Now, we’ll need to determine the values of \(\theta \) where the graphs intersect (indicated by the black lines on the graph in the previous step).

There are easy enough to find. Because they are where the graphs intersect we know they must have the same value of \(r\). So,

\[\begin{align*}6 + 2\cos \theta & = 4 - 2\cos \theta \\ \cos \theta & = - \frac{1}{2}\hspace{0.5in} \Rightarrow \hspace{0.5in}\theta = {\cos ^{ - 1}}\left( { - \frac{1}{2}} \right) = \frac{{2\pi }}{3}\end{align*}\]

This is the value for the angle in the second quadrant where the graphs intersect.

From a quick sketch of a unit circle we can quickly see two possible values for the angle in the third quadrant where the two graphs intersect.

\[\theta = 2\pi - \frac{{2\pi }}{3} = \frac{{4\pi }}{3}\hspace{0.5in}\hspace{0.5in}\theta = - \frac{{2\pi }}{3}\]

Now, we need to recall that the angles must go from smaller to larger values and as they do that they must trace out the boundary curves of the enclosed area. Keeping this in mind and we can see that we’ll need to use the positive angle for this problem. If we used the negative angle we’d be tracing out the “right” portions of our curves and we need to trace out the “left” portions of our curves.

Therefore, the ranges of \(\theta \) for this problem is then \(\frac{{2\pi }}{3} \le \theta \le \frac{{4\pi }}{3}\).

Show Step 3

From the graph we can see that \(r = 4 - 2\cos \theta \) is the “outer” graph for this region and \(r = 6 + 2\cos \theta \) is the “inner” graph.

The area then,

\[\begin{align*}A & = \int_{{\frac{{2\pi }}{3}}}^{{\frac{{4\pi }}{3}}}{{\frac{1}{2}\left[ {{{\left( {4 - 2\cos \theta } \right)}^2} - {{\left( {6 + 2\cos \theta } \right)}^2}} \right]\,d\theta }}\\ & = \int_{{\frac{{2\pi }}{3}}}^{{\frac{{4\pi }}{3}}}{{ - 10 - 20\cos \theta \,d\theta }}\\ & = \left. {\left( { - 10\theta - 20\sin \left( \theta \right)} \right)} \right|_{\frac{{2\pi }}{3}}^{\frac{{4\pi }}{3}} = \require{bbox} \bbox[2pt,border:1px solid black]{{13.6971}}\end{align*}\]

Do not get too excited about all the minus signs in the integral. Just because all the terms have minus signs in front of them does not mean that we should get a negative value from the integral!