Find the area of the region that lies inside both curves: $r = \sin(2\theta), \quad r = \cos(2\theta)$

Neetesh Kumar | January 2, 2025

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This is the solution to Math 1c
Assignment: 10.4 Question Number 13
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Step-by-step solution:

To determine the area of the region common to both polar curves $r = \sin(2\theta)$ and $r = \cos(2\theta)$, follow these steps:

1. Find the Points of Intersection:

Set the two equations equal to find the angles $\theta$ where the curves intersect:

$\sin(2\theta) = \cos(2\theta)$

Divide both sides by $\cos(2\theta)$ (where $\cos(2\theta) \neq 0$):

$\tan(2\theta) = 1 \\ 2\theta = \frac{\pi}{4} + n\pi \\ \theta = \frac{\pi}{8} + \frac{n\pi}{2}, \quad n \in \mathbb{Z}$

Thus, the points of intersection occur at:

$\theta = \frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8}, \frac{13\pi}{8}$

2. Determine the Limits of Integration:

The curves intersect at $\theta = \frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8},$ and $\frac{13\pi}{8}$.

Due to the symmetry of the curves, we can calculate the area for one sector and multiply by $8$ to obtain the total area.

3. Identify the Minimum Function in Each Sector:

In each sector between two consecutive points of intersection, determine which curve has the smaller $r$ value:

• For $0 \leq \theta \leq \frac{\pi}{8}$: $r = \sin(2\theta) \leq r = \cos(2\theta)$
• For $\frac{\pi}{8} \leq \theta \leq \frac{\pi}{4}$: $r = \cos(2\theta) \leq r = \sin(2\theta)$

This pattern repeats in each subsequent sector due to symmetry.

4. Set Up the Integral for One Sector:

The area $A_{\text{sector}}$ for one sector is the sum of the areas where each function is the minimum:

$A_{\text{sector}} = \frac{1}{2} \int_{0}^{\frac{\pi}{8}} [\sin(2\theta)]^2 \, d\theta + \frac{1}{2} \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} [\cos(2\theta)]^2 \, d\theta$

5. Compute the Integrals:

a. Integral of $[\sin(2\theta)]^2$:

$\int [\sin(2\theta)]^2 \, d\theta = \int \frac{1 - \cos(4\theta)}{2} \, d\theta = \frac{\theta}{2} - \frac{\sin(4\theta)}{8} + C$

Evaluating from $0$ to $\frac{\pi}{8}$:

$\left[ \frac{\theta}{2} - \frac{\sin(4\theta)}{8} \right]_0^{\frac{\pi}{8}} = \left( \frac{\pi}{16} - \frac{\sin\left(\frac{\pi}{2}\right)}{8} \right) - (0 - 0) = \frac{\pi}{16} - \frac{1}{8}$

b. Integral of $[\cos(2\theta)]^2$:

$\int [\cos(2\theta)]^2 \, d\theta = \int \frac{1 + \cos(4\theta)}{2} \, d\theta = \frac{\theta}{2} + \frac{\sin(4\theta)}{8} + C$

Evaluating from $\frac{\pi}{8}$ to $\frac{\pi}{4}$:

$\left[ \frac{\theta}{2} + \frac{\sin(4\theta)}{8} \right]_{\frac{\pi}{8}}^{\frac{\pi}{4}} = \left( \frac{\pi}{8} + \frac{\sin(\pi)}{8} \right) - \left( \frac{\pi}{16} + \frac{\sin\left(\frac{\pi}{2}\right)}{8} \right) = \frac{\pi}{16} - \frac{1}{8}$

c. Total Area for One Sector:

$A_{\text{sector}} = \frac{1}{2} \left( \frac{\pi}{16} - \frac{1}{8} \right) + \frac{1}{2} \left( \frac{\pi}{16} - \frac{1}{8} \right) = \frac{\pi}{16} - \frac{1}{8}$

6. Calculate the Total Area:

Since there are 8 such sectors in the full circle:

$A = 8 \times \left( \frac{\pi}{16} - \frac{1}{8} \right ) = \frac{8\pi}{16} - 1 = \frac{\pi}{2} - 1$

Final Answer:

The area of the region that lies inside both curves $r = \sin(2\theta)$ and $r = \cos(2\theta)$ is:

$A = \boxed{\frac{\pi}{2} - 1 \quad \text{square units}}$

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