15. Polar Coordinates

d. Area Enclosed by a Polar Graph

2. Area Between Polar Graphs

So far, we have seen how to compute the area between a polar curve and the origin. We now look at computing the area between two polar curves.

The area enclosed by \(r=f(\theta)\) is \(\displaystyle A=\int_\alpha^\beta \dfrac{1}{2}f(\theta)^2\,d\theta\), while the area enclosed by \(r=g(\theta)\) is \(\displaystyle A=\int_\alpha^\beta \dfrac{1}{2}g(\theta)^2\,d\theta\). We simply subtract to obtain:

The area bounded by the graphs of \(r=f(\theta)\) and \(r=g(x)\) between the rays \(\theta=\alpha\) and \(\theta=\beta\) is \[ A=\int_\alpha^\beta \dfrac{1}{2}[f(\theta)^2-g(\theta)^2]\,d\theta \] provided \(f(\theta) \ge g(\theta) \ge 0\).

The only difficulty is finding the limits of integration. These are where the two curves intersect. So we equate the functions and solve: \[ 2+2\sin\theta=1 \qquad \Rightarrow \qquad 2\sin\theta=-1 \qquad \Rightarrow \qquad \sin\theta=-\,\dfrac{1}{2} \] The intersection in the \(4^\text{th}\) quadrant is at \(\theta=-\,\dfrac{\pi}{6}\). The intersection in the \(3^\text{rd}\) quadrant is at \(\theta=\dfrac{7\pi}{6}\). So the area is \[\begin{aligned} A &=\dfrac{1}{2}\int_{-\pi/6}^{7\pi/6} [(2+2\sin\theta)^2-1^2]\,d\theta \\ &=\dfrac{1}{2}\int_{-\pi/6}^{7\pi/6} [3+8\sin\theta+4\sin^2]\,d\theta \\ &=\dfrac{1}{2}\int_{-\pi/6}^{7\pi/6} [3+8\sin\theta+2(1-\cos(2\theta))]\,d\theta \\ &=\dfrac{1}{2}\int_{-\pi/6}^{7\pi/6} [5+8\sin\theta-2\cos(2\theta)]\,d\theta \\ &=\dfrac{1}{2}[5\theta-8\cos\theta-\sin(2\theta)]_{-\pi/6}^{7\pi/6} =\dfrac{10\pi}{3}+\dfrac{7\sqrt{3}}{2} \end{aligned}\]

It is important to integrate from \(\theta=-\,\dfrac{\pi}{6}\) to \(\theta=\dfrac{7\pi}{6}\) which covers the area between the curves, including \(\theta=0\), \(\dfrac{\pi}{2}\) and \(\pi\). If one instead integrates from \(\theta=\dfrac{7\pi}{6}\) to \(\theta=\dfrac{11\pi}{6}\), then one gets the area at the bottom inside the circle outside the cardioid.

Caution: It is not always possible to find the points of intersection of two polar graphs by simply equating the functions as was done in the preceeding example. Recall the discussion on Intersections of Polar Graphs.