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.

Find the area bounded by the graph of \(r=f(\theta)\) and the rays \(\theta=\alpha\) and \(\theta=\beta\). Assume \(f(\theta) \ge 0\).

areageneral

Recall that in rectangular coordinates, to find the area under a curve \(y=f(x)\), we divided the \(x\)-interval into subintervals and approximated the area as the sum of a rectangle over each subinterval. Similarly, to find the area enclosed by a polar curve \(r=f(\theta)\), we divide the \(\theta\)-interval into subintervals and approximate the area as the sum of a circular pie wedge for each subinterval.

areawedges

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

Find the area of the upper half of the cardioid, \(r=1+\cos\theta\).

halfcardioid

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

Find the area bounded by one petal of the 3-leaf rose \(r=\cos(3\theta)\).

area3leaf

The most difficult part of the problem is determining the limits of integration. However, these occur at the angles where the curve passes through the origin. So solve \(r=\cos(3\theta)=0\) and select two adjacent solutions.

\(A=\dfrac{\pi}{12}\)

The solutions of \(r=\cos(3\theta)=0\) occur when \[ 3\theta=\pm\dfrac{\pi}{2},\pm\dfrac{3\pi}{2},\pm\dfrac{5\pi}{2},\ldots \] or \[ \theta=\pm\dfrac{\pi}{6},\pm\dfrac{3\pi}{6},\pm\dfrac{5\pi}{6},\ldots \] The petal on the right occurs for \(-\,\dfrac{\pi}{6} \le \theta \le \dfrac{\pi }{6}\). Thus the area is: \[\begin{aligned} A&=\dfrac{1}{2}\int_{-\pi/6}^{\pi/6} \cos^2(3\theta)\,d\theta \\ &=\dfrac{1}{2}\int_{-\pi/6}^{\pi/6} \dfrac{1+\cos(6\theta)}{2}\,d\theta \\ &=\dfrac{1}{4}\left[\theta+\dfrac{\sin(6\theta)}{6}\right]_{-\pi/6}^{\pi/6} =\dfrac{\pi }{12} \end{aligned}\]

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