We want to find the area of the region
enclosed by the curve given by the parametric equations
x equals t minus t to the fourth
and y equals t minus t to the seventh
when t is on the closed interval from zero to one.
We want to use Green's Theorem to determine this area.
So the first thing we should recognize is that
the area of a region in a plane is given by this
double integral, where we have the double integral
over the region r of one differential A.
So if we recognize that this will give us
area of a region r in a plane, then we want to explain
why this is equal to this ln integral here where we have
1/2 times ln integral along the curve C
of negative y differential x plus x differential y.
If we can explain why this makes sense
then we'll understand why Green's Theorem can be applied
to determine this area.
So again, first recognizing that this double integral
would give us the area of region r
and then comparing this to Green's Theorem shown here.
If we want this double integral
to be equal to this double integral,
the first thing we want to recognize
is that we would need the difference of these two
partial derivatives to be equal to one.
The next thing to notice is looking at the ln integral,
because we have negative ydx plus xdy
the vector field F would have
an x component of negative y and a y component of x.
So if we determine the partial derivatives
using this vector field, notice a partial of Q
with respect to x would be one.
The partial of P with respect to y would be negative one.
So the difference of the two partial derivatives
doesn't give us one, which we want.
It actually equals two.
And this is where the 1/2 comes into play in the formula.
Because 1/2 times two equals one,
this is where we get a formula.
The area is equal to
1/2 times ln integral of negative ydx plus xdy.
Again, using the vector field shown here,
the difference of the partial derivatives gives us two,
and because we need a one as the integrand function,
we have this extra factor of 1/2 here.
Now before we apply the formula though,
we do want to make sure the curve C,
given by the parametric equations,
does satisfy Green's Theorem.
Where for review, the curve C must be a simply connected
piecewise smooth curve with a positive or counterclockwise
orientation that encloses a region R.
Remember if the curve C is a positive orientation,
then the region R would always be on the left
of the curve C.
And I've already checked this,
the curve given by the parametric equations
would be traced out in this direction
and therefore it does have a positive orientation.
So Green's Theorem is satisfied.
The curve C is a simply connected piecewise smooth curve
with a positive orientation.
Before we set up our ln integral though,
let's go and write down the vector field F.
It would have an x component of negative y
and a y component of x.
Let's also write the curve given by
the parametric equations as a vector-valued function
r of t.
So r of t
would have an x component of
t minus t to the fourth
and a y component of t minus t to the seventh.
And we're also going to need r prime of t
so let's go ahead and find that now.
R prime of t is equal to
the x component would be one minus four t to the third
and the y component would be one minus seven t to the sixth.
So now we know the area
is equal to
1/2 times ln integral
along the curve C of negative ydx plus xdy.
And now to evaluate this ln integral in differential form
we can also think of this as
1/2 times ln integral of F dot differential r
which can be written as a single integral
as integral from a to b of F of x of t comma y of t
dot r prime of t dt.
So this is equal to 1/2 times integral
and now F of x of t comma y of t
would be, well the x component of F is negative y
which is the opposite of t minus t to the seventh
so that'd be negative t plus t to the seventh.
The y component is x, which is t minus t to the fourth.
Dotted with r prime of t which we already found.
R prime of t has an x component
of one minus four t to the third
and the y component is one minus seven t to the sixth.
And we have differential t and limits of integration for t
are from zero to one.
Let's evaluate this on the next slide.
And now we'll find the dot product
so we have 1/2 times integral from zero to one
of, we'd have the quantity negative t plus t to the seventh
times the quantity one minus four t to the third
plus the quantity t minus t to the fourth
times the quantity one minus seven t to the sixth.
And now we multiply.
So here we'll have negative t plus four t to the fourth
plus t to the seventh and then minus four t to the tenth
plus the quantity, here we'll have
t minus seven t to the seventh
minus t to the fourth
and plus seven t to the tenth.
And now I'll combine like terms
so we'd have 1/2 times integral from zero to one
of, starting with the highest degree terms,
we have three t to the tenth,
and then minus six t to the seventh
plus three t to the fourth
and the t terms simplify out.
And now we integrate with respect to t,
so we have 1/2, antiderivative is going to be
3/11 t to the 11th minus 6/8 t to the eighth
plus 3/5 t to the fifth.
And when t is one
we'd have
3/11 minus 6/8 or minus 3/4 plus 3/5.
And then when t is zero
all terms would be zero so we'd have
zero minus zero plus zero.
So here we end up with 1/2 times 27/220
so the exact area is 27/440
which is approximately zero point zero six one four.
So going back to our graph just for a moment,
we just found the area of the region r
where the area bounded by the curve C.
I hope you found this helpful.