Welcome to the second example
of determining area using double
integrals with polar coordinates.
Again, just as we found area using double
integrals in rectangular coordinates,
we can find area using polar coordinates.
We replace differential A with an extra
factor R,
dr, theta,
and then we integrate with respects to r
and respects to theta.
Let's go and take a look at our example.
We want to determine the bounded area
by r equals one plus cosine two theta.
It's always nice to take a look at the graph,
using some technology.
Our goal here is to determine the area
here, as well as here.
Let's see if we can set this up.
Of course, we're going to have r,
dr, theta,
and we know the radius is controlled by
r equals one plus cosine two theta.
So limits to integration for the radius will
be from zero to one plus cosine two theta.
But the challenging part here is how to determine
the limits of integration for theta.
We're actually going to use some symmetry here.
We're going to trace out 1/4th of this,
and multiply this by four.
And to do this, we're going to use some symmetry.
We're going to try to trace out one piece
of this in the first quadrant, and then
multiply it by four.
So, we're going to focus on, just,
this piece here.
And it looks like it's going to be in the first
quadrant, so we might think it's going to
just be from zero to Pi over two,
but that's not always the case,
we need to be careful about this.
One way to determine this would to be to use
the graphing calculator.
So let's go ahead and do that.
Again,
I put my calculator in degree mode,
because it's easier to read that way,
and also polar mode.
And I typed in our function,
r equals one plus cosine two theta.
Let's go ahead and press graph.
And if we press trace, we'll easily
be able to see, what interval for theta
will trace out this piece of the curve
in the first quadrant.
So, right now we're at theta equals zero.
And notice that when theta is equal to,
90 degrees, it does trace out that
curve in the first quadrant.
As expected, so our interval for theta
is from zero to pi over two.
And don't forget we have to multiply this by four
because we're finding this area here,
but we want the total area.
Now, let's go ahead and integrate with
respects to r.
We're going to have r squared over two.
Here we have four times one-half,
that's going to be two.
Now we're going to have to square one plus
cosine two theta.
And then when r is zero,
we just have zero.
So, we're going to have one plus two cosine
two theta, plus cosine squared two theta.
Let's go to the next page.
Now, we'll have to apply the power reducing
formula here, so let's go ahead and break
this up into three different integrals.
Let's go re-write this one more time,
applying the power reducing formula
to cosine squared two theta.
So this two and this two are going to simplify out.
So we're left with one plus cosine four theta.
And now we can integrate.
Here we'll have two theta.
The anti-derivative of cosine two theta,
is going to be 1/2 sine two theta.
So this will end up being sine two theta.
Keep leaving off the limits of integration.
Plus, here we're going to have theta plus
we'll have 1/4th sine theta.
Again, we had to perform use substitution
here for four theta.
So here we'll have two times pi over two,
minus zero, plus, okay, when theta
is pi over two we'll have pi, which is zero.
So, we'll have zero, minus sine of zero,
which is also zero.
Plus, this will be pi over two.
Plus, this will be zero,
and then we'll have minus zero plus zero.
Looks like we have pi plus pi over two,
that's going to give us three pi over two.
And that'll do it.
Thank you for watching.