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In this section we are going to take a look at a theorem
that is a higher dimensional version of Green’s
Theorem. In Green’s Theorem we
related a line integral to a double integral over some region. In this section we are going to relate a line
integral to a surface integral. However,
before we give the theorem we first need to define the curve that we’re going
to use in the line integral.
Let’s start off with the following surface with the
indicated orientation.
Around the edge of this surface we have a curve C.
This curve is called the boundary
curve. The orientation of the
surface S will induce the positive orientation of C.
To get the positive orientation of C
think of yourself as walking along the curve.
While you are walking along the curve if your head is pointing in the
same direction as the unit normal vectors while the surface is on the left then
you are walking in the positive direction on C.
Now that we have this curve definition out of the way we can
give Stokes’ Theorem.
Stokes’ Theorem
In this theorem note that the surface S can actually be any surface so long as its boundary curve is
given by C. This is something that can be used to our
advantage to simplify the surface integral on occasion.
Let’s take a look at a couple of examples.
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Example 1 Use
Stokes’ Theorem to evaluate  where and S
is the part of above the plane . Assume that S is oriented upwards.
Solution
Let’s start this off with a sketch of the surface.
In this case the boundary curve C will be where the surface intersects the plane and so will be the curve
So, the boundary curve will be the circle of radius 2 that
is in the plane . The parameterization of this curve is,
The first two components give the circle and the third
component makes sure that it is in the plane .
Using Stokes’ Theorem we can write the surface integral as
the following line integral.
So, it looks like we need a couple of quantities before we
do this integral. Let’s first get the
vector field evaluated on the curve.
Remember that this is simply plugging the components of the
parameterization into the vector field.
Next, we need the derivative of the parameterization and
the dot product of this and the vector field.
We can now do the integral.
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Example 2 Use
Stokes’ Theorem to evaluate where and C
is the triangle with vertices ,
and with counter-clockwise rotation.
Solution
We are going to need the curl of the vector field
eventually so let’s get that out of the way first.
Now, all we have is the boundary curve for the surface
that we’ll need to use in the surface integral. However, as noted above all we need is any
surface that has this as its boundary curve.
So, let’s use the following plane with upwards orientation for the
surface.
Since the plane is oriented upwards this induces the
positive direction on C as
shown. The equation of this plane is,
Now, let’s use Stokes’ Theorem and get the surface
integral set up.
Okay, we now need to find a couple of quantities. First let’s get the gradient. Recall that this comes from the function of
the surface.
Note as well that this also points upwards and so we have
the correct direction.
Now, D is the region
in the xy-plane shown below,
We get the equation of the line by plugging in into the equation of the plane. So based on this the ranges that define D are,
The integral is then,
Don’t forget to plug in for z since we are doing the surface integral on the plane. Finishing this out gives,
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In both of these examples we were able to take an integral
that would have been somewhat unpleasant to deal with and by the use of Stokes’
Theorem we were able to convert it into an integral that wasn’t too bad.