Stokes' Theorem

The Essentials

Stokes' Theorem lets you convert a line integral to the surface integral of the curl:

\[ \int_C \vec{F}\cdot d\vec{r}=\iint_S\text{curl}\vec{F}\cdot d\vec{S} \]

When using Stokes' Theorem an integral can be converted from a line integral to a surface integral and from a surface integral to a line integral. The line C has to be a closed line which means that it has to enclose an area. The surface S is that area with the boundaries of the surface being the line C. Therefore, surface integrals of curled functions are surface independent, as long as the surfaces have the same openings. To illustrate this look as Figure 6.84 in the Openstax calculus III book.

Example

Use Stokes' Theorem to evaluate the integral where \( \vec{F}=\langle xy, -z, y\rangle \) and S is the surface defined by the cube with \( 0\leq x\leq1 \), \( 0\leq y\leq1 \), and \( 0\leq z\leq1 \) but missing the side where \( z=1 \):

\[ \iint_S (\text{curl}\vec{F}\cdot\vec{N})dS \]

We will use Stokes' Theorem to convert the surface integral of the curl of a function to a line integral of that function:

\[ \int_C \vec{F}(\vec{r}(t))\cdot\vec{r'}(t)dt \]

To evaluate this integral, we will integrate over the closed line that is the square on the plane \( z=1 \). We will parameterize four different line segments and add four integral together:

\[ \vec{r}_1=\langle t,0,1\rangle\hspace{36pt}\vec{r}_2=\langle t,1,1\rangle\hspace{36pt}\vec{r}_3=\langle 0,t,1\rangle\hspace{36pt}\vec{r}_4=\langle 1,t,1\rangle\hspace{36pt}0\leq t\leq1 \]
\[ \int_0^1\langle 0,-1,0\rangle\cdot\langle 1,0,0\rangle dt + \int_0^1\langle t,-1,1\rangle\cdot\langle 1,0,0\rangle dt + \int_0^1\langle 0,-1,t\rangle\cdot\langle 0,1,0\rangle dt + \int_0^1\langle t,-1,t\rangle\cdot\langle 0,1,0\rangle dt \]
\[ \int_0^1 0 dt + \int_0^1 t dt + \int_0^1 -1 dt + \int_0^1 -1 dt= -\frac{3}{2} \]

Practice

Use Stokes' Theorem to evaluate the integral where \( \vec{F}=\langle y^2,1,x^2\rangle \) and S is the surface defined by the intersection of the cone \( z^2=x^2+y^2 \) and the plane \( z=\frac{1}{2}x+2 \):

\[ \iint_S \text{curl}\vec{F}\cdot d\vec{S} \]

Solution:

0

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