Paul's Online Notes
Paul's Online Notes
Home / Calculus II / Parametric Equations and Polar Coordinates / Area with Polar Coordinates
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 9.8 : Area with Polar Coordinates

5. Find the area that is inside \(r = 4 - 2\cos \theta \) and outside \(r = 6 + 2\cos \theta \).

Show All Steps Hide All Steps

Start Solution

First, here is a quick sketch of the graph of the region we are interested in.

Show Step 2

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!