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# 16.7: Stokes' Theorem

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In this section we see the generalization of a familiar theorem, Green’s Theorem. Just as before we are interested in an equality that allows us to go between the integral on a closed curve to the double integral of a surface. Some important definitions to know before proceeding are: simple closed curve, divergence, flux, curl, and normal vector. Knowing how to calculate the determinant of 2x2 and 3x3 matrices will also help deepen your understanding of divergence and curl.

## Theoretical Discussion

Curl: Let

$\mathbf{F} = M(x,y,z)\hat{i} + N(x,y,z)\hat{j} + P(x,y,z)\hat{k} \nonumber$

and

$\nabla = \hat{i} \dfrac{\partial }{\partial x} + \hat{j} \dfrac{\partial }{\partial y} + \hat{k} \dfrac{\partial }{\partial z} \nonumber$

then the curl of $$\mathbf{F}$$ is simply the determinant of the 3 x 3 matrix $$\nabla \times \mathbf{F}$$. There are many ways to take the determinant, but the following is an example of cofactor expansion.

\begin{align*} \nabla \times \mathbf{F} &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\[4pt] \dfrac{\partial }{\partial x} & \dfrac{\partial }{\partial y} & \dfrac{\partial }{\partial z}\ \\ M & N & P \end{vmatrix} \\ &= \hat{i} \begin{vmatrix} \dfrac{\partial }{\partial y} & \dfrac{\partial }{\partial z}\ \\ N & P \end{vmatrix} - \hat{j} \begin{vmatrix} \dfrac{\partial }{\partial x} & \dfrac{\partial }{\partial z}\ \\[4pt] M & P \end{vmatrix} + \hat{k} \begin{vmatrix} \dfrac{\partial }{\partial x} & \dfrac{\partial }{\partial y}\ \\ M & N \end{vmatrix} \\[4pt] &= \hat{i}\left(\dfrac{\partial P}{\partial y} - \dfrac{\partial N}{\partial z}\right) - \hat{j}\left(\dfrac{\partial P}{\partial x} - \dfrac{\partial M}{\partial z}\right) + \hat{k}\left(\dfrac{\partial N}{\partial x} - \dfrac{\partial M}{\partial y}\right) \\[4pt] &= \hat{i}\left(\dfrac{\partial P}{\partial y} - \dfrac{\partial N}{\partial z}\right) + \hat{j}\left(\dfrac{\partial M}{\partial z} - \dfrac{\partial P}{\partial x}\right) + \hat{k}\left(\dfrac{\partial N}{\partial x} - \dfrac{\partial M}{\partial y}\right) \\[4pt] &= curl \ \mathbf{F} \end{align*} \nonumber

Stokes' Theorem

Let $$\mathbf{n}$$ be a normal vector (orthogonal, perpendicular) to the surface S that has the vector field $$\mathbf{F}$$, then the simple closed curve C is defined in the counterclockwise direction around $$\mathbf{n}$$. The circulation on C equals surface integral of the curl of $$\mathbf{F} = \nabla \times \mathbf{F}$$ dotted with $$\mathbf{n}$$.

$\oint _C \mathbf{F} \cdot d\mathbf{r} = \iint_{S} \nabla \times \mathbf{F \cdot n} \ d\sigma \nonumber$

This theorem fails when a function, vector field, or derivative is not continuous.

## Green's Theorem out of Stokes

If the counterclockwise circulation C is only in x-y plane, and it defines a region, call it R, with the vector field $$\mathbf{F }$$ then the z direction is normal to the plane. Thus

\begin{align*} \oint _C \mathbf{F} \cdot d\mathbf{r} &= \iint_{S} \nabla \times \mathbf{F \cdot n} \ d\sigma \\ & = \iint_{R} \nabla \times \mathbf{F \cdot k} \ dx \ dy \\ & = \iint_{R} \dfrac{\partial N}{\partial x} - \dfrac{\partial M}{\partial y}\ dx \ dy \end{align*}\nonumber

As a note,

$\dfrac{\partial N}{\partial x} - \dfrac{\partial M}{\partial y}\nonumber$

is the determinant of the 2x2 matrix

$\begin{vmatrix} \dfrac{\partial }{\partial x} & \dfrac{\partial }{\partial y} \\ M & N \end{vmatrix}.\nonumber$

Example $$\PageIndex{1}$$

Evaluate the equation for Stokes' Theorem for the hemisphere $$S: x^2+y^2+z^2=9, z \geq 0$$, its bounding circle $$C:x^2+y^2=0, z=0$$ and the field $$\textbf{F}=y\hat{\textbf{i}}-x\hat{\textbf{j}}$$. Hints: Remember that a simple way to parameterize a circle is if $$x^2+y^2=r^2$$ then $$r (\theta) = r \cos \theta + r \sin \theta$$ for $$\theta \in [0, 2 \pi]$$. Also, try drawing a picture of the hemisphere and its bounding circle to understand the theory behind the problem. Should know how to normalize a vector and what $$|\nabla f|$$ means. Find the counterclockwise circulation by using the left-hand side of Stokes' Theorem, then find the curl integral by using the right-hand side of Stokes' Theorem and compare your results.

Solution

The hemisphere looks much like the image below, with the circumference of the pink bottom being the bounding circle $$C$$ in the $$xy$$ - plane . We can calculate the counterclockwise circulation around $$C$$ (viewed from above) using the parametrization $$r (\theta ) = ( 3 \cos \theta ) \hat{\textbf{i}} + (3 \sin \theta ) \hat{\textbf{j}}, 0 \leq \theta \leq 2 \pi$$:

$d\textbf{r} = (-3 \sin \theta d \theta ) \hat{\textbf{i}} + (3 \cos \theta d\theta ) \hat{\textbf{j}} \nonumber$

$\textbf{F} = y\hat{\textbf{i}} - x\hat{\textbf{j}} = (3 \sin \theta ) \hat{\textbf{i}} - (3 \cos \theta )\hat{\textbf{j}}\nonumber$

$\textbf{F} \cdot d \textbf{r} = -9 \sin^2 \theta d \theta - 9 \cos^2 \theta d \theta = -9 d\theta \nonumber$

$\oint_C \textbf{F} \cdot d \textbf{r} = \int_0^2\pi -9 d\theta = -18 \theta. \nonumber$

This is the evaluated left-hand side of Stokes' Theorem. Now we want to show that the right-hand side is equal by evaluting the curl integral.

For the curl integral of $$\textbf{F}$$, we have

$\nabla \times \textbf{F} = \left( \dfrac{\partial P}{\partial y} - \dfrac{\partial N}{\partial z} \right) \hat{\textbf{i}} + \left( \dfrac{\partial M}{\partial z} - \dfrac{\partial P}{\partial x} \right) \hat{\textbf{j}}+\left( \dfrac{\partial N}{\partial x} - \dfrac{\partial M}{\partial y} \right) \hat{\textbf{k}} \nonumber$

from taking the determinant of the 3X3 matrix of the curl (explained in theoretical discussion). If we look at that 3X3 matrix:

$\begin{vmatrix} \hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\ \dfrac{\partial }{\partial x} & \dfrac{\partial }{\partial y} & \dfrac{\partial }{\partial z} \\ y & -x & 0 \end{vmatrix} \nonumber$

evaluating the curl, we see that

$\nabla \times \textbf{F} = (0-0)\hat{\textbf{i}}+(0-0)\hat{\textbf{j}}+(-1-1)\hat{\textbf{k}} = -2\hat{\textbf{k}} .\nonumber$

Our outer unit normal vector will be

\begin{align*} \textbf{n} &= \dfrac{ x \hat{\textbf{i}} + y \hat{\textbf{j}} + z \hat{\textbf{k}}}{|x\hat{\textbf{i}} + y \hat{\textbf{j}} + z\hat{\textbf{k}}|} \\ &=\dfrac{x\hat{\textbf{i}} + y \hat{\textbf{j}} + z\hat{\textbf{k}}}{\sqrt{x^2+y^2+z^2}} \\ &=\dfrac{x\hat{\textbf{i}}+y\hat{\textbf{j}}+z\hat{\textbf{k}}}{\sqrt{9 \cos^2 \theta + 9 \sin^2 \theta}} \\ &= \dfrac{x\hat{\textbf{i}}+y\hat{\textbf{j}}+z\hat{\textbf{k}}}{3}. \end{align*}\nonumber

Then

$d \sigma = \dfrac{|\nabla f |}{|\nabla f \cdot \textbf{k}|} dA = \dfrac{3}{z} dA.\nonumber$

Finally, we can put everything together to find that:

$\nabla \times \textbf{F} \cdot \textbf{n} d \sigma = - \dfrac{2z}{3} \dfrac{3}{z} dA = -2dA \nonumber$

and

$\int \int_S \nabla \times \textbf{F} \cdot \textbf{n} d \sigma = \int \int_{x^2+y^2 \leq 9} -2dA=-18 \pi \nonumber$

and we see that the circulation around the circle equals the integral of the curl over the hemisphere, as it should.