Question

Find the area of the region that lies inside the first curve and outside the second curve.
r2 = 18 cos 2θ, r = 3

Answer 1

Answer on Question #53419 – Math – Integral Calculus

Question

Find the area of the region that lies inside the first curve and outside the second curve. $r2 = 18\cos 2\theta, r = 3$

Solution

Definition

Let $G$ be the region bounded by curves with polar equations $r = f(\vartheta)$ and $r = g(\vartheta)$ (fig.1), $\vartheta = a$ and $\vartheta = b$, where $f(\vartheta) \geq g(\vartheta) > 0$, $0 < b - a \leq 2\pi$. Then area $A$ of $G$ is

$$A(G) = \frac{1}{2} \int_{a}^{b} \left( [f(\vartheta)]^2 - [g(\vartheta)]^2 \right) d\vartheta.$$

Fig.1

First we describe the given curves in polar coordinates $\{r,\vartheta\}$:

1) $r^2 = 18\cos(2\vartheta)$ is the lemniscate of Bernoulli;

2) $r = 3$ is a circle with radius 3 and the center at the pole (origin).

Further we sketch these curves (fig.2). The required area is represented by the shaded regions.

Fig.2

As we can see,

$$A(G) = A(G_1) + A(G_2).$$

As the figures in the fig.2 are symmetrical, then we can write

$$A(G) = 2A(G_1).$$

Let's find points of intersection of the curves 1) and 2) for region $G_1$:

$$\pm \sqrt{18 \cos(2\vartheta)} = 3,$$ $$18 \cos(2\vartheta) = 9,$$ $$\cos(2\vartheta) = \frac{9}{18} = \frac{1}{2},$$ $$2\vartheta = \pm \arccos\left(\frac{1}{2}\right) = \pm \frac{\pi}{3},$$ $$\theta_{1,2} = \pm \frac{\pi}{6}.$$

Therefore, using (1), (2) and (3) we obtain

$$\begin{aligned} A(G) &= 2 \cdot \frac{1}{2} \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} (18 \cos(2\vartheta) - 3^2) d\vartheta \\ &= \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} (18 \cos(2\vartheta) - 9) d\vartheta \\ &= \left( \sin\left(2 \cdot \frac{\pi}{6}\right) - \left( \sin\left(2 \cdot \left(-\frac{\pi}{6}\right)\right) \right) - \left( \frac{\pi}{6} - \left(-\frac{\pi}{6}\right) \right) \right) \\ &= 9 \left( 2 \sin\left(2 \cdot \frac{\pi}{6}\right) - \frac{2\pi}{6} \right) \\ &= 18 \left( \sin\left(\frac{\pi}{3}\right) - \frac{\pi}{6} \right) = 18 \left( \frac{\sqrt{3}}{2} - \frac{\pi}{6} \right) \approx 6.16 \text{ square units}. \end{aligned}$$

Answer: $A(G) = 6.16$ square units.

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