Welcome to part one on Stoke's Theorem.
In this video we'll define Stoke's Theorem
and then use Stoke's Theorem to evaluate a line integral.
In part two, we'll use Stoke's Theorem
to evaluate a surface integral.
The idea of Stoke's Theorem is similar to
Green's Theorem stated here.
Remember, that this tells us the net rotation
along the curve C is equal to the
net rotation of the vector field in the
region R where R is in the X, Y plane.
However, in Stoke's Theorem,
the region R becomes the surface S in 3D or in space.
Under the proper conditions, we have this
line integral here equal to the surface integral
of the curl of F dotted with n.
So, Stoke's Theorem states the relationship between
line integrals and surface integrals of a vector field,
given specific conditions.
And it tells us the net rotation over the surface
is equal to the net rotation on the boundary C,
given certain conditions.
So, let's now talk about the necessary
conditions to apply Stoke's Theorem.
If we have our surface here in grey,
the vector field n, the unit normal vectors,
must point in the same direction as your right thumb,
as your right fingers curl and point in the
direction of the path of the boundary C.
Looking at this diagram here,
if our thumb is pointing upward
in the same direction as the unit normal vectors,
notice the fingers would point counter clockwise
giving this the correct orientation.
And this is called a positive orientation,
which is required by Stoke's Theorem.
Let's go ahead and formalize Stoke's Theorem now.
If S is an oriented surface with unit normal vectors n
and also having a smooth, closed boundary C
with a positive orientation and the vector field F
has components with continuous partial derivatives
then we have the line integral of
F dotted with dr
along C,
equal to the surface integral of
the curl of F
dotted with ndS.
Remember the curl of F can be written like this
or we can also think of it as del
crossed with the vector field F,
because this form reminds us of
how to determine the curl of F.
Then remember from the video on the
surface integral of a vector field,
n differential S is equal to this vector field here
times differential A,
given we have a positive orientation.
Remember, this really is just the
gradient of big G,
or big G
is equal to Z minus g of x and y.
And because we have a vector field
dotted with n, this is called a flux integral.
And so, we could say that the line integral
around the boundary C equals the flux of the curl
over the surface S.
Let's go and take a look at our first example,
and as mentioned before, in this video
we're going to use Stoke's Theorem
to evaluate a line integral.
So we'll be given a line integral
and we'll convert it to a surface integral.
Let's go and take a look at our first example.
In this example, we want to evaluate the
line integral of F dotted with dr,
with the given vector field and C is
the intersection of the plane y plus z equals four,
and the cylinder x squared plus y squared equals one.
We'll assume all the conditions of
Stoke's Theorem have been met.
Let's start by taking a look at the graph of this.
So, our surface is the inner section of
the yellow cylinder and the green plane.
So, we'll use Stoke's Theorem to
evaluate the given line integral,
which will tell us the net rotation
caused by the vector field, which is the
intersection of the cylinder and the plane
with a counterclockwise orientation,
which is also equal to the net rotation of
the vector field over this surface.
And if it's positive, then that means
the net rotation would be counterclockwise
and if it's negative, the net rotation would be clockwise,
based upon the given orientation.
So, to evaluate this line integral,
we'll apply Stoke's Theorem
and evaluate this surface integral instead.
So, we'll start by determining the
curl of the vector field F.
So, we can use a three by three determinant
where the first row will be i j k.
The second row will be the partial derivative operators.
The third row will be the components of our vector field.
We'll go ahead and use the
cofactor expansion method this time.
So, for our first two 2 by 2 determinant
will eliminate row one and column one.
For our second determinant we'll
eliminate row one and column two.
Then for our third determinant we'll
eliminate row one and column three.
So, here are all the partials of z squared
with respects to y that'll be zero,
minus the partial of x with respects to z,
that's also going to be zero.
Here we have the partial of z squared
with respects to x, that's zero,
minus the partial of negative 1/2 y squared
with respects to z, that's also zero.
And then for our z component we'll have
the partial of x, with respects to x,
that's going to be one minus
the partial of negative 1/2 y squared with respects to y.
Well, that's going to be negative y.
So, minus a negative y will be plus y.
So, here's the curl of F.
Now to evaluate this surface integral,
we're going to convert it to a
double integral over the region R,
and so we'll have the curl of F dotted with,
this is the shortcut for the gradient of big G.
Where big G is equal to z minus g of x, y,
or we can say that z is equal to g of x, y.
Which in this case, given by the plane,
is going to be equal to four minus y.
So, we'll use this
for the shortcut of our second vector field.
So, we're going to have the curl of F
dotted with the opposite of the partial of g
with respects to x, well that's going to be zero.
The opposite of the partial of g with respects to y.
Well the partial will be negative one
and then the opposite of that will be positive one.
Then the z component is always equal to one
given our orientation.
Let's go determine our dot product.
So, we have zero plus zero plus one plus y times one,
and that's going to be one plus y.
Now for differential A, let's go ahead and
sketch our x, y trace.
Let's go back and look at our graph, real quick.
The x, y trace of our surface, as you can see here,
would just be this circle here
determined by the cylinder.
Let's go ahead and sketch that.
Since our x, y trace is a circle,
we're probably going to want to use polar coordinates,
but let's go ahead and sketch it.
So, for differential A, let's go ahead and
rewrite this as rdrd theta,
and then we'll use our sketch here
to determine our limits of integration.
So, we'll have one plus, y is equal to r sine theta.
Remember that differential A in
polar form would be rdrd theta.
So, limits of integration for r
would be from zero to one,
and theta would be from zero
all the way around to two pi.
Let's ahead and evaluate this on the next page.
Let's go and distribute here, so we have
r plus r squared sine theta.
So, we're going to have r squared over two
plus r cubed over three sine theta.
Subbing in r equals one will have 1/2 plus 1/3 sine theta,
and then when r is zero we have zero.
So, we'll have 1/2 theta
minus 1/3 cosine theta.
So, when theta is equal to two pi
we'd have pi
minus 1/3 times cosine two pi,
well cosine two pi is one.
So I'll have minus 1/3,
minus, and then when theta is zero,
we'd have zero minus 1/3 cosine zero,
we'd have 1/3 again.
So, this simplifies to pi.
Again, this is measuring the net
rotation of the vector field across the surface,
but is also equal to the net rotation
along the boundary curve C.
Going back to our graph just for a moment,
since it's positive pi, that means
the rotation would be counterclockwise.
The same way as the orientation of the curve,
and it's kind of hard to tell
based upon this vector field,
but that's what that value is telling us.
In part two, we'll take a look at
using Stoke's Theorem to evaluate a
surface integral using a line integral.
I hope you found this helpful.