1.4 Area and arc length in polar coordinates  (Page 2/3)

[link] involved finding the area inside one curve. We can also use [link] to find the area between two polar curves. However, we often need to find the points of intersection of the curves and determine which function defines the outer curve or the inner curve between these two points.

Finding the area between two polar curves

Find the area outside the cardioid $r=2+2\text{sin}\theta$ and inside the circle $r=6\text{sin}\theta .$

First draw a graph containing both curves as shown.

To determine the limits of integration, first find the points of intersection by setting the two functions equal to each other and solving for $\theta \text{:}$

$\begin{array}{ccc}\hfill 6\text{sin}\theta & =\hfill & 2+2\text{sin}\theta \hfill \\ \hfill 4\text{sin}\theta & =\hfill & 2\hfill \\ \hfill \text{sin}\theta & =\hfill & \frac{1}{2}.\hfill \end{array}$

This gives the solutions $\theta =\frac{\pi }{6}$ and $\theta =\frac{5\pi }{6},$ which are the limits of integration. The circle $r=3\text{sin}\theta$ is the red graph, which is the outer function, and the cardioid $r=2+2\text{sin}\theta$ is the blue graph, which is the inner function. To calculate the area between the curves, start with the area inside the circle between $\theta =\frac{\pi }{6}$ and $\theta =\frac{5\pi }{6},$ then subtract the area inside the cardioid between $\theta =\frac{\pi }{6}$ and $\theta =\frac{5\pi }{6}\text{:}$

$\begin{array}{cc}\hfill A& =\text{circle}-\text{cardioid}\hfill \\ & =\frac{1}{2}{\displaystyle {\int }_{\pi \text{/}6}^{5\pi \text{/}6}{\left[6\text{sin}\theta \right]}^{2}d\theta }-\frac{1}{2}{\displaystyle {\int }_{\pi \text{/}6}^{5\pi \text{/}6}{\left[2+2\text{sin}\theta \right]}^{2}d\theta }\hfill \\ & =\frac{1}{2}{\displaystyle {\int }_{\pi \text{/}6}^{5\pi \text{/}6}36{\text{sin}}^2\theta d\theta }-\frac{1}{2}{\displaystyle {\int }_{\pi \text{/}6}^{5\pi \text{/}6}4+8\text{sin}\theta +4{\text{sin}}^2\theta d\theta }\hfill \\ & =18{\displaystyle {\int }_{\pi \text{/}6}^{5\pi \text{/}6}\frac{1-\text{cos}\left(2\theta \right)}{2}d\theta }-2{\displaystyle {\int }_{\pi \text{/}6}^{5\pi \text{/}6}1+2\text{sin}\theta +\frac{1-\text{cos}\left(2\theta \right)}{2}d\theta }\hfill \\ & =9{\left[\theta -\frac{\text{sin}\left(2\theta \right)}{2}\right]}_{\pi \text{/}6}^{5\pi \text{/}6}-2{\left[\frac{3\theta }{2}-2\text{cos}\theta -\frac{\text{sin}\left(2\theta \right)}{4}\right]}_{\pi \text{/}6}^{5\pi \text{/}6}\hfill \\ & =9\left(\frac{5\pi }{6}-\frac{\text{sin}2(5\pi \text{/}6)}{2}\right)-9\left(\frac{\pi }{6}-\frac{\text{sin}2(\pi \text{/}6)}{2}\right)\hfill \\ & \text{−}\left(3\left(\frac{5\pi }{6}\right)-4\text{cos}\frac{5\pi }{6}-\frac{\text{sin}2(5\pi \text{/}6)}{2}\right)+\left(3\left(\frac{\pi }{6}\right)-4\text{cos}\frac{\pi }{6}-\frac{\text{sin}2(\pi \text{/}6)}{2}\right)\hfill \\ & =4\pi .\hfill \end{array}$

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In [link] we found the area inside the circle and outside the cardioid by first finding their intersection points. Notice that solving the equation directly for $\theta$ yielded two solutions: $\theta =\frac{\pi }{6}$ and $\theta =\frac{5\pi }{6}.$ However, in the graph there are three intersection points. The third intersection point is the origin. The reason why this point did not show up as a solution is because the origin is on both graphs but for different values of $\theta .$ For example, for the cardioid we get

so the values for $\theta$ that solve this equation are $\theta =\frac{3\pi }{2}+2n\pi ,$ where n is any integer. For the circle we get

The solutions to this equation are of the form $\theta =n\pi$ for any integer value of n. These two solution sets have no points in common. Regardless of this fact, the curves intersect at the origin. This case must always be taken into consideration.

Arc length in polar curves

Here we derive a formula for the arc length of a curve defined in polar coordinates.

In rectangular coordinates, the arc length of a parameterized curve $\left(x\left(t\right),y\left(t\right)\right)$ for $a\le t\le b$ is given by

In polar coordinates we define the curve by the equation $r=f\left(\theta \right),$ where $\alpha \le \theta \le \beta .$ In order to adapt the arc length formula for a polar curve, we use the equations

and we replace the parameter t by $\theta .$ Then

We replace $dt$ by $d\theta ,$ and the lower and upper limits of integration are $\alpha$ and $\beta ,$ respectively. Then the arc length formula becomes