Find the area of the region that lies inside both curves: $r = 7 \sin(\theta), \quad r = 7 \cos(\theta)$

Neetesh Kumar | January 1, 2025

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This is the solution to Math 1c
Assignment: 10.4 Question Number 11
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Step-by-step solution:

We are tasked with finding the area of the region common to both curves given by $r = 7 \sin(\theta)$ and $r = 7 \cos(\theta)$.

Step 1: Intersection Points:

To find the intersection points, equate the two curves:

$7 \sin(\theta) = 7 \cos(\theta)$

Cancel out the common factor of $7$:

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

The equality holds when:

$\theta = \frac{\pi}{4}$

This means the curves intersect at $\theta = \frac{\pi}{4}$ and are symmetric across the line $\theta = \frac{\pi}{4}$.

Since the curves are symmetric, we can calculate the area of one section and then double it.

Step 2: Area Formula:

The area common to two polar curves is given by:

$A = 2 \cdot \frac{1}{2} \int_{0}^{\pi/4} (7 \sin(\theta))^2 d\theta$

Simplify:

$A = \int_{0}^{\pi/4} 49 \sin^2(\theta) d\theta$

Step 3: Use the Double-Angle Identity:

Using the identity:

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

Substituting:

$A = \int_{0}^{\pi/4} 49 \cdot \frac{1 - \cos(2\theta)}{2} d\theta$

Simplify:

$A = \frac{49}{2} \int_{0}^{\pi/4} (1 - \cos(2\theta)) d\theta$

Evaluate the integral:

$A = \frac{49}{2} \left[\theta - \frac{\sin(2\theta)}{2}\right]_{0}^{\pi/4}$

Substitute the limits:

$A = \frac{49}{2} \left[\frac{\pi}{4} - \frac{\sin\left(\frac{\pi}{2}\right)}{2}\right]$

We know $\sin\left(\frac{\pi}{2}\right) = 1$:

$A = \frac{49}{2} \left[\frac{\pi}{4} - \frac{1}{2}\right]$

Simplify:

$A = \frac{49}{2} \cdot \frac{\pi - 2}{4}$

Simplify further:

$A = \frac{49 (\pi - 2)}{8}$

Final Answer:

The area of the region common to both curves is:

$A = \boxed{\frac{49 (\pi - 2)}{8}}$

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