We're asked to use parametric equations
of an ellipse to find the area enclosed by the ellipse.
We have x equals two cosine theta
and y equals three sine theta
where theta is on the closed interval from zero to two pi.
So the graph of our ellipse would look like this
on the coordinate plane.
So to find the area using these parametric equations,
we're going to use the fact that we know
that if a function of x or f of x is nonnegative
on the closed interval from a to b on the coordinate plane,
then the area is equal to the def interval
of f of x from a to b.
If we want, we can substitute y for f of x,
giving us this definite interval.
So starting with this definite integral,
we know that y is equal to three sine theta.
And now we can find dx since we're given
x equals two cosine theta.
So differential x would be equal
to negative two sine theta d theta.
So now let's perform these two substitutions
into our def integral.
We would have the integral of three sine theta
times negative two sine theta d theta.
Now we have an integral in terms of theta,
but we still need to determine the limits of integration.
So we might be thinking we could just integrate
from zero to two pi, but that's actually not going
to work in this case.
So what we're going to do is consider the symmetry
of the ellipse on the coordinate plane.
Notice how, if we found the area
in the first quadrant here,
we could then just multiply by four
because of the symmetry.
But we know on the coordinate plane,
if we wanted to find this area,
we'd have to integrate from x equals zero
to x equals two.
If you notice in this case, we make a table of values
for theta and x, when theta equals zero,
notice how x would be equal to two cosine zero
which would be equal to two.
And we don't want to start integrating at x equals two.
We want to start integrating at x equals zero.
But when theta is pi over two,
notice how x should be two times cosine pi over two
or two times zero which is zero.
So in this case, we're actually going to integrate
from pi over two to zero which would be the same
as integrating from x equals zero to two.
So our lower limit of integration is pi over two,
and the upper limit of integration would be zero.
This would give us this green shaded area.
But since we want the whole area,
we have to multiply this by four.
So now let's find this area on the next slide.
Since three times negative two equals negative six,
let's go ahead and factor out the negative six.
So we would have negative 24
and the integrand would now be sine squared theta.
And now we'll use a power reducing formula
for sine squared theta given here below in red.
So let's go ahead and factor out the one-half.
That would give us negative 24 divided by two
times the integral from pi over two to zero of one
minus cosine two theta d theta.
And now we can go ahead and integrate.
Here we would have negative 12.
The antiderivative of one with respect to theta
would just be theta.
And then integrate cosine two theta,
we'd have to perform substitution
where u would be two theta
and differential u would be two d theta.
Dividing both sides by two,
notice how d theta would be equal to one-half d u.
So we have an extra factor of one-half
when integrating cosine two theta.
So the antiderivative would be
minus one-half sine two theta.
So we would have negative 12
times, we first substitute zero for theta.
That'd be zero minus one-half sine zero.
That's zero minus zero.
And then when theta is pi over two,
we would have pi over two
and then minus one-half times sine pi.
Well, sine pi is also zero.
Notice how this simplifies to negative 12
times negative pi over two
which is equal to six pi.
So six pi square units would be the area
of the ellipse as we see here.
Since we multiply the def integral by four,
it would give us the entire area shaded here in green.
I hope you found this explanation helpful.