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 \(\theta \) where the curve goes through the origin. We can get these by setting the equation equal to zero and solving.

\[\begin{align*}0 & = 2 + 4\cos \theta \\ \cos \theta & = - \frac{1}{2}\hspace{0.5in} \Rightarrow \hspace{0.5in}\theta = \frac{{2\pi }}{3},\frac{{4\pi }}{3}\end{align*}\]

Here is the sketch of this curve with the inner loop shaded in.

Can you see why we needed to know the values of \(\theta \) 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,

\[\begin{align*}A & = \int_{{\,\frac{{2\pi }}{3}}}^{{\,\frac{{4\pi }}{3}}}{{\frac{1}{2}{{\left( {2 + 4\cos \theta } \right)}^2}\,d\theta }}\\ & = \int_{{\,\frac{{2\pi }}{3}}}^{{\,\frac{{4\pi }}{3}}}{{\frac{1}{2}\left( {4 + 16\cos \theta + 16{{\cos }^2}\theta } \right)\,d\theta }}\\ & = \int_{{\,\frac{{2\pi }}{3}}}^{{\,\frac{{4\pi }}{3}}}{{2 + 8\cos \theta + 4\left( {1 + \cos \left( {2\theta } \right)} \right)\,d\theta }}\\ & = \int_{{\,\frac{{2\pi }}{3}}}^{{\,\frac{{4\pi }}{3}}}{{6 + 8\cos \theta + 4\cos \left( {2\theta } \right)\,d\theta }}\\ & = \left. {\left( {6\theta + 8\sin \theta + 2\sin \left( {2\theta } \right)} \right)} \right|_{\frac{{2\pi }}{3}}^{\frac{{4\pi }}{3}}\\ & = 4\pi - 6\sqrt 3 = 2.174\end{align*}\]

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.

Here is a sketch of the region that we are after.

To determine this area, we’ll need to know the values of \(\theta \) for which the two curves intersect. We can determine these points by setting the two equations and solving.

\[\begin{align*}3 + 2\sin \theta & = 2\\ \sin \theta & = - \frac{1}{2}\hspace{0.5in} \Rightarrow \hspace{0.5in}\theta = \frac{{7\pi }}{6},\frac{{11\pi }}{6}\end{align*}\]

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 \(\frac{{11\pi }}{6}\) is \( - \frac{\pi }{6}\). 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 \(\frac{{7\pi }}{6}\) to \(\frac{{11\pi }}{6}\) we will not enclose the shaded area, instead we will enclose the bottom most of the three regions. However, if we use the angles \( - \frac{\pi }{6}\) to \(\frac{{7\pi }}{6}\) we will enclose the area that we’re after.

So, the area is then,

\[\begin{align*}A & = \int_{{\, - \frac{\pi }{6}}}^{{\,\frac{{7\pi }}{6}}}{{\frac{1}{2}\left( {{{\left( {3 + 2\sin \theta } \right)}^2} - {{\left( 2 \right)}^2}} \right)\,d\theta }}\\ & = \int_{{\, - \frac{\pi }{6}}}^{{\,\frac{{7\pi }}{6}}}{{\frac{1}{2}\left( {5 + 12\sin \theta + 4{{\sin }^2}\theta } \right)\,d\theta }}\\ & = \int_{{\, - \frac{\pi }{6}}}^{{\,\frac{{7\pi }}{6}}}{{\frac{1}{2}\left( {7 + 12\sin \theta - 2\cos \left( {2\theta } \right)} \right)\,d\theta }}\\ & = \left. {\frac{1}{2}\left( {7\theta - 12\cos \theta - \sin \left( {2\theta } \right)} \right)} \right|_{ - \frac{\pi }{6}}^{\frac{{7\pi }}{6}}\\ & = \frac{{11\sqrt 3 }}{2} + \frac{{14\pi }}{3} = 24.187\end{align*}\]

This time we’re looking for the following region.

So, this is the region that we get by using the limits \(\frac{{7\pi }}{6}\) to \(\frac{{11\pi }}{6}\). The area for this region is,

\[\begin{align*}A & = \int_{{\,\frac{{7\pi }}{6}}}^{{\,\frac{{11\pi }}{6}}}{{\frac{1}{2}\left( {{{\left( 2 \right)}^2} - {{\left( {3 + 2\sin \theta } \right)}^2}} \right)\,d\theta }}\\ & = \int_{{\,\frac{{7\pi }}{6}}}^{{\,\frac{{11\pi }}{6}}}{{\frac{1}{2}\left( { - 5 - 12\sin \theta - 4{{\sin }^2}\theta } \right)\,d\theta }}\\ & = \int_{{\,\frac{{7\pi }}{6}}}^{{\,\frac{{11\pi }}{6}}}{{\frac{1}{2}\left( { - 7 - 12\sin \theta + 2\cos \left( {2\theta } \right)} \right)\,d\theta }}\\ & = \left. {\frac{1}{2}\left( { - 7\theta + 12\cos \theta + \sin \left( {2\theta } \right)} \right)} \right|_{\frac{{7\pi }}{6}}^{\frac{{11\pi }}{6}}\\ & = \frac{{11\sqrt 3 }}{2} - \frac{{7\pi }}{3} = 2.196\end{align*}\]

Notice that for this area the “outer” and “inner” function were opposite!

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 \(\theta \) the area we’re after must be enclosed. However, the only two ranges for \(\theta \) 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,

\[\begin{align*}{\mbox{Area}} & = {\mbox{Area of Circle}} - {\mbox{Area from Example 3}}\\ & = \pi {\left( 2 \right)^2} - 2.196\\ & = 10.370\end{align*}\]

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,

\[\begin{align*}{\mbox{Area}} & = {\mbox{Area of Limacon}} - {\mbox{Area from Example 2}}\\ & = \int_{{\,0}}^{{\,2\pi }}{{\frac{1}{2}{{\left( {3 + 2\sin \theta } \right)}^2}\,d\theta }} - 24.187\\ & = \int_{{\,0}}^{{\,2\pi }}{{\frac{1}{2}\left( {9 + 12\sin \theta + 4{{\sin }^2}\theta } \right)\,d\theta }} - 24.187\\ & = \int_{{\,0}}^{{\,2\pi }}{{\frac{1}{2}\left( {11 + 12\sin \theta - 2\cos \left( {2\theta } \right)} \right)\,d\theta }} - 24.187\\ & = \left. {\frac{1}{2}\left( {11\theta - 12\cos \left( \theta \right) - \sin \left( {2\theta } \right)} \right)} \right|_0^{2\pi } - 24.187\\ & = 11\pi - 24.187\\ & = 10.370\end{align*}\]

A slightly longer approach, but sometimes we are forced to take this longer approach.