Welcome to a lesson on determining area
using line integrals.
If C is a simply connected piecewise smooth curve
with a counterclockwise orientation
in the plane region R, then the area of the region
bounded by the curve is equal to one-half
times the line integral of x dy minus y dx
along curve c.
And this is actually an application
from Green's Theorem.
So let's take a look at where this formula comes from.
We should now know Green's Theorem
stated here in red, and we should also remember
from our studies of double integrals
that the double integral over the region R
integrated with respects to A
would give us the area of region R.
So what we're going to do here is look
at a special case of Green's Theorem
when this difference here is equal to one.
And then we'll write this double integral here
as a line integral in differential form.
So if our vector field F has components
little f and little g, then if we let f equal
negative y over two and g equal x over two,
then the partial derivative of g with respects to x
would be positive one-half
and the partial derivative of f with respects to y
would be negative one-half.
And that would give us the integrand of one
because we'd have one-half minus negative one-half.
So if we use these functions and write the line integral
in differential form, f dx would be
negative one-half y dx.
And g dy would be one-half x dy.
So now if we factor out the one-half
and then put the positive differential first,
we'd have x dy minus y dx.
And this is a line integral that will give us
the area of the region as long as it satisfies
the conditions of Green's Theorem.
Let's take a look at our example.
We're going to determine the area
of the ellipse given below using a line integral.
Notice how the orientation is counterclockwise.
The curve is smooth, continuous, and connected.
And the equation of our path is x-squared
divided by 16 plus y-squared divided by four
is equal to one.
So, to set up this line integral,
we'll parametrize our curve in terms of t
and then perform substitutions for x dy and y dx.
So looking at our equation of the ellipse,
if we let x equal four cosine t
and y equal two sine t,
we would satisfy the equation of the curve.
So this also tells us that dx would be equal
to negative four sine t dt
and that dy would be equal to two cosine t dt.
So we can use all of this to perform substitution
into our line integral.
So the area of this region would be equal
to one-half times the line integral of x dy.
That's going to give us four cosine t
times dy which is two cosine t dt.
I'm going to go ahead
and put the dt out here.
Minus y, which is two sin t
times dx which is negative four sine t.
And our limits of integration in terms of t
would be from zero all the way to two pi.
Looks like we'll have eight cosine-squared t
plus eight sine-squared t as our integrand.
Let's go ahead and evaluate that.
So if we factor out the eight,
we're going to have cosine-squared t
plus sine-squared t.
That's equal to one.
So this simplifies nicely.
So this is going to be eight t times one-half.
That's four t.
So we're going to have four times two pi.
That's eight pi minus zero.
So the area of that ellipse would be eight pi.
Let's go back and take a look at that.
Remember, the geometric formula
to determine the area of an ellipse
is a times b times pi.
Well, a is equal to four and b is equal to two.
So that verifies our area should be eight pi
which is what we found using a line integral.
And that'll do it for this video.
Thank you for watching.