Find the area of the region that lies inside the first curve and outside the second curve: $r^2 = 72 \cos(2\theta), \quad r = 6$

Neetesh Kumar | January 1, 2025

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This is the solution to Math 1c
Assignment: 10.4 Question Number 8
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Step-by-step solution:

To calculate the area of the region inside the first curve $r^2 = 72 \cos(2\theta)$ and outside the second curve $r = 6$, we proceed step by step.

Step 1: Express the Area Formula:

The area in polar coordinates is given by:

$A = \frac{1}{2} \int_{\theta_1}^{\theta_2} \left(r_1^2 - r_2^2\right) \, d\theta$

Here:

• $r_1^2$ corresponds to the first curve $r^2 = 72 \cos(2\theta)$, which simplifies to $r = \sqrt{72 \cos(2\theta)}$.
• $r_2$ corresponds to the second curve $r = 6$.

Step 2: Determine the Limits of Integration:

To find the limits of integration, solve for $\theta$ where the two curves intersect.

Equating $r^2$ of the first curve and $r^2$ of the second curve:

$r^2 = 72 \cos(2\theta) = 6^2$

This simplifies to:

$72 \cos(2\theta) = 36$

$\cos(2\theta) = \frac{1}{2}$

The solutions for $\cos(2\theta) = \frac{1}{2}$ are:

$2\theta = \pm \frac{\pi}{3} + 2n\pi$

$\theta = \pm \frac{\pi}{6} + n\pi$

Within one period $(0 \leq \theta \leq 2\pi)$, the relevant limits are:

$\theta_1 = -\frac{\pi}{6}, \quad \theta_2 = \frac{\pi}{6}$

Step 3: Set Up the Integral:

The area is:

$A = \frac{1}{2} \int_{-\pi/6}^{\pi/6} \left( r_1^2 - r_2^2 \right) \, d\theta$

Substitute $r_1^2 = 72 \cos(2\theta)$ and $r_2^2 = 36$:

$A = \frac{1}{2} \int_{-\pi/6}^{\pi/6} \left( 72 \cos(2\theta) - 36 \right) \, d\theta$

Step 4: Simplify the Integral:

Separate the integral into two parts:

$A = \frac{1}{2} \left[ \int_{-\pi/6}^{\pi/6} 72 \cos(2\theta) \, d\theta - \int_{-\pi/6}^{\pi/6} 36 \, d\theta \right]$

Part 1: Evaluate $\int 72 \cos(2\theta) \, d\theta$:

The integral of $\cos(2\theta)$ is:

$\int \cos(2\theta) \, d\theta = \frac{1}{2} \sin(2\theta)$

Thus:

$\int_{-\pi/6}^{\pi/6} 72 \cos(2\theta) \, d\theta = 72 \cdot \frac{1}{2} \left[ \sin(2\theta) \right]_{-\pi/6}^{\pi/6}$

At the limits:

$\sin(2 \cdot \pi/6) = \sin(\pi/3) = \frac{\sqrt{3}}{2}, \quad \sin(2 \cdot -\pi/6) = \sin(-\pi/3) = -\frac{\sqrt{3}}{2}$

So:

$\int_{-\pi/6}^{\pi/6} 72 \cos(2\theta) \, d\theta = 36 \left[ \frac{\sqrt{3}}{2} - \left(-\frac{\sqrt{3}}{2}\right) \right] = 36 \cdot \sqrt{3}$

Part 2: Evaluate $\int 36 \, d\theta$:

The integral of $36$ is:

$\int_{-\pi/6}^{\pi/6} 36 \, d\theta = 36 \cdot \left[ \theta \right]_{-\pi/6}^{\pi/6}$

At the limits:

$\theta = \frac{\pi}{6} - \left(-\frac{\pi}{6}\right) = \frac{\pi}{3}$

So:

$\int_{-\pi/6}^{\pi/6} 36 \, d\theta = 36 \cdot \frac{\pi}{3} = 12\pi$

Step 5: Combine Results:

Combine the two parts:

$A = \frac{1}{2} \left[ 36\sqrt{3} - 12\pi \right]$

Simplify:

$A = 18\sqrt{3} - 6\pi$

Final Answer:

The area of the region is:

$A = \boxed{18\sqrt{3} - 6\pi}$

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