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Section 9.8 : Area with Polar Coordinates

4. Find the area that is inside \(r = 2\) and outside \(r = 3 + 3\sin \theta \).

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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,

\[\begin{align*}3 + 3\sin \theta & = 2\\ \sin \theta & = - \frac{1}{3}\hspace{0.5in} \Rightarrow \hspace{0.5in}\theta = {\sin ^{ - 1}}\left( { - \frac{1}{3}} \right) = - 0.3398\end{align*}\]

This is one possible value for the angle in the fourth quadrant where the graphs intersect.

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

\[\theta = \pi + 0.3398 = 3.4814\]

Now, we’ll have a problem if we use these two angles for our area integral. 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. These two clearly will not do that. In fact, they trace out the area from the previous problem.

To fix this problem it is probably easiest to use a quick sketch of a unit circle to see that another value for the angle in the fourth quadrant is,

\[\theta = 2\pi - 0.3398 = 5.9434\]

Using this angle along with the angle we already have in the third quadrant will trace out the area we are interested in.

Therefore, the ranges of \(\theta \) for this problem is then \(3.4814 \le \theta \le 5.9434\).

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From the graph we can see that \(r = 2\) is the “outer” graph for this region and \(r = 3 + 3\sin \theta \) is the “inner” graph.

The area is then,

\[\begin{align*}A & = \int_{{3.4814}}^{{5.9434}}{{\frac{1}{2}\left[ {{{\left( 2 \right)}^2} - {{\left( {3 + 3\sin \theta } \right)}^2}} \right]\,d\theta }}\\ & = \frac{1}{2}\int_{{3.4814}}^{{5.9434}}{{ - 5 - 18\sin \theta - 9{{\sin }^2}\left( \theta \right)\,d\theta }}\\ & = \frac{1}{2}\int_{{3.4814}}^{{5.9434}}{{ - 5 - 18\sin \theta - \frac{9}{2}\left( {1 - \cos \left( {2\theta } \right)} \right)\,d\theta }}\\ & = \frac{1}{2}\int_{{3.4814}}^{{5.9434}}{{ - \frac{{19}}{2} - 18\sin \theta + \frac{9}{2}\cos \left( {2\theta } \right)\,d\theta }}\\ & = \left. {\frac{1}{2}\left( { - \frac{{19}}{2}\theta + 18\cos \left( \theta \right) + \frac{9}{4}\sin \left( {2\theta } \right)} \right)} \right|_{3.4814}^{5.9434} = \require{bbox} \bbox[2pt,border:1px solid black]{{3.8622}}\end{align*}\]

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