Area Between Polar Curves: Interactive Visualizer

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 $pi/4$):

• 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 $pi/4$): See which function bounds the area; in this case, it is the red sine graph.

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

Formula

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

Easy Example Problem

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

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