Welcome to an example
on how to use Green's Theorem
to evaluate a given line integral.
Here want to to evaluate a given line integral
along the curve C, where the curve C is a rectangle
with vertices zero comma zero, two comma zero,
two comma four, and zero comma four.
Let's being by reviewing Green's Theorem.
Green's Theorem states that if we let C
be a simply connected piecewise smooth curve
with a positive or counterclockwise orientation
that encloses a region R.
So if the curve C has a positive
or counter-clockwise orientation,
region R is always going to be to the left of the curve.
And we let the vector field F have components P comma Q,
where have P and Q 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 P with respect to y,
differential A.
Where again Q is the y component of f
and P is the x component of f.
When the path of integration satisfies Green's Theorem,
we denote the line integral as shown here
where we have this circle in the middle
of the integral symbol.
Which in our example, we had.
And before I go back to our example though,
let's talk more about what a simply connected,
piecewise smooth curve with a positive
or counter-clockwise orientation is.
Let's first talk about a simply connected curve.
Here are two examples of simply connect curves
that enclose the region R.
And notice how both of these also have a positive
or counter-clockwise orientation.
And here a two examples of curves
that are not simply connected.
And in some special cases, Green's Theorem
can still be applief, if the curve is not simply connected.
For example in this case here, which we'll talk about later.
But the specific wording of Green's Theorem
states that the curve C must be a simply connected,
piecewise smooth curve with a positive orientation
and here are three examples.
Each of these curves are simply connected,
piecewise, smooth, and have a positive orientation.
Notice how in this example, the curve C is made up
of four pieces, which make it piecewise smooth
and we can also see it's connected and encloses the region R
Here the curve C is made up of three curves
to make it piecewise smooth,
and again we can see it's connected
and encloses the region R.
And this third example is similar to the second example.
Notice in each case, the curves have a positive
or counter-clockwise orientation,
which is indicated by the arrows along the curve.
Going back to our example now.
Again we are told because of the circle here
in the integral symbol that the curve C
does satisfy Green's Theorem.
But just to show this, if we were to plot these points
on the Cartesian plane as we see here
and form the rectangle,
the orientation would be in this direction here,
a positive orientation or a counter-clockwise orientation.
And notice how the region R is always on the left side
of the curve.
So beginning with the given line integral,
we have the line integral along the curve C
where the curve C does satisfy Green's Theorem
of x y squared dx
plus x to the fifth dy.
So this tells us that p is equal to x y squared
and Q is equal to x to the fifth.
So now let's find the partial of Q with respect to x
and the partial P with respect to y.
So the partial of q with respect to x,
is equal to the derivative of x to the fifth
with respect to x, which would be five x to the fourth.
And the partial of P with respect to y,
is equal to the derivative x y squared with respect to y,
which would two x y.
So now applying Green's Theorem,
this line integral is equal to the double integral
over the region R of the partial of Q with respect to x,
which is five x to the fourth
minus the partial of P with respect to y,
which is two x y.
And then for differential A, we can use dx/dy or dy/dx.
Because the region R is a rectangle,
it really doesn't matter.
Let's first integrate with respect to x,
then with respect to y.
And now we'll determine the limits of integration
first with respect to x.
The region R is bounded to the left by x equals zero
and to the right by x equals two.
So the limits of integration with respect to x are
from zero to two,
and the region R is bounded below by y equals zero
and above by y equals four.
So the limits of integration for y are from zero to four.
And now we integrate with respect to x first.
So the anti-derivative would five times x to the fifth
divided by five minus two y times x squared divided by two.
Let's go ahead and simplify the anti-derivative.
So we'd have x to the fifth,
minus x squared y.
Now we'll perform substitution for x to find
big F of b minus big F of a.
So when x is two, we'd have two to the fifth
minus two squared times y minus,
notice both terms contain at least one factor of x.
So we'd have zero minus zero.
Simplifying we have the integral from zero to four
of 32 minus four y,
differential y.
Continuing on the next slide,
integrating with respect to y we'd have,
an anti-derivative of 32 y minus four times y squared
divided by two, which would be minus two y squared.
So big F of b minus big F of A is going to be,
32 times four minus two times four squared
minus, when y is zero both terms are zero.
So here we have 128 minus 32,
which equals 96.
So because the given line integral,
satisfied the conditions of Green's Theorem,
we were able to evaluate the line integral
by writing it as a double integral and then evaluating.
I hope you found this helpful.