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Mathematics LibreTexts

16.7: Stokes' Theorem

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 \( \mathbf{F} = M(x,y,z)\hat{i} + N(x,y,z)\hat{j} + P(x,y,z)\hat{k} \) and \( \nabla = \hat{i} \frac{\partial }{\partial x} + \hat{j} \frac{\partial }{\partial y} + \hat{k} \frac{\partial }{\partial z} \) then the curl of \(\mathbf{F}\) is simply the determinant of the 3x3 matrix \(\nabla \times \mathbf{F}\). There are many ways to take the determinant, but the following is an example of cofactor expansion.

\[\begin{align} \nabla \times \mathbf{F}  &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\  \frac{\partial }{\partial x}  & \frac{\partial }{\partial y} & \frac{\partial }{\partial z}\ \\ M &  N & P \end{vmatrix}     \\     &= \hat{i} \begin{vmatrix} \frac{\partial }{\partial y} & \frac{\partial }{\partial z}\  \\ N & P \end{vmatrix} - \hat{j} \begin{vmatrix} \frac{\partial }{\partial x} & \frac{\partial }{\partial z}\ \\ M & P \end{vmatrix} + \hat{k} \begin{vmatrix} \frac{\partial }{\partial x} & \frac{\partial }{\partial y}\ \\ M & N \end{vmatrix}     \\     &= \hat{i}(\frac{\partial P}{\partial y} - \frac{\partial N}{\partial z}) - \hat{j}(\frac{\partial P}{\partial x} - \frac{\partial M}{\partial z}) + \hat{k}(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y})    \\       &= \hat{i}(\frac{\partial P}{\partial y} - \frac{\partial N}{\partial z}) + \hat{j}(\frac{\partial M}{\partial z} - \frac{\partial P}{\partial x}) + \hat{k}(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y})     \\      &= curl \ \mathbf{F} \end{align} \]

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 \(\mathbf{n}\).

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

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

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

As a note,

\[\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\]

is the determinant of the 2x2 matrix

\[\begin{vmatrix} \frac{\partial }{\partial x}  & \frac{\partial }{\partial y} \\ M &  N \end{vmatrix}\]

Example 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 \( r (\theta ) = ( 3 \cos \theta ) \hat{\textbf{i}} + (3 \sin \theta ) \hat{\textbf{j}}, 0 \leq \theta \leq 2 \pi \): 

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

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

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

\[ \oint_C \textbf{F} \cdot d \textbf{r} = \int_0^2\pi -9 d\theta = -18 \theta \] 

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

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

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

\[ \begin{vmatrix} \hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ y & -x & 0 \end{vmatrix} \]

Evaluating the curl, we see that

\[ \nabla \times \textbf{F} = (0-0)\hat{\textbf{i}}+(0-0)\hat{\textbf{j}}+(-1-1)\hat{\textbf{k}} = -2\hat{\textbf{k}} \]

Our outer unit normal vector will be

\[ \begin{align} \textbf{n} &= \frac{ x \hat{\textbf{i}} + y \hat{\textbf{j}} + z \hat{\textbf{k}}}{|x\hat{\textbf{i}} + y \hat{\textbf{j}} + z\hat{\textbf{k}}|} \\ &=\frac{x\hat{\textbf{i}} + y \hat{\textbf{j}} + z\hat{\textbf{k}}}{\sqrt{x^2+y^2+z^2}} \\ &=\frac{x\hat{\textbf{i}}+y\hat{\textbf{j}}+z\hat{\textbf{k}}}{\sqrt{9 \cos^2 \theta + 9 \sin^2 \theta}} \\ &= \frac{x\hat{\textbf{i}}+y\hat{\textbf{j}}+z\hat{\textbf{k}}}{3}  \end{align}\]

Then

\[d \sigma = \frac{|\nabla f |}{|\nabla f \cdot \textbf{k}|} dA = \frac{3}{z} dA\]

Finally, we can put everything together to find that:

\[ \nabla \times \textbf{F} \cdot \textbf{n} d \sigma = - \frac{2z}{3} \frac{3}{z} dA = -2dA \]

and 

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

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

Contributors

  • Emily Javan (UCD), Melody Molander (UCD)