Find the area of the region that lies inside both curves: $r = 4 + \cos(\theta), \quad r = 4 - \cos(\theta)$

Neetesh Kumar | January 2, 2025

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This is the solution to Math 1c
Assignment: 10.4 Question Number 12
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Step-by-step solution:

To determine the area of the region common to both polar curves $r = 4 + \cos(\theta)$ and $r = 4 - \cos(\theta)$, follow these steps:

1. Find the Points of Intersection:

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

$4 + \cos(\theta) = 4 - \cos(\theta)$

Solving for $\theta$:

$2\cos(\theta) = 0 \\ \cos(\theta) = 0$

Thus, the points of intersection occur at:

$\theta = \frac{\pi}{2}, \frac{3\pi}{2}$

2. Determine the Common Area:

The area common to both curves is where $r$ is less than or equal to both $4 + \cos(\theta)$ and $4 - \cos(\theta)$. This can be expressed as:

$r \leq \min(4 + \cos(\theta), 4 - \cos(\theta)) = 4 - |\cos(\theta)|$

3. Set Up the Integral for the Area:

The area $A$ in polar coordinates is given by:

$A = \frac{1}{2} \int_{0}^{2\pi} [4 - |\cos(\theta)|]^2 \, d\theta$

Expanding the integrand:

$[4 - |\cos(\theta)|]^2 = 16 - 8|\cos(\theta)| + \cos^2(\theta)$

Thus, the integral becomes:

$A = \frac{1}{2} \int_{0}^{2\pi} (16 - 8|\cos(\theta)| + \cos^2(\theta)) \, d\theta$

4. Compute the Integral:

Evaluate each term separately:

1. Integral of 16:

   $\int_{0}^{2\pi} 16 \, d\theta = 16 \times 2\pi = 32\pi$

2. Integral of $8|\cos(\theta)|$:

   $\int_{0}^{2\pi} 8|\cos(\theta)| \, d\theta = 8 \times 4 = 32$

   *(Since $\int_{0}^{2\pi} |\cos(\theta)| \, d\theta = 4)$

3. Integral of $\cos^2(\theta)$:

   $\int_{0}^{2\pi} \cos^2(\theta) \, d\theta = \pi$

   (Using the identity $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$)

Combining these results:

$A = \frac{1}{2} (32\pi - 32 + \pi) = \frac{1}{2} (33\pi - 32) = \frac{33\pi - 32}{2}$

Final Answer:

The area of the region that lies inside both curves $r = 4 + \cos(\theta)$ and $r = 4 - \cos(\theta)$ is:

$A = \boxed{\frac{33\pi - 32}{2} \quad \text{square units}}$

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