 \, \sin t \, \cos^2 t ) \, dt \\[4pt] &= \left[ \cos t - \dfrac{2 \, \cos^3 t}{3} \right]_0^{2\pi} \\[4pt] &= \cos (2\pi) - \dfrac{2 \, \cos^3 (2\pi)}{3} - \left(\cos (0) - \dfrac{2 \, \cos^3 (0)}{3} \right) \\[4pt] &= 0. \end{align*} \nonumber \]

An amazing consequence of Stokes’ theorem is that if \(S'\) is any other smooth surface with boundary \(C\) and the same orientation as \(S\), then \[\iint_S curl \, \vecs F \cdot d\vecs S = \int_C \vecs F \cdot d\vecs r = 0 \nonumber \] because Stokes’ theorem says the surface integral depends on the line integral around the boundary only.

In Example \(\PageIndex{2}\), we calculated a surface integral simply by using information about the boundary of the surface. In general, let \(S_1\) and \(S_2\) be smooth surfaces with the same boundary \(C\) and the same orientation. By Stokes’ theorem,

\[\iint_{S_1} curl \, \vecs{F} \cdot d\vecs{S} = \int_C \vecs{F} \cdot d\vecs{r} = \iint_{S_2} curl \, \vecs{F} \cdot d\vecs{S}. \label{20} \]

Therefore, if

\[\iint_{S_1} curl \, \vecs{F} \cdot d\vecs{S} \nonumber \]

is difficult to calculate but

\[\iint_{S_2} curl \, \vecs{F} \cdot d\vecs S \nonumber \]

is easy to calculate, Stokes’ theorem allows us to calculate the easier surface integral. In Example \(\PageIndex{2}\), we could have calculated

\[\iint_S curl \, \vecs{F} \cdot d \vecs{S} \nonumber \]

by calculating

\[\iint_{S'} curl \, \vecs{F} \cdot d\vecs{S}, \nonumber \]

where \(\vecs{S}'\) is the disk enclosed by boundary curve \(C\) (a much more simple surface with which to work).

Equation \ref{20} shows that flux integrals of curl vector fields are surface independent in the same way that line integrals of gradient fields are path independent. Recall that if \(\vecs{F}\) is a two-dimensional conservative vector field defined on a simply connected domain, \(f\) is a

potential functionfor \(\vecs{F}\), and \(C\) is a curve in the domain of \(\vecs{F}\), then\[\int_C \vecs{F} \cdot d\vecs{r} \nonumber \]

depends only on the endpoints of \(C\). Therefore if \(C'\) is any other curve with the same starting point and endpoint as \(C\) (that is, \(C'\) has the same orientation as \(C\)), then

\[\int_C \vecs{F} \cdot d\vecs{r} = \int_{C'} \vecs{F} \cdot d\vecs{r} \nonumber \]

In other words, the value of the integral depends on the boundary of the path only; it does not really depend on the path itself.

Analogously, suppose that \(S\) and \(S'\) are surfaces with the same boundary and same orientation, and suppose that \(\vecs{G}\) is a three-dimensional vector field that can be written as the curl of another vector field \(\vecs{F}\) (so that \(\vecs{F}\) is like a “potential field” of \(\vecs{G}\)). By Equation \ref{20},

\[ \begin{align*} \iint_S \vecs G \cdot d\vecs S = \iint_S curl \, \vecs F \cdot d\vecs S = \int_C \vecs F \cdot d\vecs r = \iint_{S'} curl \, \vecs F \cdot d\vecs S = \iint_{S'} \vecs G \cdot d\vecs S.\end{align*} \nonumber \]

Therefore, the flux integral of \(\vecs{G}\) does not depend on the surface, only on the boundary of the surface. Flux integrals of vector fields that can be written as the curl of a vector field are surface independent in the same way that line integrals of vector fields that can be written as the gradient of a scalar function are path independent.

## Exercise \(\PageIndex{1}\)

Use Stokes’ theorem to calculate surface integral \[\iint_S curl \, \vecs{F} \cdot d\vecs{S}, \nonumber \] where \(\vecs{F} = \langle x,y,z \rangle\) and \(S\) is the surface as shown in the following figure.

HintParameterize the boundary of \(S\) and translate to a line integral.

Answer\(-\pi\)