Professor Dave here, I wanna tell you about Stokes’s Theorem.
Previously, we went over Green’s Theorem, which let us turn the line integral for a
closed curve C in two dimensions into a double integral. Stokes’s Theorem can be thought
of as a higher dimensional version, when we have a closed curve in three dimensions, where
the curve is the boundary of a surface S. Just as with Green’s theorem, we assume
that the curve is positively oriented, which means the path of the curve goes in the counterclockwise
direction with respect to the normal vector of the surface. Now if we have a vector field
permeating space, to find the amount of the field that goes in the direction of the curve
we would need to take the line integral of F along C. Stokes’s Theorem gives us another
method of finding this line integral, which can lead to simpler calculations. The theorem
states that the line integral of F along C is equal to the surface integral of the curl
of F dot n over the surface S. We discussed how to calculate surface integrals in the
previous tutorial, so we have seen that they aren’t too difficult to solve, it’s just
that now we have the extra step of taking a curl. Remember that the curl of a vector
field is calculated by taking del cross F, where del was the vector .
While we can use Stokes’s theorem to calculate the line integrals, it can also be used in
reverse if we want to find the surface integral of a curl, and the boundary is simple enough
to calculate. Now let’s look at an example. Let’s say
we have a vector field F = and we want to find the line integral of F over
C, where our curve C is given by the triangle that forms when we connect the points (1,0,0),
(0,1,0), and (0,0,1). These points define the plane x + y + z = 1, which will be useful
information in our calculation. Using Stokes’s theorem we will calculate the line integral
as the surface integral of the curl of F dot the normal vector of our plane surface. With
the knowledge that the plane is given by x + y + z = 1, we can solve for z to get the
parametric representation of the surface. Since z = 1 – x – y, our surface can be
represented by r = . Now recall that when our surface has z expressed
as a function of x and y, we can use a simplified expression. The surface integral of a vector
field, in this case the curl of F, dot n dS will equal the double integral of the curl
of F dot the cross product of rx and ry, dx dy. To get rx and ry, we must take the x and
y derivatives of the r vector. So rx = <1, 0, -1> and ry = <0, 1, -1>. Now we must calculate
the curl of F and the cross product rx cross ry. The curl of F is del cross F, which means
we must take the determinant of this matrix. We end up getting i(0 - 1) - j(0 - 0) + k(0
- x) which simplifies to <-1, 0, -x>. Now for rx cross ry, we take the determinant of
this matrix. Here we get i(0 + 1) - j(-1 - 0) + k(1 - 0) which is simply <1, 1, 1>. Plugging
both of these in we end up with the double integral of <-1,0,-x> dot <1,1,1> dxdy. Calculating
the dot product we get the double integral of (-1 - x) dx dy. Now we just need to determine
our bounds of integration. If we look at our surface again, we see that it is over this
triangular region in the xy plane, the border of which is the line between (1,0) and (0,1).
This line is given by y = 1 – x, and we must return to our thought process of setting
up bounds for double integrals. We will choose to integrate over y first, which means our
range of y goes from the lower curve to the higher curve. So in this case y runs from
0 to 1 – x. Meanwhile, x will simply run from 0 to 1. Now our integral is completely
set up and we can easily evaluate it. Integrating over y, we see that there is no current dependence
on the variable, so the antiderivative is (-1 – x)y, which runs from 0 to 1 – x.
Plugging in the upper bound, we get the integral of (-1 – x) times (1 – x) dx. Multiplying
these binomials, we get the integral of -1 + x – x + x^2 dx, which simplifies to the
integral of (x^2 – 1) dx. Taking the antiderivative gives us (x^3/3 – x) where x runs from 0
to 1. Plugging in these values gives us the quantity (1/3 – 1) minus the quantity (0
– 0) which is simply negative 2/3. This is our line integral which we used Stokes’s
theorem to help calculate. Stokes’s theorem is a powerful tool that
allows us to choose the simplest way to calculate line integrals and some surface integrals.
Naturally, many problems will require much more thought and consideration than the simple
case we’ve gone over here, but we’ve still laid down the ground work for future endeavors,
so let’s check comprehension.