9.9 Finding the Area of the Region Bounded by Two Polar Curves

Learning objectives

—CHA-5.D—Calculate areas of regions defined by polar curves using definite integrals—Topic 9.9• CHA-5.D Calculate areas of regions defined by polar curves using definite integrals.• CHA-5.D.2 Areas of regions bounded by polar curves can be calculated with definite integrals.• Enduring understanding CHA-5: Definite integrals allow us to solve problems involving the accumulation of change in area or volume over an interval.

For a fixed angle $\theta$ , the region between two polar curves runs radially from an inner radius to an outer radius. Its thin-slice area is therefore the outer sector minus the inner sector, which produces a difference of squared radii.

A=\frac12\int_{\alpha}^{\beta}\left([r_{\text{outer}}(\theta)]^2-[r_{\text{inner}}(\theta)]^2\right),d\theta

Solve $r_1(\theta)=r_2(\theta)$ to find candidate intersection angles.
Use the desired region and a test angle to decide which curve is farther from the pole.
Integrate outer squared minus inner squared between consecutive relevant boundaries.
Split the integral wherever the curves exchange outer and inner roles, then add the positive region areas.

Example: find the area inside $r=2\cos\theta$ but outside $r=1$ . Intersections satisfy $2\cos\theta=1$ , giving $\theta=\pm\pi/3$ . On $[-\pi/3,\pi/3]$ , $2\cos\theta$ is the outer radius. Thus
$A=\frac12\int_{-\pi/3}^{\pi/3}(4\cos^2\theta-1)\,d\theta=\frac{\pi}{3}+\frac{\sqrt3}{2}$
square units. The integrand is nonnegative on this interval, consistent with the chosen outer and inner curves.

Do not subtract the radii first and then square: $(r_{\text{outer}}-r_{\text{inner}})^2$ is not a sector-area difference. Also, one curve may not remain outer across the whole region; an unsplit integral can create negative contributions or cancel genuine area.

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—AP Calculus BCRead