Area Between Polar Curves: Interactive Visualizer

The area between two polar curves equals one-half times the integral of the outer radius squared minus the inner radius squared, taken over the angle interval where they bound the region; first set the curves equal to find the intersection angles that mark the limits.

To find the area between two curves is quite tricky, and the best way to do it is just to zoom in and look at the quadrant where the graphs intersect.

For example, at the intersection between sine and cosine, we can see that when we start from the far right at zero and move up to their intersection of 45 (which is π/4):

• The beginning of the graph: If we ignore the part of the cosine graph that curves and only look at the sine graph, we can see that's what contains our area. That's why we know the sine graph is our first function—it is the lower bound of that region at the start.

• After the intersection: Do we switch to the cosine function? Yes! Even though the sine function keeps going past the intersection point, that is not where the area is contained between the two. That's why we switch to the cosine function in this case, which is what makes it feel a bit counterintuitive.

You just have to keep track of the intersection and see which function bounds the region:

1. For the bottom bit (0 to π/4): See which function bounds the area; in this case, it is the red sine graph.

2. After the intersection (π/4 to π/2): See what function bounds the top part of the region; in this case, it is the blue cosine graph.

How do you find the area between two polar curves?

A = \frac{1}{2} \int_{\alpha}^{\beta} \left( [\textcolor{#ef4444}{r_2(\theta)}]^2 - [\textcolor{#38bdf8}{r_1(\theta)}]^2 \right) d\theta

Polar area worked example: inside one curve, outside another

Find the area of the region inside the circle r = 3cos(θ) and outside the cardioid r = 1 + cos(θ).

1. Find intersection angles: 3cos(θ) = 1 + cos(θ) ⟹ 2cos(θ) = 1 ⟹ cos(θ) = ½ ⟹ θ = -π/3, π/3.

2. Set up the area integral:

A = \frac{1}{2} \int_{-\pi/3}^{\pi/3} \left( (3\cos\theta)^2 - (1 + \cos\theta)^2 \right) d\theta

A = \frac{1}{2} \int_{-\pi/3}^{\pi/3} (9\cos^2\theta - 1 - 2\cos\theta - \cos^2\theta) d\theta

A = \frac{1}{2} \int_{-\pi/3}^{\pi/3} (8\cos^2\theta - 2\cos\theta - 1) d\theta = \pi - \frac{9\sqrt{3}}{8} \approx 1.19

Another worked example

Find the area inside r = 2 and outside r = 1 over a full turn. Step 1: the outer radius is 2 and the inner radius is 1, so set up one-half times the integral of (2 squared minus 1 squared), which is 3, from 0 to 2 pi. Step 2: integrate the constant 3 to get one-half times 3 times 2 pi. Step 3: that equals 3 pi, the area of the ring between the two circles.

Aligned with the College Board AP Calculus BC CED — Unit 9 (Parametric, Polar, and Vector-Valued Functions), Topic 9.9 (Area Bounded by Two Polar Curves). College Board ↗

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