In this section we are going to look at areas enclosed by
polar curves. Note as well that we said
“enclosed by” instead of “under” as we typically have in these problems. These problems work a little differently in
polar coordinates. Here is a sketch of
what the area that we’ll be finding in this section looks like.
We’ll be looking for the shaded area in the sketch
above. The formula for finding this area
is,
Notice that we use r
in the integral instead of so make sure and substitute accordingly when
doing the integral.
Let’s take a look at an example.
Example 1 Determine
the area of the inner loop of .
Solution
We graphed this function back when we first started
looking at polar coordinates. For this problem we’ll also need to know
the values of where the curve goes through the
origin. We can get these by setting
the equation equal to zero and solving.
Here is the sketch of this curve with the inner loop
shaded in.
Can you see why we needed to know the values of where the curve goes through the
origin? These points define where the
inner loop starts and ends and hence are also the limits of integration in
the formula.
So, the area is then,
You did follow the work done in this integral didn’t
you? You’ll run into quite a few
integrals of trig functions in this section so if you need to you should go
back to the Integrals Involving Trig
Functions sections and do a quick review.

So, that’s how we determine areas that are enclosed by a
single curve, but what about situations like the following sketch were we want
to find the area between two curves.
In this case we can use the above formula to find the area
enclosed by both and then the actual area is the difference between the
two. The formula for this is,
Let’s take a look at an example of this.
Example 2 Determine
the area that lies inside and outside .
Solution
Here is a sketch of the region that we are after.
To determine this area we’ll need to know the values of θ for which the two curves intersect. We can determine these points by setting
the two equations and solving.
Here is a sketch of the figure with these angles added.
Note as well here that we also acknowledged that another
representation for the angle is . This is important for this problem. In order to use the formula above the area
must be enclosed as we increase from the smaller to larger angle. So, if we use to we will not enclose the shaded area, instead
we will enclose the bottom most of the three regions. However if we use the angles to we will enclose the area that we’re after.
So, the area is then,

Let’s work a slight modification of the previous example.
Example 3 Determine
the area of the region outside and inside .
Solution
This time we’re looking for the following region.
So, this is the region that we get by using the limits to . The area for this region is,
Notice that for this area the “outer” and “inner” function
were opposite!

Let’s do one final modification of this example.
Example 4 Determine
the area that is inside both and .
Solution
Here is the sketch for this example.
We are not going to be able to do this problem in the same
fashion that we did the previous two.
There is no set of limits that will allow us to enclose this area as
we increase from one to the other.
Remember that as we increase the area we’re after must be enclosed. However, the only two ranges for that we can work with enclose the area from
the previous two examples and not this region.
In this case however, that is not a major problem. There are two ways to do get the area in
this problem. We’ll take a look at
both of them.
Solution 1
In this case let’s notice that the circle is divided up
into two portions and we’re after the upper portion. Also notice that we found the area of the
lower portion in Example 3. Therefore,
the area is,
Solution 2
In this case we do pretty much the same thing except this
time we’ll think of the area as the other portion of the limacon than the
portion that we were dealing with in Example 2. We’ll also need to actually compute the
area of the limacon in this case.
So, the area using this approach is then,
A slightly longer approach, but sometimes we are forced to
take this longer approach.

As this last example has shown we will not be able to get
all areas in polar coordinates straight from an integral.