Welcome to Part One of Green's Theorem.
Green's Theorem states that the relationship
between a line integral and a double integral
over a simply connected plane region R.
The double integral over the region is equal
to the line integral around the region counterclockwise.
So the region needs to be
simply connected and piecewise smooth,
and the orientation must be counterclockwise.
Looking at this first region, R,
notice the curve around the region
is oriented counterclockwise and it's smooth.
Looking at the second region here,
the region is piecewise smooth,
meaning each of the four pieces of the curve are smooth,
and therefore, because the orientation is counterclockwise,
we could also apply Green's Theorem.
And then it won't apply to us in the video,
but you can extend Green's Theorem
where you have a simply connection region,
as we see here, with a hole in it.
But again, this won't be addressed in this video.
So let's see what Green's Theorem says.
If C is a simply connected, piecewise, smooth curve
with a counterclockwise orientation in a plane region R,
and our vector field with components f and g
have continuous first partial derivatives,
then the line integral, written here as
F dotted with d r, or in differential form,
is equal to the double integral over the region
of the partial derivative of g with respects to x,
minus the partial derivative of f
with respects to the y, d A.
And since we just talked about the
fundamental theorem of line integrals,
notice that if this difference was zero,
the vector field would be conservative.
So the vector field does not have to be
conservative to apply Green's Theorem.
When the path of integration satisfies Green's Theorem
we can denote the line integral as we see here,
which tells us we have a closed path line integral.
So in short, if Green's Theorem is satisfied,
it often gives us an easier method
to evaluate a line integral,
and again, the vector field F does not need to be
conservative to apply Green's Theorem.
So instead of parameterizing everything
in terms of t as we did several sections ago,
if the region is piecewise smooth connected,
or even counterclockwise, we can apply Green's Theorem.
Let's take a look at our first example.
We want to evaluate this line integral,
where C is the circle x-squared plus y-squared equals four
transversed counterclockwise.
So our regions satisfies the requirement of Green's Theorem,
so we're going to rewrite this line integral
as a double integral over the region R.
So the x component of vector field F
is equal to little f, and the y component is equal to g.
So the partial derivative of g
with respects to x would be one,
minus the partial derivative of f
with respects to y would be negative one.
So here, we're just going to have the double integral
over the region of two, differential A,
well this is just two times the area of this region,
this would be two times,
well the area of a circle is pi r-squared,
and the radius here is two,
so we have pi times two squared,
that's going to give us eight pi.
However I do want to go ahead and show how
we can evaluate this if we didn't recognize
that this is just two times the area of the region.
So to evaluate this over this circle,
would be much easier if we converted this to polar form,
and remember when we do that, we have an extra factor of r,
and then d r d theta,
where r would be from zero to two,
and theta would be from zero to two pi.
Okay we know the answer should be eight pi,
but let's go and just verify it using the longer method.
And let's go ahead and evaluate this on the next page.
So we have two times r-squared over two,
which would give us r-squared.
This would just be four minus zero,
so we have the integral from zero to two pi of four d theta,
so we have four theta, and then you can see here,
we'd have eight pi minus zero, or eight pi.
So the result is the same,
regardless of whether we take the shortcut,
by just multiplying two times the area of the region,
or we convert to polar form and evaluate the long way.
Let's go and take a look at another example.
So we have the same type of problem,
we're going to evaluate the line integral,
where C is given below along y equal x-squared over two,
that would be the curve,
and y equals x, which is the line,
and this is transversed counterclockwise,
so the region does satisfy Green's Theorem,
so to evaluate this, let's go ahead
and convert this to a double integral
over the region R using Green's Theorem.
So again, little f would be the x component
of the vector field F, and g would be the y component.
So our line integral is going to be equal to the
double integral over the region R of
the partial derivative of g with respects to x,
that's going to be two x, minus the partial derivative of f
with respects to y, is going to be two y, differential A.
Now need to decide the order of integration,
meaning do we want differential A to be dy/dx or dx/dy.
Since our equations are functions of x,
I'm going to go integrate with respects to y first and then x.
So limits integration for y must be expressed
as functions of x, so lower limit will come from the curve,
that will be x-squared over two,
upper limit for y would be the line, which is y equals x,
and if the limit integration for x,
it'll go from zero to two.
So we'll first integrate with respects to y,
that'll give us two x y,
minus two times y-squared over two, or just minus y-squared.
Evaluate this and our limits of integration,
remember we're replacing y with our limits of integration.
So we're going to have two x times x,
that's two x-squared, minus x-squared,
we're have minus two x times x-squared over two,
that's going to give us x-cubed,
minus x-squared over two to the second power,
that's going to give us x to the fourth over four.
Combining our like terms, we'll have x-squared minus x-cubed
plus x to the fourth over four.
Let's go ahead and finish this on the next page.
So we'll have x to the third over three
minus x to the fourth over four plus
x to the fifth over five times four,
x to the fifth over 20.
So when x is two, we're going to have 8/3 minus
this would be four, here we'll have 32 over 20,
and then when x is zero, we'll have zero,
this works out to 4/15.
And we'll stop here for this video.
In part two, we'll take a look at
some additional examples of Green's Theorem.
Thank you for watching.