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 $$\frac{1}{2} \int_{-\pi/6}^{\pi/6} \cos^2 (3\theta) \, d\theta = \frac{1}{2} \left(\frac{\theta}{2} + \frac{\sin(6x)}{12} \right)\Big\vert_{-\pi/6}^{\pi/6} = \frac{\pi}{12}.$$