# 16.7: Stokes' Theorem - Mathematics

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 ( abla = 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 ( abla imes mathbf{F}). There are many ways to take the determinant, but the following is an example of cofactor expansion.

[egin{align} abla imes mathbf{F} &= egin{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} egin{vmatrix} frac{partial }{partial y} & frac{partial }{partial z} N & P end{vmatrix} - hat{j} egin{vmatrix} frac{partial }{partial x} & frac{partial }{partial z} M & P end{vmatrix} + hat{k} egin{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} = abla imes mathbf{F}) dotted with (mathbf{n}).

[oint _C mathbf{F} cdot dmathbf{r} = iint_{S} abla imes mathbf{F cdot n} dsigma ]

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

[egin{align} oint _C mathbf{F} cdot dmathbf{r} &= iint_{S} abla imes mathbf{F cdot n} dsigma & = iint_{R} abla imes 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

[egin{vmatrix} frac{partial }{partial x} & frac{partial }{partial y} M & N end{vmatrix}.]

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 ( extbf{F}=yhat{ extbf{i}}-xhat{ extbf{j}}). Hints: Remember that a simple way to parameterize a circle is if ( x^2+y^2=r^2) then ( r ( heta) = r cos heta + r sin heta ) for ( heta 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 (| abla 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 ( heta ) = ( 3 cos heta ) hat{ extbf{i}} + (3 sin heta ) hat{ extbf{j}}, 0 leq heta leq 2 pi ):

[ d extbf{r} = (-3 sin heta d heta ) hat{ extbf{i}} + (3 cos heta d heta ) hat{ extbf{j}} ]

[ extbf{F} = yhat{ extbf{i}} - xhat{ extbf{j}} = (3 sin heta ) hat{ extbf{i}} - (3 cos heta )hat{ extbf{j}}]

[ extbf{F} cdot d extbf{r} = -9 sin^2 heta d heta - 9 cos^2 heta d heta = -9 d heta ]

[ oint_C extbf{F} cdot d extbf{r} = int_0^2pi -9 d heta = -18 heta. ]

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 ( extbf{F} ), we have

[ abla imes extbf{F} = left( frac{partial P}{partial y} - frac{partial N}{partial z} ight) hat{ extbf{i}} + left( frac{partial M}{partial z} - frac{partial P}{partial x} ight) hat{ extbf{j}}+left( frac{partial N}{partial x} - frac{partial M}{partial y} ight) hat{ extbf{k}} ]

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

[ egin{vmatrix} hat{ extbf{i}} & hat{ extbf{j}} & hat{ extbf{k}} frac{partial }{partial x} & frac{partial }{partial y} & frac{partial }{partial z} y & -x & 0 end{vmatrix} ]

evaluating the curl, we see that

[ abla imes extbf{F} = (0-0)hat{ extbf{i}}+(0-0)hat{ extbf{j}}+(-1-1)hat{ extbf{k}} = -2hat{ extbf{k}} .]

Our outer unit normal vector will be

[ egin{align} extbf{n} &= frac{ x hat{ extbf{i}} + y hat{ extbf{j}} + z hat{ extbf{k}}}{|xhat{ extbf{i}} + y hat{ extbf{j}} + zhat{ extbf{k}}|} &=frac{xhat{ extbf{i}} + y hat{ extbf{j}} + zhat{ extbf{k}}}{sqrt{x^2+y^2+z^2}} &=frac{xhat{ extbf{i}}+yhat{ extbf{j}}+zhat{ extbf{k}}}{sqrt{9 cos^2 heta + 9 sin^2 heta}} &= frac{xhat{ extbf{i}}+yhat{ extbf{j}}+zhat{ extbf{k}}}{3}. end{align}]

Then

[d sigma = frac{| abla f |}{| abla f cdot extbf{k}|} dA = frac{3}{z} dA.]

Finally, we can put everything together to find that:

[ abla imes extbf{F} cdot extbf{n} d sigma = - frac{2z}{3} frac{3}{z} dA = -2dA ]

and

[ int int_S abla imes extbf{F} cdot extbf{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.