Polar Equations - Area

The area under a curve can be determined both using Cartesian plane with rectangular \((x,y)\) coordinates, and polar coordinates. For instance the polar equation \(r = f(\theta)\) describes a curve. The formula for the area under this polar curve is given by the formula below:

Consider the arc of the polar curve \(r = f(\theta)\) traced as \(\theta\) varies from \(\theta_1\) to \(\theta_2\). If this arc bounds a closed region of the plane, the area of this region is \[\frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d\theta = \frac{1}{2} \int_{\theta_1}^{\theta_2} f(\theta)^2 \, d\theta.\]

Consider a circle of radius \(r\); a sector of this circle subtending an angle of \(\theta \) will have area \[\pi r^2 \cdot \frac{\theta}{2\pi} = \frac{1}{2} r^2 \theta.\]

For an infinitesimal change in the angle \(\theta\), from \(\theta \) to \(\theta + d\theta\), the region of the plane swept out by the polar curve \(r = f(\theta)\) is approximately a circular sector, hence has area \[\frac{1}{2} r^2 d\theta.\] Integrating this factor over the interval \(\theta_1 \le \theta \le \theta_2\) gives the area swept by the curve as \[\frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d\theta, \] which is the desired formula.

What is the area of a circle with radius \(R\)?

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A circle of radius \(R\) is swept out by the polar equation \(r = R\) as \(\theta\) varies from \(0\) to \(2\pi\). Thus, the area of the circle is \[\frac{1}{2} \int_{0}^{2\pi} R^2 \, d\theta = \frac{1}{2} \cdot 2\pi R^2 = \pi R^2.\]

A rose curve is defined by the polar equation \(r = \cos(3\theta)\). This curve has three "petals"; what is the area bounded by one of the petals?

A plot of the rose curve with 3 petals.

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As \(\theta\) varies from \(-\pi/6\) to \(\pi/6\), the curve traces out one petal. Thus, the area of a petal is \[\frac{1}{2} \int_{-\pi/6}^{\pi/6} \cos^2 (3\theta) \, d\theta = \frac{1}{2} \left(\frac{\theta}{2} + \frac{\sin(6x)}{12} \right)\Big\vert_{-\pi/6}^{\pi/6} = \frac{\pi}{12}.\]