Welcome to an example
on how to use Green's Theorem
to evaluate a line integral.
We want to evaluate the given line integral
where the curve C is the boundary of the region
between the circles given by x squared
plus y squared equals four,
and x squared plus y squared equals nine.
Let's begin by reviewing Green's Theorem.
Green's Theorem states that if we have a simply connected
piecewise smooth curve with a positive,
or counterclockwise orientation that encloses a region, R,
meaning the region R is always on the left of the curve,
then if the vector field F has components P and Q,
where both have continuous first order partial derivatives,
then the line integral along the curve C
of F dot differential r,
or the line integral in differential form,
is equal to the double integral over the region R
of the partial of Q with respect to x
minus the partial of P with respect to y,
differential A.
When the path of integration satisfies Green's Theorem
we can denote the line integral in this form here
where we put a little circle
in the middle of the integral symbol.
Going back to our example notice how
because the integral symbol does have this circle,
we know the curve C does satisfy
the conditions of Green's Theorem.
But I do want to spend a little bit more time explaining
how this region R would satisfy Green's Theorem
based upon the definition.
Remember earlier we discussed how Green's Theorem states,
the curve C must be simply connected
and piecewise smooth with a positive orientation
and here were three examples that we looked at.
We said this region here
which looks somewhat like our region is not simply connected
and therefore it might seem Green's Theorem does not apply.
But it still does and let's see why.
If we can trace out this region
with a piecewise smooth curve with a positive orientation,
meaning the region R is always
on the left side of the curve
then Green's Theorem would apply.
Notice how I designated four curves here.
The outer curve along the larger circle is C sub one.
Then we have C sub two,
the curve from the outer circle to the inner circle.
Then we have curve C sub three around the inner circle
and then we have the opposite of C sub two
going back out from the inner circle to the outer circle.
I labeled this negative C sub two
because notice how it's the opposite direction
of curve C sub two.
Notice how if we were to start here
and trace around the outer circle
this is a positive orientation
because the region R is on the left of the curve.
Then we stop here and go to the inner circle.
Then notice how we're going clockwise
around the inner circle
but still the region R is on the left side of the curve
and therefore we still have a positive orientation.
Then we go back over to here
and back out to the outer circle
to make a simply connected piecewise smooth curve
that traces out the region R.
Which means our curve C could be defined
as C sub one plus C sub two plus C sub three
plus the opposite of C sub two.
Simplifying we get the curve C is equal to C sub one
plus C sub three
This is one way we can trace out the region R
to satisfy Green's Theorem.
Notice how this also shows us
that if we have a region with a hole in it,
as long as the outer region
has a counter clockwise orientation
and the inner region has a clockwise orientation
both curves still have a positive orientation
because the region R is on the left
and therefore the curve C would just be the sum
of the outer curve and the inner curves.
Also notice how if the inner curve
had a counter clockwise orientation
the curve C would have to be C sub one minus C sub three.
Again all this isn't necessary for this example
because we're told by this little circle
in the integral symbol the condition
of Green's Theorem for curve C has already been met.
So starting with the given line integral
along the curve C that satisfies Green's Theorem
we have y cubed dx
minus x cubed dy.
But this tells us that P is equal to y cubed
and Q is equal to negative x cubed.
Now let's go ahead and find
the partial of Q with respect to x
and the partial of P with respect to y.
The partial of Q with respect to x
is the derivative of negative x to the third
with respect to x, which is negative three x squared.
Now let's find the partial of P with respect to y.
Which is the derivative of y cubed with respect to y,
which would be three y squared.
Now applying Green's Theorem the line integral
is equal to the double integral
over the region R of the partial of Q with respect to x,
which is negative three x squared
minus the partial of P with respect to y
which is three y squared, differential A.
But now because the region R is bounded between two circles
the [mumbles] double integral in polar form
where we know differential A is equal to rdrd theta.
Notice how here if we factor out negative three
we would have negative three times the quantity
x squared plus y squared.
x squared plus y squared is equal to r squared.
So in polar form we'd have the double integral
over the region R of negative three r squared
and again differential A is equal to rdrd theta.
Now let's determine the limits
of integration for r and theta.
Because the region R is bounded between these two circles
where the inner circle has a radius of two
and the outer circle has a radius of three
the limits of integration for r
are going to be from two to three.
Then trace out the region between these two circles.
The limits of integration for theta would be
from zero to two pi all the way around the circle.
Now let's evaluate this on the next slide.
Let's rewrite the integrand function
as negative three r cubed.
Integrate with respect to r first,
we would have
an antiderivative of negative three times
r to the fourth divided by four,
negative 3/4 r to the fourth.
Now to find big F of b minus big F of a
let's factor out the negative 3/4
and then we'd have three to the fourth
minus two to the fourth.
So three to the fourth minus two to the fourth
is equal to 65.
Negative three times 65 is equal to negative 195.
We have the integral from zero to two pi
of negative 195/4 d theta.
Integrate with respect to theta,
we have negative 195/4 theta.
Again big F of b minus big F of a,
so we're going to have negative 195/4
times two pi minus zero.
Notice two into four simplify.
So the exact value is negative 195/2 pi,
which as a decimal would be approximately negative 306.3053.
I hope you found this helpful.