15. Polar Coordinates

d.1. Area Enclosed by a Polar Graph

We have seen how to plot a polar curve. Since any polar curve is periodic, it usually encloses a region. We now want to compute the area of such a region.

If we divide the \(\theta\)-interval into \(n\) intervals, then the angular width of each wedge is \(\Delta\theta=\dfrac{\beta-\alpha}{n}\) and the radius of the \(i^\text{th}\) wedge is \(r_i=f\left(\theta_i^*\right)\) for some angle \(\theta_i^*\) in the \(i^\text{th}\) interval. Now the area of a pie wedge of angle \(\Delta\theta\) (in radians) and radius \(r_i\) is \[ \Delta A =\left(\dfrac{\Delta\theta}{2\pi}\right)\pi r_i^2 =\dfrac{1}{2}r_i^2\,\Delta\theta \] So the total area is obtained by adding up the areas of the wedges and taking the limit as the number of wedges becomes large, (\(n\to\infty\)): \[ A =\lim_{n\to\infty} \sum_{i=1}^n \dfrac{1}{2}r_i^2\,\Delta\theta =\lim_{n\to\infty} \sum_{i=1}^n \dfrac{1}{2}f\left(\theta_i^*\right)^2\,\Delta\theta \] This limit of a sum is recognized as the integral: \(\displaystyle A=\int_\alpha^\beta \dfrac{1}{2}f(\theta)^2\,d\theta\).

The area bounded by the graph of \(r=f(\theta)\) and the rays \(\theta=\alpha\) and \(\theta=\beta\) is: \[ A=\int_\alpha^\beta \dfrac{1}{2}f(\theta)^2\,d\theta \] We often remove the reference to \(f\) and write this formula as \[ A=\int_\alpha^\beta \dfrac{1}{2}r^2\,d\theta \]

It is clear that the upper half of the cardioid is swept out as \(\theta\) increases from \(0\) to \(\pi\). Thus the area is: \[\begin{aligned} A&=\int_0^\pi \dfrac{1}{2}(1+\cos\theta)^2\,d\theta \\ &=\dfrac{1}{2}\int_0^\pi (1+2\cos\theta+\cos^2\theta)\,d\theta \\ &=\dfrac{1}{2}\int_0^\pi \left(1+2\cos\theta+\dfrac{1+\cos(2\theta)}{2}\right)\,d\theta \\ &=\dfrac{1}{2}\left[\theta+2\sin\theta+\dfrac{\theta}{2}+\dfrac{\sin(2\theta)}{4}\right]_0^\pi =\dfrac{3\pi}{4} \end{aligned}\]