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Section 3-8 : Area with Polar Coordinates

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.

This is a sketch that looks a lot like a slice of pie that is in the 1st quadrant.  The “point” of the piece of pie is in the origin and is between the two angles from $\theta = \alpha$ to $\theta = \beta$ and the “crust” of the piece of pie is given by $r=f\left( \theta  \right)$.

We’ll be looking for the shaded area in the sketch above. The formula for finding this area is,

\[A = \int_{{\,\alpha }}^{{\,\beta }}{{\frac{1}{2}{r^2}\,d\theta }}\]

Notice that we use \(r\) in the integral instead of \(f\left( \theta \right)\) 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 \(r = 2 + 4\cos \theta \).
Show 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 \(\theta \) 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.

This is the graph of $r=2+4\cos \theta $.  The heart shaped portion of this graph goes through the points (in Cartesian coordinates to make it a little easier to visualize the graph) (6,0), (0,2), (0,2) and the origin.  The inner loop is in the 1st and 4th quadrant and goes up to the point (2,0).  Also included are two lines moving out of the origin.  The first is in the 2nd quadrant and forms an angle of $\frac{2\pi}{3}$ with the positive x-axis.  The second is in the 3rd quadrant and forms an angle of $\frac{4\pi}{3}$ with the positive x-axis.

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.

So, that’s how we determine areas that are enclosed by a single curve, but what about situations like the following sketch where we want to find the area between two curves.

This is a sketch that looks a lot like an upper portion of a slice of pie that is in the 1st quadrant.  The region is between the two angles from $\theta = \alpha$ to $\theta = \beta$.  The “outer” edge of the region is given by $r_{o}=g\left( \theta  \right)$.  The “inner” edge of the region is given by $r_{i}=f\left( \theta  \right)$.

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,

\[A = \int_{{\,\alpha }}^{{\,\beta }}{{\frac{1}{2}\left( {r_o^2 - r_i^2} \right)\,d\theta }}\]

Let’s take a look at an example of this.

Example 2 Determine the area that lies inside \(r = 3 + 2\sin \theta \) and outside \(r = 2\).
Show Solution

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

This is the graph of $r=3+2\sin \theta $.  This is a vaguely heart shaped portion of this graph goes through the points (in Cartesian coordinates to make it a little easier to visualize the graph) (3,0), (0,5), (-3,0) and (0,-2).  The “crease” in the heart is at (0,-2).  Also included on the graph is the circle give by r=2.  The crease of the first graph is inside the circle and the upper portion of the circle is inside the heart.  The circle and heart intersect in the 3rd and 4th quadrant.  The area from the heart down to the circle that is in the 1st and 2nd quadrant is shaded.

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.

This is the same graph as above except two lines have been added to the graph.  The first line comes out of the origin and goes through the intersection in the 4th quadrant makes angles of $-\frac{\pi}{6}$ or $\frac{11\pi}{6}$ (depending on direction of rotation) with the positive x-axis.  The second line comes out of the origin and goes through the intersection in the 3rd quadrant makes angles of $\frac{7\pi}{6}$ with the positive x-axis.

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,

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

Let’s work a slight modification of the previous example.

Example 3 Determine the area of the region outside \(r = 3 + 2\sin \theta \) and inside \(r = 2\).
Show Solution

This time we’re looking for the following region.

This is the graph of $r=3+2\sin \theta $.  This is a vaguely heart shaped portion of this graph goes through the points (in Cartesian coordinates to make it a little easier to visualize the graph) (3,0), (0,5), (-3,0) and (0,-2).  The “crease” in the heart is at (0,-2).  Also included on the graph is the circle give by r=2.  The crease of the first graph is inside the circle and the upper portion of the circle is inside the heart.  The circle and heart intersect in the 3rd and 4th quadrant.  Two lines have been added to the graph.  The first line comes out of the origin and goes through the intersection in the 4th quadrant makes angles of $-\frac{\pi}{6}$ or $\frac{11\pi}{6}$ (depending on direction of rotation) with the positive x-axis.  The second line comes out of the origin and goes through the intersection in the 3rd quadrant makes angles of $\frac{7\pi}{6}$ with the positive x-axis.  The area from the heart down to the circle that is in the 3rd and 4th quadrant is shaded.

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!

Let’s do one final modification of this example.

Example 4 Determine the area that is inside both \(r = 3 + 2\sin \theta \) and \(r = 2\).
Show Solution

Here is the sketch for this example.

This is the graph of $r=3+2\sin \theta $.  This is a vaguely heart shaped portion of this graph goes through the points (in Cartesian coordinates to make it a little easier to visualize the graph) (3,0), (0,5), (-3,0) and (0,-2).  The “crease” in the heart is at (0,-2).  Also included on the graph is the circle give by r=2.  The crease of the first graph is inside the circle and the upper portion of the circle is inside the heart.  The circle and heart intersect in the 3rd and 4th quadrant.  Two lines have been added to the graph.  The first line comes out of the origin and goes through the intersection in the 4th quadrant makes angles of $-\frac{\pi}{6}$ or $\frac{11\pi}{6}$ (depending on direction of rotation) with the positive x-axis.  The second line comes out of the origin and goes through the intersection in the 3rd quadrant makes angles of $\frac{7\pi}{6}$ with the positive x-axis.  The area that is inside both graphs has been shaded.

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.

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