Calculate Area in Polar Coordinates

Note that the circle is swept by the rays \( \theta = \theta_1 \) and \( \theta = \theta_2 \), and we need to find \( \theta_1 \) and \( \theta_2 \).

The circle starts at the origin (\( r(\theta) = 0 \)) and finishes at the origin (\( r(\theta) = 0 \)). Hence, the angles \( \theta_1 \) and \( \theta_2 \) are found by solving \( r(\theta) = 0 \):

\[ \sin(\theta) = 0 \]

Which gives the solutions:

\[ \theta_1 = 0 \quad \text{and} \quad \theta_2 = \pi \]

The area \( A \) of the circle is given by:

\[ A = \dfrac{1}{2} \int_{0}^{\pi} (2 \sin \theta)^2 \, d\theta = 2 \int_{0}^{\pi} (\sin \theta)^2 \, d\theta \]

Use the trigonometric identity \( \sin^2 \theta = \dfrac{1}{2}(1 - \cos(2\theta)) \):

\[ A = \int_{0}^{\pi} (1 - \cos(2\theta)) \, d\theta = \left[ \theta - \dfrac{1}{2}\sin(2\theta) \right]_0^{\pi} = \pi \]

The area of the given circle whose radius is \( 1 \) matches the standard formula \( \pi \cdot r^2 = \pi \cdot 1^2 = \pi \).

Because of the symmetry of the graph, we find the area enclosed by one loop and multiply the result by 3.

The loop on the right starts and finishes at the origin (\( r(\theta) = 0 \)). Solving \( \sin(3\theta) = 0 \):

\[ 3\theta = n\pi \quad (n = 0, \pm 1, \pm 2, \dots) \]

For the first loop in quadrant I:

\[ \theta_1 = 0 \quad \text{and} \quad \theta_2 = \dfrac{\pi}{3} \]

The area of the single loop is:

\[ A = \dfrac{1}{2} \int_{0}^{\frac{\pi}{3}} (\sin 3\theta)^2 \, d\theta = \dfrac{1}{4} \int_{0}^{\frac{\pi}{3}} (1 - \cos(6\theta)) \, d\theta = \dfrac{1}{4} \left[ \theta - \dfrac{1}{6}\sin(6\theta) \right]_0^{\frac{\pi}{3}} = \dfrac{\pi}{12} \]

The total area of all three loops is:

\[ 3A = \dfrac{\pi}{4} \]

Due to symmetry, we find the area of the surface enclosed by the two curves on the right and multiply by 2.

The area is made of two parts, \( A_1 \) and \( A_2 \):

\[ A_1 = \dfrac{1}{2} \int_{\theta_1}^{\theta_2} r_1^2(\theta) \, d\theta, \quad A_2 = \dfrac{1}{2} \int_{\theta_2}^{\theta_3} r_2^2(\theta) \, d\theta \]

Limits of integration:

• \( \theta_1 = 0 \) (where \( r_1(\theta) = 0 \))
• \( \theta_3 = \dfrac{\pi}{2} \) (where \( r_2(\theta) = 0 \))
• \( \theta_2 \) is the intersection point: \( \sin(\theta) = 0.5(1 + \cos(2\theta)) \). Using \( \cos(2\theta) = 1 - 2\sin^2\theta \), we get \( \sin^2\theta + \sin\theta - 1 = 0 \), yielding \( \sin\theta = \dfrac{-1+\sqrt{5}}{2} \approx 0.61803 \), so \( \theta_2 = \arcsin\left(\dfrac{\sqrt{5}-1}{2}\right) \approx 0.66624 \).

Evaluating the integrals:

\[ A_1 = \dfrac{1}{4} \int_{0}^{0.66624} (1 - \cos(2\theta)) \, d\theta \approx 0.0450915 \] \[ A_2 = \dfrac{0.5^2}{2} \int_{0.66624}^{\frac{\pi}{2}} \left(1 + \dfrac{1}{2}(1 + \cos(4\theta)) + 2\cos(2\theta)\right) d\theta \approx 0.0409645 \]

The total common area \( A = 2(A_1 + A_2) \approx 2(0.0450915 + 0.0409645) = 0.172112 \).