1 Stokes’ theorem

This is a theorem that equates a line integral to a surface integral. For any vector field $\underset{̲}{F}$ and a contour $C$ which bounds an area $S$ ,

$\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left(\underset{̲}{\nabla }\times \underset{̲}{F}\right)\cdot \underset{̲}{dS}={\oint }_C\underset{̲}{F}\cdot \underset{̲}{dr}$

Figure 17:

Notes

1. $\underset{̲}{dS}$ is a vector perpendicular to the surface and $\underset{̲}{dr}$ is the line element along the contour $C$ .
2. Both sides of the equation are scalars.
3. The theorem is often a useful way of calculating a line integral along a contour composed of several distinct parts (e.g. a square or other figure).
4. $\underset{̲}{\nabla }\times \underset{̲}{F}$ is a vector field representing the curl of the vector field $\underset{̲}{F}$ and may, alternatively, be written as curl $\underset{̲}{F}$ .

1.1 Justification of Stokes’ theorem

Imagine that the surface is divided into a set of infinitesimally small rectangles $ABCD$ where the axes are adjusted so that $AB$ and $CD$ lie parallel to the new $x$ -axis i.e. $AB=\delta x$ and $BC$ and $AD$ lie parallel to the new $y$ -axis i.e. $BC=\delta y$ .

Now, ${\oint }_C\underset{̲}{F}\cdot \underset{̲}{dr}$ is calculated.

The contributions along $AB$ , $BC$ , $CD$ and $DA$ are

$\underset{̲}{F}\left(x,y,0\right)\cdot \underset{̲}{\delta x}=F_x\left(x,y,z\right)\delta x$ ,

$\underset{̲}{F}\left(x+\delta x,y,0\right)\cdot \underset{̲}{\delta y}=F_y\left(x+\delta x,y,z\right)\delta y$ ,

$\underset{̲}{F}\left(x,y+\delta y,0\right)\cdot \left(-\underset{̲}{\delta x}\right)=-F_x\left(x,y+\delta y,z\right)\delta x$

$\underset{̲}{F}\left(x,y,0\right)\cdot \left(-\underset{̲}{\delta x}\right)=-F_y\left(x,y,z\right)\delta y$ .

Thus,

as $\underset{̲}{dS}$ is perpendicular to the $x$ - and $y$ - axes.

Thus, for each small rectangle, ${\oint }_C\underset{̲}{F}\cdot \underset{̲}{dr}\approx \left(\underset{̲}{\nabla }\times \underset{̲}{F}\right)\cdot \underset{̲}{dS}$

When the contributions over all the rectangles are summed, the line integrals for the inner parts of the rectangles cancel and all that remains is the line integral around the outside of the shape. The surface integrals sum. Hence, the theorem applies for the area $S$ bounded by the contour $C$ .

While the above does not comprise a formal proof of Stokes’ theorem, it gives an appreciation of where the theorem comes from.

Figure 18:

Example 31

Verify Stokes’ theorem for the vector function $\underset{̲}{F}=y^2\underset{̲}{i}-\left(x+z\right)\underset{̲}{j}+yz\underset{̲}{k}$ and the unit square $0\le x\le 1$ , $0\le y\le 1$ for $z=0$ .

Solution

If $\underset{̲}{F}=y^2\underset{̲}{i}-\left(x+z\right)\underset{̲}{j}+yz\underset{̲}{k}$ then $\underset{̲}{\nabla }\times \underset{̲}{F}=\left(z+1\right)\underset{̲}{i}+\left(-1-2y\right)\underset{̲}{k}=\underset{̲}{i}+\left(-1-2y\right)\underset{̲}{k}$ (as $z=0$ ). Note that $\underset{̲}{dS}=dxdy\underset{̲}{k}$ so that $\left(\underset{̲}{\nabla }\times \underset{̲}{F}\right)\cdot \underset{̲}{dS}=\left(-1-2y\right)dydx$

To evaluate ${\oint }_C\underset{̲}{F}\cdot \underset{̲}{dr}$ , consider it separately on the four sides.

When $y=0$ , $\underset{̲}{F}=-x\underset{̲}{j}$ and $\underset{̲}{dr}=dx\underset{̲}{i}$ so $\underset{̲}{F}\cdot \underset{̲}{dr}=0$ so the contribution to the integral is zero.

When $x=1$ , $\underset{̲}{F}=y^2\underset{̲}{i}-\underset{̲}{j}$ and $\underset{̲}{dr}=dy\underset{̲}{j}$ so $\underset{̲}{F}\cdot \underset{̲}{dr}=-dy$ so the contribution to the integral is

${\int \; <!--nolimits\; -->}_{y=0}^1\left(-dy\right)={\left[-y\right]}_0^1=-1$ .

When $y=1$ , $\underset{̲}{F}=\underset{̲}{i}-x\underset{̲}{j}$ and $\underset{̲}{dr}=-dx\underset{̲}{i}$ so $\underset{̲}{F}\cdot \underset{̲}{dr}=-dx$ so the contribution to the integral is

${\int \; <!--nolimits\; -->}_{x=0}^1\left(-dx\right)={\left[-x\right]}_0^1=-1$ .

When $x=0$ , $\underset{̲}{F}=y^2\underset{̲}{i}$ and $\underset{̲}{dr}=-dy\underset{̲}{j}$ so $\underset{̲}{F}\cdot \underset{̲}{dr}=0$ so the contribution to the integral is zero.

The integral ${\oint }_C\underset{̲}{F}\cdot \underset{̲}{dr}$ is the sum of the contributions i.e. $0-1-1+0=-2$ .

Thus $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left(\underset{̲}{\nabla }\times \underset{̲}{F}\right)\cdot \underset{̲}{dS}={\oint }_C\underset{̲}{F}\cdot \underset{̲}{dr}=-2$

Example 32

Using cylindrical polar coordinates verify Stokes’ theorem for the function $\underset{̲}{F}={\rho }^2\underset{̲}{\stackrel{̂}{\varphi }}$ and the circle $\rho =a$ and the surface $\rho \le a$ . (It is effectively plane-polar coordinates here as this example just considers the plane $z=0$ .)

Solution

Firstly, find ${\oint }_C\underset{̲}{F}\cdot \underset{̲}{dr}$ . This can be done by integrating along the contour $\rho =a$ from $\varphi =0$ to $\varphi =2\pi$ . Here $\underset{̲}{F}=a^2\underset{̲}{\stackrel{̂}{\varphi }}$ (as $\rho =a$ ) and $\underset{̲}{dr}=ad\varphi \underset{̲}{\stackrel{̂}{\varphi }}$ (remembering the scale factor) so $\underset{̲}{F}\cdot \underset{̲}{dr}=a^{3}d\varphi$ and hence

${\oint }_C\underset{̲}{F}\cdot \underset{̲}{dr}={\int \; <!--nolimits\; -->}_0^{2\pi }{a}^{3}d\varphi =2\pi a^3$

As $\underset{̲}{F}={\rho }^2\underset{̲}{\stackrel{̂}{\varphi }}$ , $\underset{̲}{\nabla }\times \underset{̲}{F}=3\rho \underset{̲}{ẑ}$ and $\left(\underset{̲}{\nabla }\times \underset{̲}{F}\right)\cdot \underset{̲}{dS}=3\rho$ as $\underset{̲}{dS}=\underset{̲}{ẑ}$ .

Thus

Hence

${\oint }_C\underset{̲}{F}\cdot \underset{̲}{dr}=\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left(\underset{̲}{\nabla }\times \underset{̲}{F}\right)\cdot \underset{̲}{dS}=2\pi a^3$

Example 33

Find the closed line integral  ${\oint }_C\underset{̲}{F}\cdot \underset{̲}{dr}$  for the vector field $\underset{̲}{F}=y^2\underset{̲}{i}+\left(x^2-z\right)\underset{̲}{j}+2xy\underset{̲}{k}$ and for the contour $ABCDEFGHA$ in Figure 19.

Figure 19:

Solution

To find the line integral directly would require eight line integrals i.e. along $AB$ , $BC$ , $CD$ , $DE$ , $EF$ , $FG$ , $GH$ and $HA$ . It is easier to carry out a surface integral to find $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left(\underset{̲}{\nabla }\times \underset{̲}{F}\right)\cdot \underset{̲}{dS}$ which is equal to the required line integral ${\oint }_C\underset{̲}{F}\cdot \underset{̲}{dr}$ by Stokes’ theorem.

As $\underset{̲}{F}=y^2\underset{̲}{i}+\left(x^2-z\right)\underset{̲}{j}+2xy\underset{̲}{k}$ , $\underset{̲}{\nabla }\times \underset{̲}{F}=\left|\begin{array}{ccc}\hfill \underset{̲}{i}\hfill & \hfill \underset{̲}{j}\hfill & \hfill \underset{̲}{k}\hfill \\ \hfill \hfill \\ \hfill \hfill \\ \hfill \frac{\partial }{\partial x}\hfill & \hfill \frac{\partial }{\partial y}\hfill & \hfill \frac{\partial }{\partial z}\hfill \\ \hfill \hfill \\ \hfill \hfill \\ \hfill y^2\hfill & \hfill x^2-z\hfill & \hfill 2xy\hfill \end{array}\right|=\left(2x+1\right)\underset{̲}{i}-2y\underset{̲}{j}+\left(2x-2y\right)\underset{̲}{k}$

As the contour lies in the $x$ - $y$ plane, the unit normal is $\underset{̲}{k}$ and $\underset{̲}{dS}=dxdy\underset{̲}{k}$

Hence $\left(\underset{̲}{\nabla }\times \underset{̲}{F}\right)\cdot \underset{̲}{dS}=\left(2x-2y\right)dxdy$ .

To work out $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left(\underset{̲}{\nabla }\times \underset{̲}{F}\right)\cdot \underset{̲}{dS}$ , it is necessary to divide the area inside the contour into two smaller areas i.e. the rectangle $ABCDGH$ and the trapezium $DEFG$ . On $ABCDGH$ , the integral is

On $DEFG$ , the integral is

So the full integral is, $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left(\underset{̲}{\nabla }\times \underset{̲}{F}\right)\cdot \underset{̲}{dS}=48-51=-3$ .

By Stokes’ theorem, $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left(\underset{̲}{\nabla }\times \underset{̲}{F}\right)\cdot \underset{̲}{dS}={\oint }_C\underset{̲}{F}\cdot \underset{̲}{dr}=-3$

From Stokes’ theorem, it can be seen that surface integrals of the form $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left(\underset{̲}{\nabla }\times \underset{̲}{F}\right)\cdot \underset{̲}{dS}$ depend only on the contour bounding the surface and not on the internal part of the surface.

Task!

Verify Stokes’ theorem for the vector field $\underset{̲}{F}=x^2\underset{̲}{i}+2xy\underset{̲}{j}+z\underset{̲}{k}$ and the triangle with vertices at $\left(0,0,0\right)$ , $\left(3,0,0\right)$ and $\left(3,1,0\right)$ .

First find the normal vector $\underset{̲}{dS}$ :

Answer

Then find the vector $\underset{̲}{\nabla }\times \underset{̲}{F}$ :

Answer

Now evaluate the double integral ${\int \; <!--nolimits\; -->}_{x=0}^3{\int \; <!--nolimits\; -->}_{y=0}^{x∕3}\left(\underset{̲}{\nabla }\times \underset{̲}{F}\right)\cdot \underset{̲}{dS}$ :

Answer

Finally find the integral $\int \; <!--nolimits\; -->\underset{̲}{F}\cdot \underset{̲}{dr}$ along the 3 sides of the triangle and so verify that the two sides of the equation in the theorem are equal:

Answer

Exercises

1. Using plane-polar coordinates (or cylindrical polar coordinates with $z=0$ ), verify Stokes’ theorem for the vector field $\underset{̲}{F}=\rho \underset{̲}{\stackrel{̂}{\rho }}+\rho cos\frac{\pi \rho }{2}\underset{̲}{\stackrel{̂}{\varphi }}$ and the semi-circle $\rho \le 1$ , $-\frac{\pi }{2}\le \varphi \le \frac{\pi }{2}$ .
2. Verify Stokes’ theorem for the vector field $\underset{̲}{F}=2x\underset{̲}{i}+\left(y^2-z\right)\underset{̲}{j}+xz\underset{̲}{k}$ and the contour around the rectangle with vertices at $\left(0,-2,0\right)$ , $\left(2,-2,0\right)$ , $\left(2,0,1\right)$ and $\left(0,0,1\right)$ .
3. Verify Stokes’ theorem for the vector field $\underset{̲}{F}=-y\underset{̲}{i}+x\underset{̲}{j}+z\underset{̲}{k}$ and for the contour starting from the origin and going to $\left(1,0,0\right)$ , $\left(0,0,0\right)$ , $\left(1,1,0\right)$ and $\left(1,1,1\right)$ before returning to the origin.
   1. Find the surface integral over the triangle $\left(0,0,0\right),\left(1,0,0\right),\left(1,1,0\right)$ .
   2. Find the surface integral over the triangle $\left(1,0,0\right),\left(1,1,0\right),\left(1,1,1\right)$ .
   3. Find the line integrals along the four parts of the contour.
   4. Show that the two sides of the equation of the theorem are equal.
4. Use Stokes’ theorem to evaluate the integral

   ${\oint }_C\underset{̲}{F}\cdot \underset{̲}{dr}$ where $\underset{̲}{F}=\left(sin\left(\frac{1}{x}+1\right)+5y\right)\underset{̲}{i}+\left(2x-e^{y^2}\right)\underset{̲}{j}$

   and $C$ is the contour starting at $\left(0,0\right)$ and going to $\left(5,0\right)$ , $\left(5,2\right)$ , $\left(6,2\right)$ , $\left(6,5\right)$ , $\left(3,5\right)$ , $\left(3,2\right)$ , $\left(0,2\right)$ and returning to $\left(0,0\right)$ .

Answer