Find the area outside $r = 3\cos\Theta$ and inside $r = 1 + \cos\Theta$

calculuschange-of-variabledefinite integralsintegrationpolar coordinates

Two polar curves.

I am looking for the area outside the circle $r = 3\cos\Theta$ and inside the limaçon $r = 1 + \cos\Theta$:

$$A = \frac{1}{2}\int_{\frac{\pi }{3}}^{\frac{5\pi }{3}}\left[(1+\cos\Theta )^{2}-(3\cos\Theta )^{2}\right]\,\mathrm{d}\Theta = -2\pi \,\text{units}^{2} $$

If the opposite situation is used, I simply reverse the limits and curves. But since the lower limit must be smaller, I use coterminal angles to change one of the limits. Here, after the reversal I changed 5pi/3 to -pi/3

$$A = \frac{1}{2}\int_{-\frac{\pi }{3}}^{\frac{\pi }{3}}\left[(3\cos\Theta )^{2}-(1+\cos\Theta )^{2}\right]\,\mathrm{d}\Theta = pi \,\text{units}^{2} $$

I can't understand why the area I'm getting is negative for the first situation. I get that maybe I have the limits or the curves in reverse but I'm following the correct limits and placement of the curves. What is the general rule for determining the curves and limits so that the result is always positive?

I followed the convention that the outer curve in the graph is the first radius used and the inner curve is the second radius. I've also made sure that the limits start and end appropriately based on the graph. It worked for the opposite situation, as shown in this question.

Find the area inside $r=3\cos\Theta $ and outside $r = 1+\cos\Theta $

I've seen other problems where that rule works. I'm just wondering why it doesn't work here.

Best Answer

EDIT: Original answer didn't address radial function taking on negative values.

The positive direction for the radial polar variable $r$ is outward so if you want the area of the region outside the polar curve $r = f(\theta) \geq 0$ and inside the curve $r = g(\theta)$, i.e. $$ f(\theta) \leq r \leq g(\theta) $$ for all $\alpha \leq \theta \leq \beta$, then you calculate the definite integral $$ A = \frac12 \int_\alpha^\beta \bigl[ g(\theta)^2 - f(\theta)^2 \bigr] \, \mathrm{d}\theta. $$

Let $f(\theta) = 3\cos\theta$ and $g(\theta) = 1 + \cos\theta$.

Both functions exhibit the symmetry $f(2\pi - \theta) = f(\theta)$ and $g(2\pi - \theta) = g(\theta)$ for all $\theta$, which is equivalent to the fact that their polar graphs have a reflection symmetry across $\theta = \pi$, the $x$-axis.

Thus, we can calculate the total area for $\frac\pi3 \leq \theta \leq \frac{5\pi}3$ by calculating the area for $\frac\pi3 \leq \theta \leq \pi$ and doubling the result.

However, there's still an issue: $f(\theta) \leq 0$ for $\frac\pi2 \leq \theta \leq \pi$, so the points wrap around the bottom half of the circle. The naive application of the integral formula gives negative values over this domain. See this picture:

Cardioid and circle with radial lines.

Instead, we have to calculate the integral in two pieces: \begin{array}{ccr} \tfrac\pi3 \leq{} \theta \leq \tfrac\pi2 & \quad\leadsto & f(\theta) \leq{} r \leq g(\theta) \\ \tfrac\pi2 \leq{} \theta \leq \pi & \quad\leadsto & 0 \leq{} r \leq g(\theta) \end{array}

Cardioid and circle, closeup.

Thus, the total area is \begin{align} A &= 2 \cdot \frac12 \biggl( \int_{\pi/3}^{\pi/2} \bigl[ g(\theta)^2 - f(\theta)^2 \bigr] \, \mathrm{d}\theta \;+\; \int_{\pi/2}^{\pi} g(\theta)^2 \, \mathrm{d}\theta \biggr) \\ &= \int_{\pi/3}^{\pi/2} \bigl[ (1 + \cos\theta)^2 - (3\cos\theta)^2 \bigr] \, \mathrm{d}\theta \;+\; \int_{\pi/2}^{\pi} (3\cos\theta)^2 \, \mathrm{d}\theta \end{align}

You can probably take it from here.

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